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Redox Reactions and Electrochemistry — JEE Previous Year Questions

Every Redox Reactions and Electrochemistry question asked in JEE Main and JEE Advanced across the last 186 papers — 265 questions, each with its correct answer. Free to read, no account needed.

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All 265 Redox Reactions and Electrochemistry questions

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Q1·ChemistrySingle correctJEE Advanced 2026
At 300 K, the molar conductivities of the aqueous solutions of three salts at two different concentrations are given below: The conductivity of a saturated aqueous solution of AgCl is 1.40×10−61.40 \times 10^{-6}1.40×10−6 S cm−1cm^{-1}cm−1 at 300 K. If the solubility of AgCl in water at 300 K is XXX mol L−1L^{-1}L−1, then log⁡10(X−1)\log_{10}(X^{-1})log10​(X−1) is (Assume that AgCl dissolved in water ionizes completely and that the molar conductivity of saturated AgCl solution is equal to its limiting molar conductivity.)
SaltConcentration (M)Molar conductivity (S cm2^{2}2 mol−1^{-1}−1)
NaNO3NaNO_{3}NaNO3​0.01111
NaNO3NaNO_{3}NaNO3​0.04101
NaClNaClNaCl0.01117
NaClNaClNaCl0.04107
AgNO3AgNO_{3}AgNO3​0.01125
AgNO3AgNO_{3}AgNO3​0.04116
  1. (A)3
  2. (B)4
  3. (C)5
  4. (D)6

Correct answer: (C)

Step-by-step solution →
Q2·ChemistrySingle correctJEE Main 2026
Given at 298 K: EFe2+/Fe⊖=XE^{\ominus}_{Fe^{2+}/Fe} = XEFe2+/Fe⊖​=X Volt; EFe3+/Fe⊖=YE^{\ominus}_{Fe^{3+}/Fe} = YEFe3+/Fe⊖​=Y Volt. The EFe3+/Fe2+⊖E^{\ominus}_{Fe^{3+}/Fe^{2+}}EFe3+/Fe2+⊖​ in Volt at 298 K is given by:
  1. (A)2X−3Y2X - 3Y2X−3Y
  2. (B)3Y−2X3Y - 2X3Y−2X
  3. (C)3Y+2X3Y + 2X3Y+2X
  4. (D)Y+XY + XY+X

Correct answer: (B)

Step-by-step solution →
Q3·ChemistrySingle correctJEE Main 2026
Consider the following data. BaSO₄ is sparingly soluble in water. If the conductivity of the saturated BaSO₄ solution is x S cm⁻¹ then the solubility product of BaSO₄ can be given as (Here Λm = Λ°m)
ElectrolyteΛ°m (S cm² mol⁻¹)
BaCl₂x₁
H₂SO₄x₂
HClx₃
  1. (A)106x2α2(x1+x2−2x3)2\frac{10^{6}x^{2}}{\alpha^{2}(x_1 + x_2 - 2x_3)^{2}}α2(x1​+x2​−2x3​)2106x2​
  2. (B)x2(x1+x2−2x3)2\frac{x^{2}}{(x_1 + x_2 - 2x_3)^{2}}(x1​+x2​−2x3​)2x2​
  3. (C)α2(x1+x2−2x3)2106x2\frac{\alpha^{2}(x_1 + x_2 - 2x_3)^{2}}{10^{6}x^{2}}106x2α2(x1​+x2​−2x3​)2​
  4. (D)x2(x1+x2+2x3)2\frac{x^{2}}{(x_1 + x_2 + 2x_3)^{2}}(x1​+x2​+2x3​)2x2​

Correct answer: (A)

Step-by-step solution →
Q4·ChemistryNumericalJEE Main 2026
500 mL of 0.2 M MnO4−\mathrm{MnO_{4}^{-}}MnO4−​ solution in basic medium when mixed with 500 mL of 1.5 M KI solution, oxidises iodide ions to liberate molecular iodine. This liberated iodine is then titrated with a standard xxx M thiosulphate solution in presence of starch till the end point. If 300 mL of thiosulphate was consumed, then the value of xxx is __________.

Correct answer: 1

Step-by-step solution →
Q5·ChemistrySingle correctJEE Main 2026
For a general redox reaction Anode : Red1→Ox1n1++n1e−\mathrm{Red_{1}} \rightarrow \mathrm{Ox_{1}^{n_{1}^{+}}} + n_{1}e^{-}Red1​→Ox1n1+​​+n1​e− Cathode : Ox2+n2e−→Red2n2−\mathrm{Ox_{2}} + n_{2}e^{-} \rightarrow \mathrm{Red_{2}^{n_{2}^{-}}}Ox2​+n2​e−→Red2n2−​​ Which of the following statement is incorrect ?
  1. (A)The overall reaction can be written as n2Red1+n1Ox2⇌n2Ox1n1++n1Red2n2−n_{2}\mathrm{Red_{1}} + n_{1}\mathrm{Ox_{2}} \rightleftharpoons n_{2}\mathrm{Ox_{1}^{n_{1}^{+}}} + n_{1}\mathrm{Red_{2}^{n_{2}^{-}}}n2​Red1​+n1​Ox2​⇌n2​Ox1n1+​​+n1​Red2n2−​​
  2. (B)The electrons do not appear in the overall reaction because electrons produced at the anode are consumed at the cathode.
  3. (C)Here nnn is the number of electrons transferred in redox reaction.
  4. (D)If the reaction is carried out reversibly, the electrical work done is equal to the ratio of charge and potential difference through which charge is moved.

Correct answer: (D)

Step-by-step solution →
Q6·ChemistryNumericalJEE Main 2026
At 298 K, the molar conductivity of x% (w/w) MX solution (aqueous) is 123.5 S cm² mol⁻¹. The conductance of same solution is 1.9 × 10⁻³ S. The value of x is _______ ×10⁻². (Given : cell constant = 1.3 cm⁻¹; molar mass of MX is 75 g mol⁻¹, density of aqueous solution of MX at 298 K is 1.0 g mL⁻¹)

Correct answer: 15

Step-by-step solution →
Q7·ChemistrySingle correctJEE Main 2026
One half cell in a voltaic cell is constructed by dipping silver rod in AgNO₃ solution of unknown concentration, other half cell is Zn rod dipped in 1 molar solution of ZnSO₄. A voltage of 1.60 V is measured at 298 K for this cell. What is the concentration of Ag⁺ ions used in terms of log x (x = [Ag⁺]) ? EZn2+/Zn∘E^{\circ}_{Zn^{2+}/Zn}EZn2+/Zn∘​ = −0.76 V, EAg+/Ag∘E^{\circ}_{Ag^{+}/Ag}EAg+/Ag∘​ = +0.80 V, 2.303RTF\frac{2.303RT}{F}F2.303RT​ = 0.059 V
  1. (A)23.9\frac{2}{3.9}3.92​
  2. (B)45.9\frac{4}{5.9}5.94​
  3. (C)2.92\frac{2.9}{2}22.9​
  4. (D)5.94\frac{5.9}{4}45.9​

Correct answer: (B)

Step-by-step solution →
Q8·ChemistryNumericalJEE Main 2026
An electrochemical cell, consist of the following two redox couples, Mx+(aq)/M(s)M^{x+}(aq)/M(s)Mx+(aq)/M(s) [Ered⊖[E^{\ominus}_{red}[Ered⊖​ = +0.15 V] and Fe3+(aq)/Fe(s)\mathrm{Fe^{3+}}(aq)/\mathrm{Fe}(s)Fe3+(aq)/Fe(s) [Ered⊖[E^{\ominus}_{red}[Ered⊖​ = −0.036 V]. The cell EMF (EcellE_{cell}Ecell​) is recorded to be 0.2057 V. If the reaction quotient of the electrochemical reaction is found to be 10−2^{-2}−2, then the value of xxx is ______. (Nearest integer) [Given: M is a p-block metal and 2.303RTF\frac{2.303RT}{F}F2.303RT​ = 0.059 V]

Correct answer: 2

Step-by-step solution →
Q9·ChemistrySingle correctJEE Main 2026
In order to oxidise a mixture of 1 mole each of FeC2O4FeC_2O_4FeC2​O4​, Fe2(C2O4)3Fe_2(C_2O_4)_3Fe2​(C2​O4​)3​, FeSO4FeSO_4FeSO4​ and Fe2(SO4)3Fe_2(SO_4)_3Fe2​(SO4​)3​ in acidic medium, the number of moles of KMnO4KMnO_4KMnO4​ required is
  1. (A)3
  2. (B)2
  3. (C)5
  4. (D)7

Correct answer: (B)

Step-by-step solution →
Q10·ChemistrySingle correctJEE Main 2026
An electrochemical cell is constructed using half cells in the direction of spontaneous change Fe(OH)2(s)+2e−→Fe(s)+2OH−(aq)\mathrm{Fe(OH)_{2}}(s) + 2e^{-} \rightarrow \mathrm{Fe}(s) + 2\mathrm{OH^{-}}(aq)Fe(OH)2​(s)+2e−→Fe(s)+2OH−(aq) Eθ=−0.88E^{\theta} = -0.88Eθ=−0.88 V and AgBr(s)+e−→Ag(s)+Br−(aq)\mathrm{AgBr}(s) + e^{-} \rightarrow \mathrm{Ag}(s) + \mathrm{Br^{-}}(aq)AgBr(s)+e−→Ag(s)+Br−(aq) Eθ=+0.07E^{\theta} = +0.07Eθ=+0.07 V Which of the following option is correct?
  1. (A)Overall reaction Fe(s)+2OH−(aq)+2AgBr(s)⇌Fe(OH)2(s)+2Ag(s)+2Br−(aq)\mathrm{Fe}(s) + 2\mathrm{OH^{-}}(aq) + 2\mathrm{AgBr}(s) \rightleftharpoons \mathrm{Fe(OH)_{2}}(s) + 2\mathrm{Ag}(s) + 2\mathrm{Br^{-}}(aq)Fe(s)+2OH−(aq)+2AgBr(s)⇌Fe(OH)2​(s)+2Ag(s)+2Br−(aq)
  2. (B)Ecellθ=−0.95E^{\theta}_{cell} = -0.95Ecellθ​=−0.95 V
  3. (C)Fe is reduced in the electrochemical cell
  4. (D)EcellθE^{\theta}_{cell}Ecellθ​ is an extensive property

Correct answer: (A)

Step-by-step solution →
Q11·ChemistryNumericalJEE Main 2026
Consider the following two half-cell reactions along with the standard reduction potential given: CO2+6H++6e−→CH3OH+H2OCO_2 + 6H^+ + 6e^- \rightarrow CH_3OH + H_2OCO2​+6H++6e−→CH3​OH+H2​O Ered∘=0.02E^\circ_{red} = 0.02Ered∘​=0.02 V 12O2+2H++2e−→H2O\frac{1}{2}O_2 + 2H^+ + 2e^- \rightarrow H_2O21​O2​+2H++2e−→H2​O Ered∘=1.23E^\circ_{red} = 1.23Ered∘​=1.23 V A fuel cell was set up using the above two reactions such that the cell operates under the standard condition of 1 bar pressure and 298 K temperature. The fuel cell works with 80% efficiency. If the work derived from the cell using 1 mol of CH3OHCH_3OHCH3​OH is used to compress an ideal gas isothermally against a constant pressure of 1 kPa, then the change in the volume of the gas, ΔV=\Delta V =ΔV= ______ m3^33. (nearest integer) Given: F=96500F = 96500F=96500 C mol−1^{-1}−1

Correct answer: 560

Step-by-step solution →
Q12·ChemistryNumericalJEE Main 2026
500 mL of 1.2 M KI solution is mixed with 500 mL of 0.2 M KMnO4_{4}4​ solution in basic medium. The liberated iodine was titrated with standard 0.1 M Na2_{2}2​S2_{2}2​O3_{3}3​ solution in the presence of starch indicator till the blue color disappeared. The volume (in L) of Na2_{2}2​S2_{2}2​O3_{3}3​ consumed is________. (Nearest integer)

Correct answer: 3

Step-by-step solution →
Q13·ChemistryNumericalJEE Main 2026
Consider the following redox reaction taking place in acidic medium BH4−_{4}^{-}4−​(aq) +++ ClO3−_{3}^{-}3−​(aq) →\rightarrow→ H2_{2}2​BO3−_{3}^{-}3−​(aq) +++ Cl−^{-}−(aq) If the Nernst equation for the above balanced reaction is Ecell_{\mathrm{cell}}cell​ = Ecello_{\mathrm{cell}}^{\mathrm{o}}cello​ −-− R TnF\frac{\mathrm{R\,T}}{\mathrm{nF}}nFRT​ ℓ\ellℓn Q, Then the value of n is _________. (Nearest integer)

Correct answer: 24

Step-by-step solution →
Q14·ChemistryNumericalJEE Main 2026
For strong electrolyte Λm\Lambda_mΛm​ increases slowly with dilution and can be represented by the equation Λm=Λm∘−Ac1/2\Lambda_m = \Lambda^\circ_m - Ac^{1/2}Λm​=Λm∘​−Ac1/2 Molar conductivity values of the solutions of strong electrolyte AB at 18°C are given below : c [mol L−1L^{-1}L−1]: 0.04 | 0.09 | 0.16 | 0.25 Λm\Lambda_mΛm​ [S cm2cm^2cm2 mol−1mol^{-1}mol−1]: 96.1 | 95.7 | 95.3 | 94.9 The value of constant A based on the above data [in S cm2cm^2cm2 mol−1mol^{-1}mol−1/(mol/L)1/2(mol/L)^{1/2}(mol/L)1/2] unit is ________.

Correct answer: 4

Step-by-step solution →
Q15·ChemistryNumericalJEE Main 2026
A volume of x mL of 5 M NaHCO3NaHCO_3NaHCO3​ solution was mixed with 10 mL of 2 M H2CO3H_2CO_3H2​CO3​ solution to make an electrolytic buffer. If the same buffer was used in the following electrochemical cell to record a cell potential of 235.3 mV, then the value of x = __________ mL (nearest integer). Sn(s)∣Sn(OH)62−Sn(s) | Sn(OH)_6^{2-}Sn(s)∣Sn(OH)62−​ (0.5 M) ∣HSnO2−| HSnO_2^-∣HSnO2−​ (0.05 M) ∣OH−∣Bi2O3(s)∣Bi(s)| OH^- | Bi_2O_3(s) | Bi(s)∣OH−∣Bi2​O3​(s)∣Bi(s) Consider upto one place of decimal for intermediate calculations [Given : EHSnO2−∣Sn(OH)62−∘=−0.9 VE^\circ_{HSnO_2^-|Sn(OH)_6^{2-}} = -0.9\,VEHSnO2−​∣Sn(OH)62−​∘​=−0.9V EBi2O3∣Bi∘=−0.44VE^\circ_{Bi_2O_3|Bi} = -0.44VEBi2​O3​∣Bi∘​=−0.44V pKa(H2CO3)=6.11pKa_{(H_2CO_3)} = 6.11pKa(H2​CO3​)​=6.11 2.303RTF=0.059V\frac{2.303RT}{F} = 0.059VF2.303RT​=0.059V Antilog⁡(1.29)=19.5Anti\log(1.29) = 19.5Antilog(1.29)=19.5 ]

Correct answer: 78

Step-by-step solution →
Q16·ChemistryNumericalJEE Main 2026
Electricity is passed through an acidic solution of Cu2+\mathrm{Cu^{2+}}Cu2+ till all the cu2+\mathrm{cu^{2+}}cu2+ was exhausted, leading to the deposition of 300 mg of Cu metal. However, a current of 600 mA was continued to pass through the same solution for another 28 minutes by keeping the total volume of the solution fixed at 200 mL. The total volume of oxygen evolved at STP during the entire process is ______ mL. (Nearest integer) [Given : Cu2+(aq)+2e−→Cu(s)  Ered0=+0.34 V\mathrm{Cu^{2+}\left(aq\right)+2e^{-} \rightarrow Cu\left(s\right)\; E^{0}_{red} = +0.34\,V}Cu2+(aq)+2e−→Cu(s)Ered0​=+0.34V O2(g)+4H++4e−→2H2O  Ered0=+1.23 V\mathrm{O_2\left(g\right)+4H^{+}+4e^{-} \rightarrow 2H_2O\; E^{0}_{red} = +1.23\,V}O2​(g)+4H++4e−→2H2​OEred0​=+1.23V Molar mass of Cu = 63.54 g mol−1^{-1}−1 Molar mass of O2\mathrm{O_2}O2​ = 32 g mol−1^{-1}−1 Faraday Constant = 96500 C mol−1^{-1}−1 Molar volume at STP = 22.4 L ]

Correct answer: 111

Step-by-step solution →
Q17·ChemistryNumericalJEE Main 2026
Molar conductivity of a weak acid HQ of concentration 0.18 M was found to be 1/30 of the molar conductivity of another weak acid HZ with concentration of 0.02 of M. If λQ−0\lambda^{0}_{\text{Q}^{-}}λQ−0​ happened to be equal with λZ−0\lambda^{0}_{\text{Z}^{-}}λZ−0​, then the difference of the pKa\text{pK}_{\text{a}}pKa​ values of the two weak acids (pKa(HQ)−pKa(HZ)\text{pK}_{\text{a}}(\text{HQ})-\text{pK}_{\text{a}}(\text{HZ})pKa​(HQ)−pKa​(HZ)) is _______ (Nearest integer). [Given : degree of dissociation (α\alphaα) << 1 for both weak acids, λ∘\lambda^{\circ}λ∘ : limiting molar conductivity of ions]

Correct answer: 2

Step-by-step solution →
Q18·ChemistryNumericalJEE Main 2026
X and Y are the number of electrons involved, respectively during the oxidation of I−\mathrm{I^{-}}I− to I2\mathrm{I_2}I2​ and S2−\mathrm{S^{2-}}S2− to S by acidified K2Cr2O7\mathrm{K_2Cr_2O_7}K2​Cr2​O7​. The value of X + Y is ______.

Correct answer: 12

Step-by-step solution →
Q19·ChemistryNumericalJEE Main 2026
200 cc of x × 10−3^{-3}−3M potassium dichromate is required to oxidise 750 cc of 0.6 M Mohr’s salt solution in acidic medium. Here x =

Correct answer: 375

Step-by-step solution →
Q20·ChemistrySingle correctJEE Main 2026
Consider the above electrochemical cell where a metal electrode (M) is undergoing redox reaction by forming M+^{+}+ (M → M+^{+}+ + e). The cation M+^{+}+ is present in two different concentrations c1_{1}1​ and c2_{2}2​ as shown above. Which of the following statement is correct for generating a positive cell potential?
  1. (A)If c1_{1}1​ is present at anode, then c1_{1}1​ = c2_{2}2​
  2. (B)If c1_{1}1​ is present at cathode, then c1_{1}1​ < c2_{2}2​
  3. (C)If c1_{1}1​ is present at cathode, then c1_{1}1​ > c2_{2}2​
  4. (D)If c1_{1}1​ is present at anode, then c1_{1}1​ > c2_{2}2​

Correct answer: (C)

Step-by-step solution →
Q21·ChemistrySingle correctJEE Main 2026
The oxidation state of chromium in the final product formed in the reaction between KI and acidified K2_{2}2​Cr2_{2}2​O7_{7}7​ solution is:
  1. (A)+4
  2. (B)+3
  3. (C)+2
  4. (D)+6

Correct answer: (B)

Step-by-step solution →
Q22·ChemistryNumericalJEE Main 2026
Consider the following electrochemical cell at 298K Pt∣\left|\right.∣HSnO2−_2^-2−​(aq)∣\left|\right.∣Sn(OH)62−_6^{2-}62−​(aq)∣\left|\right.∣Bi2_22​O3_33​(s)∣\left|\right.∣Bi(s). If the reaction quotient at a given time is 106^66, then the cell EMF (Ecell_{cell}cell​) is ________ × 10−1^{-1}−1V (Nearest integer). Given the standard half-cell reduction potential as EBi2O3/Bi,OH−0=−0.44E^{0}_{Bi_2O_3/Bi,OH^-} = -0.44EBi2​O3​/Bi,OH−0​=−0.44 V and ESn(OH)62−/HSnO2−,OH0=−0.90E^{0}_{Sn(OH)_6^{2-}/HSnO_2^-,OH} = -0.90ESn(OH)62−​/HSnO2−​,OH0​=−0.90 V

Correct answer: 4

Step-by-step solution →
Q23·ChemistrySingle correctJEE Main 2026
Consider the following reduction processes : Al3++3e−→Al(s)Al^{3+} + 3e^{-} \rightarrow Al(s)Al3++3e−→Al(s), Eo=−1.66E^{o} = -1.66Eo=−1.66 V Fe3++e−→Fe2+Fe^{3+} + e^{-} \rightarrow Fe^{2+}Fe3++e−→Fe2+, Eo=+0.77E^{o} = +0.77Eo=+0.77 V Co3++e−→Co2+Co^{3+} + e^{-} \rightarrow Co^{2+}Co3++e−→Co2+, Eo=+1.81E^{o} = +1.81Eo=+1.81 V Cr3++3e−→Cr(s)Cr^{3+} + 3e^{-} \rightarrow Cr(s)Cr3++3e−→Cr(s), Eo=−0.74E^{o} = -0.74Eo=−0.74 V The tendency to act as reducing agent decreases in the order :
  1. (A)Al>Cr>Fe2+>Co2+Al > Cr > Fe^{2+} > Co^{2+}Al>Cr>Fe2+>Co2+
  2. (B)Al>Fe2+>Cr>Co2+Al > Fe^{2+} > Cr > Co^{2+}Al>Fe2+>Cr>Co2+
  3. (C)Al>Cr>Co2+>Fe2+Al > Cr > Co^{2+} > Fe^{2+}Al>Cr>Co2+>Fe2+
  4. (D)Cr>Fe2+>Al>Co2+Cr > Fe^{2+} > Al > Co^{2+}Cr>Fe2+>Al>Co2+

Correct answer: (A)

Step-by-step solution →
Q24·ChemistryNumericalJEE Main 2026
Consider the following electrochemical cell : Pt ∣ O2(g) (1bar) ∣ HCl (aq) ∣∣ M2+(aq, 1.0 M) ∣ M(s)Pt\ |\ O_{2}(g)\ (1bar)\ |\ HCl\ (aq)\ ||\ M^{2+}(aq,\ 1.0\ M)\ |\ M(s)Pt ∣ O2​(g) (1bar) ∣ HCl (aq) ∣∣ M2+(aq, 1.0 M) ∣ M(s) The pH above which, oxygen gas would start to evolve at anode is ________ (nearest integer). [Given : EM2+/M0=0.994VE^{0}_{M^{2+}/M} = 0.994VEM2+/M0​=0.994V, EO2/H2O0=1.23VE^{0}_{O_{2}/H_{2}O} = 1.23VEO2​/H2​O0​=1.23V standard reduction potential and RTF(2.303)=0.059\frac{RT}{F}(2.303) = 0.059FRT​(2.303)=0.059 V at the given condition]

Correct answer: 4

Step-by-step solution →
Q25·ChemistryNumericalJEE Main 2026
The pH and conductance of a weak acid (HX) was found to be 5 and 4 × 10−5^{-5}−5 S, respectively. The conductance was measured under standard condition using a cell where the electrode plates having a surface area of 1 cm2^22 were at a distance of 15 cm apart. The value of the limiting molar conductivity is ……….. S m2^22 mol−1^{-1}−1. (nearest integer) (Given: degree of dissociation of the weak acid (a) << 1)

Correct answer: 6

Step-by-step solution →
Q26·ChemistryNumericalJEE Main 2026
MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K MX(s)⇌M+(aq)+X−(aq)\text{MX(s)} \rightleftharpoons \text{M}^{+}\text{(aq)} + \text{X}^{-}\text{(aq)}MX(s)⇌M+(aq)+X−(aq) ; Ksp=10−10\text{K}_{\text{sp}} = 10^{-10}Ksp​=10−10 If the standard reduction potential for M+(aq)→+e−M(s)\text{M}^{+}\text{(aq)} \xrightarrow{+e^{-}} \text{M(s)}M+(aq)+e−​M(s) is (EM+/M⊖)=0.79V\left(\text{E}^{\ominus}_{\text{M}^{+}/\text{M}}\right) = 0.79\text{V}(EM+/M⊖​)=0.79V, then the value of the standard reduction potential for the metal/metal insoluble salt electrode EX−/MX(s)/M⊖\text{E}^{\ominus}_{\text{X}^{-}/\text{MX(s)}/\text{M}}EX−/MX(s)/M⊖​ is _________ mV. (nearest integer) [Given: 2.303RTF=0.059V\frac{2.303\text{RT}}{\text{F}} = 0.059\text{V}F2.303RT​=0.059V]

Correct answer: 200

Step-by-step solution →
Q27·ChemistryNumericalJEE Advanced 2025
An electrochemical cell is fueled by the combustion of butane at 1 bar and 298 K. Its cell potential is XF×103\frac{X}{F} \times 10^{3}FX​×103 volts, where FFF is the Faraday constant. The value of X is _____. Use: Standard Gibbs energies of formation at 298 K are : ΔfGCO2o\Delta_{f}G^{o}_{CO_{2}}Δf​GCO2​o​ = −-−394 kJ mol−1^{-1}−1; ΔfGwatero\Delta_{f}G^{o}_{water}Δf​Gwatero​ = −-−237 kJ mol−1^{-1}−1; ΔfGbutaneo\Delta_{f}G^{o}_{butane}Δf​Gbutaneo​ = −-−18 kJ mol−1^{-1}−1

Correct answer: 105.50

Step-by-step solution →
Q28·ChemistryNumericalJEE Advanced 2025
In an electrochemical cell, dichromate ions in aqueous acidic medium are reduced to Cr3+Cr^{3+}Cr3+. The current (in amperes) that flows through the cell for 48.25 minutes to produce 1 mole of Cr3+Cr^{3+}Cr3+ is ______. Use: 1 Faraday = 96500 C mol−1mol^{-1}mol−1

Correct answer: 100.00

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Q29·ChemistryIntegerJEE Main 2025
Consider the following half cell reaction: Cr2O72−(aq)+6e−+14H+(aq)→2Cr3+(aq)+7H2O(l)Cr_2O_7^{2-}(aq)+6e^-+14H^+(aq)\to2Cr^{3+}(aq)+7H_2O(l)Cr2​O72−​(aq)+6e−+14H+(aq)→2Cr3+(aq)+7H2​O(l). The reaction was conducted with the ratio of [Cr3+]2[Cr2O72−]=10−4\dfrac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}]}=10^{-4}[Cr2​O72−​][Cr3+]2​=10−4. The pH value at which the EMF of the half cell will become zero is ___ (nearest integer value). [Given: standard half cell reduction potential ECr2O72−/Cr3+∘=1.33E^\circ_{Cr_2O_7^{2-}/Cr^{3+}}=1.33ECr2​O72−​/Cr3+∘​=1.33 V, 2.303RTF=0.059\dfrac{2.303RT}{F}=0.059F2.303RT​=0.059 V]

Correct answer: 10

Step-by-step solution →
Q30·ChemistryIntegerJEE Main 2025
1 Faraday electricity was passed through Cu2+^{2+}2+ (1.5 M, 1 L)/Cu and 0.1 Faraday was passed through Ag+^++ (0.2 M, 1 L)/Ag electrolytic cells. After this the two cells were connected as shown (in the figure) to make an electrochemical cell. The emf of the cell thus formed at 298 K is ______ mV. (Given: ECu2+/Cu∘=0.34E^\circ_{Cu^{2+}/Cu}=0.34ECu2+/Cu∘​=0.34 V, EAg+/Ag∘=0.8E^\circ_{Ag^+/Ag}=0.8EAg+/Ag∘​=0.8 V, 2.303RTF=0.06\dfrac{2.303RT}{F}=0.06F2.303RT​=0.06 V)

Correct answer: 400

Step-by-step solution →
Q31·ChemistrySingle correctJEE Main 2025
1 M aqueous solution of each of Cu(NO3)2Cu(NO_3)_2Cu(NO3​)2​, AgNO3AgNO_3AgNO3​, Hg2(NO3)2Hg_2(NO_3)_2Hg2​(NO3​)2​ and Mg(NO3)2Mg(NO_3)_2Mg(NO3​)2​ are electrolysed using inert electrodes. Given: EAg+/Ag⊖=0.80E^{\ominus}_{Ag^+/Ag}=0.80EAg+/Ag⊖​=0.80 V, EHg22+/Hg⊖=0.79E^{\ominus}_{Hg_2^{2+}/Hg}=0.79EHg22+​/Hg⊖​=0.79 V, ECu2+/Cu⊖=0.24E^{\ominus}_{Cu^{2+}/Cu}=0.24ECu2+/Cu⊖​=0.24 V and EMg2+/Mg⊖=−2.37E^{\ominus}_{Mg^{2+}/Mg}=-2.37EMg2+/Mg⊖​=−2.37 V. Statement (I): With increasing voltage, the sequence of deposition of metals on the cathode will be Ag, Hg and Cu. Statement (II): Magnesium will not be deposited at cathode instead oxygen gas will be evolved at the cathode. In the light of the above statements, choose the most appropriate answer from the options given below:
  1. (A)Both statement I and statement II are incorrect
  2. (B)Statement I is correct but statement II is incorrect
  3. (C)Both statement I and statement II are correct
  4. (D)Statement I is incorrect but statement II is correct

Correct answer: (B)

Step-by-step solution →
Q32·ChemistrySingle correctJEE Main 2025
Given below are two statements: Statement I: Mohr's salt is composed of only three types of ions-ferrous, ammonium and sulphate. Statement II: If the molar conductance at infinite dilution of ferrous, ammonium and sulphate ions are x1x_1x1​, x2x_2x2​ and x3x_3x3​ S cm2^22 mol−1^{-1}−1, respectively then the molar conductance for Mohr's salt solution at infinite dilution would be given by x1+x2+2x3x_1+x_2+2x_3x1​+x2​+2x3​. In the light of the given statements, choose the correct answer from the options given below:
  1. (A)Both statements I and Statement II are false
  2. (B)Statement I is false but Statement II is true
  3. (C)Statement I is true but Statement II is false
  4. (D)Both statements I and Statement II are true

Correct answer: (C)

Step-by-step solution →
Q33·ChemistrySingle correctJEE Main 2025
On charging the lead storage battery, the oxidation state of lead changes from x1x_1x1​ to y1y_1y1​ at the anode and from x2x_2x2​ to y2y_2y2​ at the cathode. The values of x1,y1,x2,y2x_1, y_1, x_2, y_2x1​,y1​,x2​,y2​ are respectively:
  1. (A)+4,+2,0,+2+4, +2, 0, +2+4,+2,0,+2
  2. (B)+2,0,+2,+4+2, 0, +2, +4+2,0,+2,+4
  3. (C)0,+2,+4,+20, +2, +4, +20,+2,+4,+2
  4. (D)+2,0,0,+4+2, 0, 0, +4+2,0,0,+4

Correct answer: (B)

Step-by-step solution →
Q34·ChemistryIntegerJEE Main 2025
KMnO4KMnO_4KMnO4​ acts as an oxidising agent in acidic medium. 'X' is the difference between the oxidation states of Mn in reactant and product. 'Y' is the number of 'd' electrons present in the brown red precipitate formed at the end of the action test with neutral ferric chloride. The value of X+YX+YX+Y is ______.

Correct answer: 10

Step-by-step solution →
Q35·ChemistryIntegerJEE Main 2025
The molar conductance of an infinitely dilute solution of ammonium chloride was found to be 185 S cm2 mol−1185\ S\,cm^2\,mol^{-1}185 Scm2mol−1 and the ionic conductance of hydroxyl and chloride ions are 170 and 70 S cm2 mol−170\ S\,cm^2\,mol^{-1}70 Scm2mol−1, respectively. If molar conductance of 0.02 M solution of ammonium hydroxide is 85.5 S cm2 mol−185.5\ S\,cm^2\,mol^{-1}85.5 Scm2mol−1, its degree of dissociation is given by x×10−1x \times 10^{-1}x×10−1. The value of x is ______ (Nearest integer).

Correct answer: 3

Step-by-step solution →
Q36·ChemistrySingle correctJEE Main 2025
The standard cell potential (Ecell⊖)(E^{\ominus}_{cell})(Ecell⊖​) of a fuel cell based on the oxidation of methanol in air that has been used to power a television relay station is measured as 1.21 V. The standard half cell reduction potential for O2O_2O2​ (EO2/H2O∘)\left(E^{\circ}_{O_2/H_2O}\right)(EO2​/H2​O∘​) is 1.229 V. Choose the correct statement:
  1. (A)The standard half cell reduction potential for the reduction of CO2CO_2CO2​ (ECO2/CH3OH∘)\left(E^{\circ}_{CO_2/CH_3OH}\right)(ECO2​/CH3​OH∘​) is 19 mV
  2. (B)Oxygen is formed at the anode.
  3. (C)Reactants are fed at one go to each electrode.
  4. (D)Reduction of methanol takes place at the cathode.

Correct answer: (A)

Step-by-step solution →
Q37·ChemistrySingle correctJEE Main 2025
Correct order of limiting molar conductivity for cations in water at 298 K is:
  1. (A)H+>Na+>K+>Ca2+>Mg2+H^+ > Na^+ > K^+ > Ca^{2+} > Mg^{2+}H+>Na+>K+>Ca2+>Mg2+
  2. (B)H+>Ca2+>Mg2+>K+>Na+H^+ > Ca^{2+} > Mg^{2+} > K^+ > Na^+H+>Ca2+>Mg2+>K+>Na+
  3. (C)Mg2+>H+>Ca2+>K+>Na+Mg^{2+} > H^+ > Ca^{2+} > K^+ > Na^+Mg2+>H+>Ca2+>K+>Na+
  4. (D)H+>Na+>Ca2+>Mg2+>K+H^+ > Na^+ > Ca^{2+} > Mg^{2+} > K^+H+>Na+>Ca2+>Mg2+>K+

Correct answer: (B)

Step-by-step solution →
Q38·ChemistryIntegerJEE Main 2025
Consider the following electrochemical cell at standard conditions: Au(s) ∣ QH2 ∣ Q ∣∣ NH4X(0.01 M) ∣∣ Ag+(1 M) ∣ Ag(s)Au(s)\,|\,QH_2\,|\,Q\,||\,NH_4X(0.01\,M)\,||\,Ag^+(1\,M)\,|\,Ag(s)Au(s)∣QH2​∣Q∣∣NH4​X(0.01M)∣∣Ag+(1M)∣Ag(s), Ecell∘=+0.4E^\circ_{cell}=+0.4Ecell∘​=+0.4 V. The couple QH2/QQH_2/QQH2​/Q represents the quinhydrone electrode, whose half-cell reaction is Q+2e−+2H+→QH2Q+2e^-+2H^+\to QH_2Q+2e−+2H+→QH2​, EQ/QH2∘=+0.7E^\circ_{Q/QH_2}=+0.7EQ/QH2​∘​=+0.7 V. Given EAg+/Ag∘=+0.8E^\circ_{Ag^+/Ag}=+0.8EAg+/Ag∘​=+0.8 V and 2.303RTF=0.06\dfrac{2.303RT}{F}=0.06F2.303RT​=0.06 V, the pKbpK_bpKb​ value of the ammonium halide salt (NH4X)(NH_4X)(NH4​X) used here is ______ (nearest integer).

Correct answer: 6

Step-by-step solution →
Q39·ChemistryIntegerJEE Main 2025
A 0.2%0.2\%0.2% (w/v) solution of NaOH is measured to have resistivity 870.0 mΩ m870.0\ m\Omega\,m870.0 mΩm. The molar conductivity of the solution will be ______ ×102 mS dm2 mol−1\times10^2\ mS\,dm^2\,mol^{-1}×102 mSdm2mol−1. (Nearest integer)

Correct answer: 23

Step-by-step solution →
Q40·ChemistrySingle correctJEE Main 2025
O2O_2O2​ gas will be evolved as a product of electrolysis of: (A) an aqueous solution of AgNO3AgNO_3AgNO3​ using silver electrodes. (B) an aqueous solution of AgNO3AgNO_3AgNO3​ using platinum electrodes. (C) a dilute solution of H2SO4H_2SO_4H2​SO4​ using platinum electrodes. (D) a high concentration solution of H2SO4H_2SO_4H2​SO4​ using platinum electrodes. Choose the correct answer from the options given below:
  1. (A)(B) and (C) only
  2. (B)(A) and (D) only
  3. (C)(B) and (D) only
  4. (D)(A) and (C) only

Correct answer: (A)

Step-by-step solution →
Q41·ChemistrySingle correctJEE Main 2025
For a Mg ∣ Mg2+(aq) ∣∣ Ag+(aq) ∣ AgMg\,|\,Mg^{2+}(aq)\,||\,Ag^{+}(aq)\,|\,AgMg∣Mg2+(aq)∣∣Ag+(aq)∣Ag the correct Nernst Equation is:
  1. (A)Ecell=Ecell∘−RT2Fln⁡[Ag+][Mg2+]E_{cell}=E^\circ_{cell}-\dfrac{RT}{2F}\ln\dfrac{[Ag^{+}]}{[Mg^{2+}]}Ecell​=Ecell∘​−2FRT​ln[Mg2+][Ag+]​
  2. (B)Ecell=Ecell∘−RT2Fln⁡[Mg2+][Ag+]2E_{cell}=E^\circ_{cell}-\dfrac{RT}{2F}\ln\dfrac{[Mg^{2+}]}{[Ag^{+}]^{2}}Ecell​=Ecell∘​−2FRT​ln[Ag+]2[Mg2+]​
  3. (C)Ecell=Ecell∘−RT2Fln⁡[Mg2+][Ag+]E_{cell}=E^\circ_{cell}-\dfrac{RT}{2F}\ln\dfrac{[Mg^{2+}]}{[Ag^{+}]}Ecell​=Ecell∘​−2FRT​ln[Ag+][Mg2+]​
  4. (D)Ecell=Ecell∘−RT2Fln⁡[Ag+]2[Mg2+]E_{cell}=E^\circ_{cell}-\dfrac{RT}{2F}\ln\dfrac{[Ag^{+}]^{2}}{[Mg^{2+}]}Ecell​=Ecell∘​−2FRT​ln[Mg2+][Ag+]2​

Correct answer: (B)

Step-by-step solution →
Q42·ChemistrySingle correctJEE Main 2025
Match List-I (Applications) with List-II (Batteries/Cell). Choose the correct answer from the options given below:
List-I (Applications)List-II (Batteries/Cell)
A.TransistorsI.Anode – Zn/Hg ; Cathode – HgO + C
B.Hearing aidsII.Hydrogen fuel cell
C.InvertorsIII.Anode – Zn ; Cathode – Carbon
D.Apollo space shipIV.Anode – Pb ; Cathode – Pb | PbO₂
  1. (A)(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
  2. (B)(A)-(III), (B)-(II), (C)-(IV), (D)-(I)
  3. (C)(A)-(IV), (B)-(III), (C)-(II), (D)-(I)
  4. (D)(A)-(II), (B)-(III), (C)-(IV), (D)-(I)

Correct answer: (A)

Step-by-step solution →
Q43·ChemistrySingle correctJEE Main 2025
0.1 M solution of KI reacts with excess of H2SO4H_2SO_4H2​SO4​ and KIO3KIO_3KIO3​ solution. According to the equation 5I−+IO3−+6H+→3I2+3H2O5I^-+IO_3^-+6H^+\rightarrow 3I_2+3H_2O5I−+IO3−​+6H+→3I2​+3H2​O. Identify the correct statements: (A) 200 mL of KI solution reacts with 0.004 mol of KIO3KIO_3KIO3​. (B) 200 mL of KI solution reacts with 0.006 mol of H2SO4H_2SO_4H2​SO4​. (C) 0.5 L of KI solution produced 0.005 mol of I2I_2I2​. (D) Equivalent weight of KIO3KIO_3KIO3​ is equal to Molecular weight5\dfrac{\text{Molecular weight}}{5}5Molecular weight​. Choose the correct answer from the options given below:
  1. (A)(A) and (D) only
  2. (B)(B) and (C) only
  3. (C)(A) and (B) only
  4. (D)(C) and (D) only

Correct answer: (A)

Step-by-step solution →
Q44·ChemistrySingle correctJEE Main 2025
The molar conductivity of a weak electrolyte when plotted against the square root of its concentration, which of the following is expected to be observed?
  1. (A)A small decrease in molar conductivity is observed at infinite dilution.
  2. (B)A small increase in molar conductivity is observed at infinite dilution.
  3. (C)Molar conductivity increases sharply with increase in concentration.
  4. (D)Molar conductivity decreases sharply with increase in concentration.

Correct answer: (D)

Step-by-step solution →
Q45·ChemistrySingle correctJEE Main 2025
Given below are two statements: Statement I: In oxalic acid v/s KMnO4_44​ (in the presence of dil H2_22​SO4_44​) titration the solution needs to be heated initially to 60°C, but no heating is required in Ferrous ammonium sulphate (FAS) vs KMnO4_44​ titration (in the presence of dil H2_22​SO4_44​). Statement II: In oxalic acid vs KMnO4_44​ titration, the initial formation of MnSO4_44​ takes place at high temperature, which then acts as catalyst for further reaction. In the case of FAS vs KMnO4_44​, heating oxidizes Fe2+^{2+}2+ into Fe3+^{3+}3+ by oxygen of air and error may be introduced in the experiment. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)Statement I is false but Statement II is true
  2. (B)Both Statement I and Statement II are true
  3. (C)Statement I is true but Statement II is false
  4. (D)Both Statement I and Statement II are false

Correct answer: (B)

Step-by-step solution →
Q46·ChemistryIntegerJEE Main 2025
Given below is the plot of the molar conductivity vs concentration\sqrt{\text{concentration}}concentration​ for KCl in aqueous solution (shown in figure). If, for the higher concentration of KCl solution, the resistance of the conductivity cell is 100Ω\OmegaΩ, then the resistance of the same cell with the dilute solution is 'x'Ω\OmegaΩ. The value of x is _______ (Nearest integer)

Correct answer: 150

Step-by-step solution →
Q47·ChemistryIntegerJEE Main 2025
Electrolysis of 600 mL aqueous solution of NaCl for 5 min changes the pH of the solution to 12. The current in Amperes used for the given electrolysis is ______. (Nearest integer)

Correct answer: 2

Step-by-step solution →
Q48·ChemistrySingle correctJEE Main 2025
Match List-I (Redox Reaction) with List-II (Type of Redox Reaction). Choose the correct answer from the options given below:
List-I (Redox Reaction)List-II (Type of Redox Reaction)
A.CH4_44​(g) + 2O2_22​(g) →Δ\xrightarrow{\Delta}Δ​ CO2_22​(g) + 2H2_22​O(l)I.Disproportionation reaction
B.2NaH(s) →Δ\xrightarrow{\Delta}Δ​ 2Na(s) + H2_22​(g)II.Combination reaction
C.V2_22​O5_55​(s) + 5Ca(s) →Δ\xrightarrow{\Delta}Δ​ 2V(s) + 5CaO(s)III.Decomposition reaction
D.2H2_22​O2_22​(aq) →Δ\xrightarrow{\Delta}Δ​ 2H2_22​O(l) + O2_22​(g)IV.Displacement reaction
  1. (A)(A)-(II), (B)-(III), (C)-(IV), (D)-(I)
  2. (B)(A)-(II), (B)-(III), (C)-(I), (D)-(IV)
  3. (C)(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
  4. (D)(A)-(IV), (B)-(I), (C)-(II), (D)-(III)

Correct answer: (A)

Step-by-step solution →
Q49·ChemistrySingle correctJEE Main 2025
For the given cell Fe2+(aq)+Ag+(aq)→Fe3+(aq)+Ag(s)Fe^{2+}(aq)+Ag^+(aq)\to Fe^{3+}(aq)+Ag(s)Fe2+(aq)+Ag+(aq)→Fe3+(aq)+Ag(s). The standard cell potential of the above reaction is. Given : Ag++e−→AgAg^++e^-\to AgAg++e−→Ag, E0=xE^0=xE0=x V; Fe2++2e−→FeFe^{2+}+2e^-\to FeFe2++2e−→Fe, E0=yE^0=yE0=y V; Fe3++3e−→FeFe^{3+}+3e^-\to FeFe3++3e−→Fe, E0=zE^0=zE0=z V.
  1. (A)x+y−zx+y-zx+y−z
  2. (B)x+2y−3zx+2y-3zx+2y−3z
  3. (C)y−2xy-2xy−2x
  4. (D)x+2yx+2yx+2y

Correct answer: (B)

Step-by-step solution →
Q50·ChemistrySingle correctJEE Main 2025
Based on the data given below: ECr2O72−/Cr3+0=1.33E^0_{Cr_2O_7^{2-}/Cr^{3+}}=1.33ECr2​O72−​/Cr3+0​=1.33 V, ECl2/Cl−0=1.36E^0_{Cl_2/Cl^-}=1.36ECl2​/Cl−0​=1.36 V, EMnO4−/Mn2+0=1.51E^0_{MnO_4^-/Mn^{2+}}=1.51EMnO4−​/Mn2+0​=1.51 V, ECr3+/Cr0=−0.74E^0_{Cr^{3+}/Cr}=-0.74ECr3+/Cr0​=−0.74 V. The strongest reducing agent is :
  1. (A)Mn2+Mn^{2+}Mn2+
  2. (B)CrCrCr
  3. (C)MnO4−MnO_4^-MnO4−​
  4. (D)Cl−Cl^-Cl−

Correct answer: (B)

Step-by-step solution →
Q51·ChemistrySingle correctJEE Main 2025
Standard electrode potentials for a few half cells are mentioned below: ECu2+/Cu∘=0.34E^\circ_{Cu^{2+}/Cu}=0.34ECu2+/Cu∘​=0.34 V, EZn2+/Zn∘=−0.76E^\circ_{Zn^{2+}/Zn}=-0.76EZn2+/Zn∘​=−0.76 V, EAg+/Ag∘=0.80E^\circ_{Ag^+/Ag}=0.80EAg+/Ag∘​=0.80 V, EMg2+/Mg∘=−2.37E^\circ_{Mg^{2+}/Mg}=-2.37EMg2+/Mg∘​=−2.37 V. Which one of the following cells gives the most negative value of ΔG∘\Delta G^\circΔG∘?
  1. (A)Zn ∣ Zn2+(1M) ∣∣ Ag+(1M) ∣ AgZn\,|\,Zn^{2+}(1M)\,||\,Ag^+(1M)\,|\,AgZn∣Zn2+(1M)∣∣Ag+(1M)∣Ag
  2. (B)Zn ∣ Zn2+(1M) ∣∣ Mg2+(1M) ∣ MgZn\,|\,Zn^{2+}(1M)\,||\,Mg^{2+}(1M)\,|\,MgZn∣Zn2+(1M)∣∣Mg2+(1M)∣Mg
  3. (C)Ag ∣ Ag+(1M) ∣∣ Mg2+(1M) ∣ MgAg\,|\,Ag^+(1M)\,||\,Mg^{2+}(1M)\,|\,MgAg∣Ag+(1M)∣∣Mg2+(1M)∣Mg
  4. (D)Cu ∣ Cu2+(1M) ∣∣ Ag+(1M) ∣ AgCu\,|\,Cu^{2+}(1M)\,||\,Ag^+(1M)\,|\,AgCu∣Cu2+(1M)∣∣Ag+(1M)∣Ag

Correct answer: (A)

Step-by-step solution →
Q52·ChemistrySingle correctJEE Main 2025
In the diagram below, the standard electrode potentials are given in volts (over the arrow). FeO42−→+2.0VFe3+→0.8VFe2+→−0.5VFe0FeO_4^{2-}\xrightarrow{+2.0V}Fe^{3+}\xrightarrow{0.8V}Fe^{2+}\xrightarrow{-0.5V}Fe^{0}FeO42−​+2.0V​Fe3+0.8V​Fe2+−0.5V​Fe0. The value of EFeO42−/Fe2+∘E^{\circ}_{FeO_4^{2-}/Fe^{2+}}EFeO42−​/Fe2+∘​ is
  1. (A)1.7 V
  2. (B)1.2 V
  3. (C)2.1 V
  4. (D)1.4 V

Correct answer: (A)

Step-by-step solution →
Q53·ChemistrySingle correctJEE Main 2025
Given below are two statements: Statement (I): Corrosion is an electrochemical phenomenon in which pure metal acts as an anode and impure metal as a cathode. Statement (II): The rate of corrosion is more in alkaline medium than in acidic medium. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)Both Statement I and Statement II are false
  2. (B)Statement I is false but Statement II is true
  3. (C)Both Statement I and Statement II are true
  4. (D)Statement I is true but Statement II is false

Correct answer: (D)

Step-by-step solution →
Q54·ChemistrySingle correctJEE Main 2025
Which of the following electrolytes can be used to obtain H₂S₂O₈ by the process of electrolysis?
  1. (A)Dilute solution of sodium sulphate
  2. (B)Dilute solution of sulphuric acid
  3. (C)Concentrated solution of sulphuric acid
  4. (D)Acidified dilute solution of sodium sulphate

Correct answer: (C)

Step-by-step solution →
Q55·ChemistrySingle correctJEE Main 2025
A solution of aluminium chloride is electrolysed for 30 minutes using a current of 2 A. The amount of aluminium deposited at the cathode is ______. (Given: molar masses of aluminium and chlorine are 27 g mol⁻¹ and 35.5 g mol⁻¹ respectively; Faraday constant = 96500 C mol⁻¹)
  1. (A)1.660 g
  2. (B)1.007 g
  3. (C)0.336 g
  4. (D)0.441 g

Correct answer: (C)

Step-by-step solution →
Q56·ChemistryMultiple correctJEE Advanced 2024
An aqueous solution of hydrazine (N2H4N_2H_4N2​H4​) is electrochemically oxidized by O2O_2O2​, thereby releasing chemical energy in the form of electrical energy. One of the products generated from the electrochemical reaction is N2(g)N_2(g)N2​(g). Choose the correct statement(s) about the above process (A) OH−OH^-OH− ions react with N2H4N_2H_4N2​H4​ at the anode to form N2(g)N_2(g)N2​(g) and water, releasing 4 electrons to the anode. (B) At the cathode, N2H4N_2H_4N2​H4​ breaks to N2(g)N_2(g)N2​(g) and nascent hydrogen released at the electrode reacts with oxygen to form water. (C) At the cathode, molecular oxygen gets converted to OH−OH^-OH−. (D) Oxides of nitrogen are major by-products of the electrochemical process.
  1. (A)OH−OH^-OH− ions react with N2H4N_2H_4N2​H4​ at the anode to form N2(g)N_2(g)N2​(g) and water, releasing 4 electrons to the anode.
  2. (B)At the cathode, N2H4N_2H_4N2​H4​ breaks to N2(g)N_2(g)N2​(g) and nascent hydrogen released at the electrode reacts with oxygen to form water.
  3. (C)At the cathode, molecular oxygen gets converted to OH−OH^-OH−.
  4. (D)Oxides of nitrogen are major by-products of the electrochemical process.

Correct answer: (A), (C)

Step-by-step solution →
Q57·ChemistrySingle correctJEE Advanced 2024
In a conductometric titration, small volume of titrant of higher concentration is added stepwise to a larger volume of titrate of much lower concentration, and the conductance is measured after each addition. The limiting ionic conductivity (Λ0)\left(\Lambda_{0}\right)(Λ0​) values (in mS m2^{2}2 mol−1^{-1}−1) for different ions in aqueous solutions are given below: Ag+^{+}+ 6.2; K+^{+}+ 7.4; Na+^{+}+ 5.0; H+^{+}+ 35.0; NO3−_{3}^{-}3−​ 7.2; Cl−^{-}− 7.6; SO42−_{4}^{2-}42−​ 16.0; OH−^{-}− 19.9; CH3_{3}3​COO−^{-}− 4.1 For different combinations of titrates and titrants given in List-I, the graphs of 'conductance' versus 'volume of titrant' are given in List-II. Match each entry List-I with the appropriate entry in List-II and choose the correct option.
List-IList-II
P.Titrate: KCl Titrant: AgNO3_{3}3​1.see figure
Q.Titrate: AgNO3_{3}3​ Titrant: KCl2.see figure
R.Titrate: NaOH Titrant: HCl3.see figure
S.Titrate: NaOH Titrant: CH3_{3}3​COOH4.see figure
5.see figure
  1. (A)(P) →\to→ (4), (Q) →\to→ (3), (R) →\to→ (2), (S) →\to→ (5)
  2. (B)(P) →\to→ (2), (Q) →\to→ (4), (R) →\to→ (3), (S) →\to→ (1)
  3. (C)(P) →\to→ (3), (Q) →\to→ (4), (R) →\to→ (2), (S) →\to→ (5)
  4. (D)(P) →\to→ (4), (Q) →\to→ (3), (R) →\to→ (2), (S) →\to→ (1)

Correct answer: (C)

Step-by-step solution →
Q58·ChemistrySingle correctJEE Main 2024
The molar conductivity for electrolytes A and B are plotted against C1/2C^{1/2}C1/2 as shown below. Electrolytes A and B respectively are :
  1. (A)A: Weak electrolyte, B: weak electrolyte
  2. (B)A: Strong electrolyte, B: strong electrolyte
  3. (C)A: Weak electrolyte, B: strong electrolyte
  4. (D)A: Strong electrolyte, B: weak electrolyte

Correct answer: (C)

Step-by-step solution →
Q59·ChemistryNumericalJEE Main 2024
The standard reduction potentials at 298 K for the following half cells are given below: Cr2O72−+14H++6e−→2Cr3++7H2OCr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2OCr2​O72−​+14H++6e−→2Cr3++7H2​O, E∘=1.33E^\circ = 1.33E∘=1.33V Fe3+(aq)+3e−→FeFe^{3+}(aq) + 3e^- \rightarrow FeFe3+(aq)+3e−→Fe, E∘=−0.04E^\circ = -0.04E∘=−0.04V Ni2+(aq)+2e−→NiNi^{2+}(aq) + 2e^- \rightarrow NiNi2+(aq)+2e−→Ni, E∘=−0.25E^\circ = -0.25E∘=−0.25V Ag+(aq)+e−→AgAg^+(aq) + e^- \rightarrow AgAg+(aq)+e−→Ag, E∘=0.80E^\circ = 0.80E∘=0.80V Au3+(aq)+3e−→AuAu^{3+}(aq) + 3e^- \rightarrow AuAu3+(aq)+3e−→Au, E∘=1.40E^\circ = 1.40E∘=1.40V Consider the given electrochemical reactions, The number of metal(s) which will be oxidized be Cr2O72−Cr_2O_7^{2-}Cr2​O72−​, in aqueous solution is _____.

Correct answer: 3

Step-by-step solution →
Q60·ChemistrySingle correctJEE Main 2024
Which out of the following is a correct equation to show change in molar conductivity with respect to concentration of a weak electrolyte, if the symbols carry their usual meaning:
  1. (A)Λm2C−Ka(Λm0)2+KaΛmΛm0=0\Lambda_m^{2}C - K_a(\Lambda_m^{0})^{2} + K_a\Lambda_m\Lambda_m^{0} = 0Λm2​C−Ka​(Λm0​)2+Ka​Λm​Λm0​=0
  2. (B)Λm=Λm0+AC1/2=0\Lambda_m = \Lambda_m^{0} + AC^{1/2} = 0Λm​=Λm0​+AC1/2=0
  3. (C)Λm=Λm0−AC1/2=0\Lambda_m = \Lambda_m^{0} - AC^{1/2} = 0Λm​=Λm0​−AC1/2=0
  4. (D)Λm2C+Ka(Λm0)2−KaΛmΛm0=0\Lambda_m^{2}C + K_a(\Lambda_m^{0})^{2} - K_a\Lambda_m\Lambda_m^{0} = 0Λm2​C+Ka​(Λm0​)2−Ka​Λm​Λm0​=0

Correct answer: (A)

Step-by-step solution →
Q61·ChemistrySingle correctJEE Main 2024
Match List-I (Cell) with List-II (Use / Property / Reaction). Choose the correct answer from the options given below:
List-I (Cell)List-II (Use / Property / Reaction)
A.Leclanche cellI.Converts energy of combustion into electrical energy
B.Ni-Cd cellII.Does not involve any ion in solution and is used in hearing aids
C.Fuel cellIII.Rechargeable
D.Mercury cellIV.Reaction at anode Zn→Zn2++2e−Zn \rightarrow Zn^{2+} + 2e^{-}Zn→Zn2++2e−
  1. (A)A-I, B-II, C-III, D-IV
  2. (B)A-III, B-I, C-IV, D-II
  3. (C)A-IV, B-III, C-I, D-II
  4. (D)A-II, B-III, C-IV, D-I

Correct answer: (C)

Step-by-step solution →
Q62·ChemistrySingle correctJEE Main 2024
Given below are two statements : Statement (I) : The oxidation state of an element in a particular compound is the charge acquired by its atom on the basis of electron gain enthalpy consideration from other atoms in the molecule. Statement (II) : pπp\pipπ-pπp\pipπ bond formation is more prevalent in second period elements over other periods. In the light of the above statements, choose the most appropriate answer from the options given below :
  1. (A)Both Statement I and Statement II are incorrect
  2. (B)Statement I is correct but Statement II is incorrect
  3. (C)Both Statement I and Statement II are correct
  4. (D)Statement I is incorrect but Statement II is correct

Correct answer: (D)

Step-by-step solution →
Q63·ChemistrySingle correctJEE Main 2024
The emf of the cell Tl ∣ Tl+(0.001 M) ∣∣ Cu2+(0.01 M) ∣ Cu\text{Tl}\,|\,\text{Tl}^+(0.001\,\text{M})\,||\,\text{Cu}^{2+}(0.01\,\text{M})\,|\,\text{Cu}Tl∣Tl+(0.001M)∣∣Cu2+(0.01M)∣Cu is 0.83 V at 298 K. It could be increased by:
  1. (A)increasing concentration of Tl+\text{Tl}^+Tl+ ions
  2. (B)increasing concentration of both Tl+\text{Tl}^+Tl+ and Cu2+\text{Cu}^{2+}Cu2+ ions
  3. (C)decreasing concentration of both Tl+\text{Tl}^+Tl+ and Cu2+\text{Cu}^{2+}Cu2+ ions
  4. (D)increasing concentration of Cu2+\text{Cu}^{2+}Cu2+ ions

Correct answer: (D)

Step-by-step solution →
Q64·ChemistrySingle correctJEE Main 2024
Thiosulphate reacts differently with iodine and bromine in the reactions given below: 2S2O32−+I2→S4O62−+2I−2S_2O_3^{2-}+I_2\to S_4O_6^{2-}+2I^-2S2​O32−​+I2​→S4​O62−​+2I−; S2O32−+5Br2+5H2O→2SO42−+4Br−+10H+S_2O_3^{2-}+5Br_2+5H_2O\to 2SO_4^{2-}+4Br^-+10H^+S2​O32−​+5Br2​+5H2​O→2SO42−​+4Br−+10H+. Which of the following statement justifies the above dual behaviour of thiosulphate?
  1. (A)Bromine undergoes oxidation and iodine undergoes reduction by iodine in these reactions
  2. (B)Bromine undergoes oxidation by bromine and reduction by iodine in these reactions
  3. (C)Bromine is a stronger oxidant than iodine
  4. (D)Bromine is a weaker oxidant than iodine

Correct answer: (C)

Step-by-step solution →
Q65·ChemistrySingle correctJEE Main 2024
The reaction 12H2(g)+AgCl(s)→H+(aq)+Cl−(aq)+Ag(s)\tfrac{1}{2}\text{H}_2(g)+\text{AgCl}(s)\to \text{H}^+(aq)+\text{Cl}^-(aq)+\text{Ag}(s)21​H2​(g)+AgCl(s)→H+(aq)+Cl−(aq)+Ag(s) occurs in which of the following galvanic cell:
  1. (A)Pt ∣ H2(g) ∣ HCl(soln.) ∣ AgCl(s) ∣ Ag\text{Pt}\,|\,\text{H}_2(g)\,|\,\text{HCl(soln.)}\,|\,\text{AgCl}(s)\,|\,\text{Ag}Pt∣H2​(g)∣HCl(soln.)∣AgCl(s)∣Ag
  2. (B)Pt ∣ H2(g) ∣ HCl(soln.) ∣ AgNO3(aq) ∣ Ag\text{Pt}\,|\,\text{H}_2(g)\,|\,\text{HCl(soln.)}\,|\,\text{AgNO}_3(aq)\,|\,\text{Ag}Pt∣H2​(g)∣HCl(soln.)∣AgNO3​(aq)∣Ag
  3. (C)Pt ∣ H2(g) ∣ KCl(soln.) ∣ AgCl(s) ∣ Ag\text{Pt}\,|\,\text{H}_2(g)\,|\,\text{KCl(soln.)}\,|\,\text{AgCl}(s)\,|\,\text{Ag}Pt∣H2​(g)∣KCl(soln.)∣AgCl(s)∣Ag
  4. (D)Ag ∣ AgCl(s) ∣ KCl(soln.) ∣ AgNO3(aq.) ∣ Ag\text{Ag}\,|\,\text{AgCl}(s)\,|\,\text{KCl(soln.)}\,|\,\text{AgNO}_3(aq.)\,|\,\text{Ag}Ag∣AgCl(s)∣KCl(soln.)∣AgNO3​(aq.)∣Ag

Correct answer: (C)

Step-by-step solution →
Q66·ChemistrySingle correctJEE Main 2024
Iron (III) catalyses the reaction between iodide and persulphate ions, in which A. Fe3+Fe^{3+}Fe3+ oxidises the iodide ion; B. Fe2+Fe^{2+}Fe2+ oxidises the persulphate ion; C. Fe2+Fe^{2+}Fe2+ reduces the iodide ion; D. Fe3+Fe^{3+}Fe3+ reduces the persulphate ion. Choose the most appropriate answer from the options given below:
  1. (A)B and C only
  2. (B)B only
  3. (C)A only
  4. (D)A and D only

Correct answer: (D)

Step-by-step solution →
Q67·ChemistrySingle correctJEE Main 2024
Match List-I (redox reactions) with List-II (the type of redox reaction). Choose the correct answer:
List-I (Reaction)List-II (Type of redox reaction)
A.N2(g)+O2(g)→2NO(g)N_{2(g)} + O_{2(g)} \rightarrow 2NO_{(g)}N2(g)​+O2(g)​→2NO(g)​I.Decomposition
B.2Pb(NO3)2(s)→2PbO(s)+4NO2(g)+O2(g)2Pb(NO_3)_{2(s)} \rightarrow 2PbO_{(s)} + 4NO_{2(g)} + O_{2(g)}2Pb(NO3​)2(s)​→2PbO(s)​+4NO2(g)​+O2(g)​II.Displacement
C.2Na(s)+2H2O(l)→2NaOH(aq.)+H2(g)2Na_{(s)} + 2H_2O_{(l)} \rightarrow 2NaOH_{(aq.)} + H_{2(g)}2Na(s)​+2H2​O(l)​→2NaOH(aq.)​+H2(g)​III.Disproportionation
D.2NO2(g)+2 −OH(aq.)→NO2(aq.)−+NO3(aq.)−+H2O(l)2NO_{2(g)} + 2\,^{-}OH_{(aq.)} \rightarrow NO_{2(aq.)}^{-} + NO_{3(aq.)}^{-} + H_2O_{(l)}2NO2(g)​+2−OH(aq.)​→NO2(aq.)−​+NO3(aq.)−​+H2​O(l)​IV.Combination
  1. (A)(A)-(I), (B)-(II), (C)-(III), (D)-(IV)
  2. (B)(A)-(III), (B)-(II), (C)-(I), (D)-(IV)
  3. (C)(A)-(II), (B)-(III), (C)-(IV), (D)-(I)
  4. (D)(A)-(IV), (B)-(I), (C)-(II), (D)-(III)

Correct answer: (D)

Step-by-step solution →
Q68·ChemistrySingle correctJEE Main 2024
How can an electrochemical cell be converted into an electrolytic cell?
  1. (A)Applying an external opposite potential greater than Ecell0E^0_{cell}Ecell0​
  2. (B)Reversing the flow of ions in salt bridge.
  3. (C)Applying an external opposite potential lower than Ecell0E^0_{cell}Ecell0​.
  4. (D)Exchanging the electrodes at anode and cathode.

Correct answer: (A)

Step-by-step solution →
Q69·ChemistrySingle correctJEE Main 2024
A conductivity cell with two electrodes (dark side) is half filled with infinitely dilute aqueous solution of a weak electrolyte (cell shown in the figure). If volume is doubled by adding more water at constant temperature, the molar conductivity of the cell will:
  1. (A)increase sharply
  2. (B)remain same or can not be measured accurately
  3. (C)decrease sharply
  4. (D)depend upon type of electrolyte

Correct answer: (B)

Step-by-step solution →
Q70·ChemistrySingle correctJEE Main 2024
For the electrochemical cell M ∣ M2+ ∣∣ X ∣ X2−M\,|\,M^{2+}\,||\,X\,|\,X^{2-}M∣M2+∣∣X∣X2−, if E(M2+/M)∘=0.46E^\circ_{(M^{2+}/M)} = 0.46E(M2+/M)∘​=0.46 V and E(X/X2−)∘=0.34E^\circ_{(X/X^{2-})} = 0.34E(X/X2−)∘​=0.34 V, which of the following is correct?
  1. (A)Ecell=−0.80E_{cell} = -0.80Ecell​=−0.80 V
  2. (B)M+X→M2++X2−M + X \rightarrow M^{2+} + X^{2-}M+X→M2++X2− is a spontaneous reaction
  3. (C)M2++X2−→M+XM^{2+} + X^{2-} \rightarrow M + XM2++X2−→M+X is a spontaneous reaction
  4. (D)Ecell=0.80E_{cell} = 0.80Ecell​=0.80 V

Correct answer: (C)

Step-by-step solution →
Q71·ChemistrySingle correctJEE Main 2024
The quantity of silver deposited when one coulomb of charge is passed through AgNO3\text{AgNO}_3AgNO3​ solution is:
  1. (A)0.1 g atom of silver
  2. (B)1 chemical equivalent of silver
  3. (C)1 g of silver
  4. (D)1 electrochemical equivalent of silver

Correct answer: (D)

Step-by-step solution →
Q72·ChemistrySingle correctJEE Main 2024
The reaction at the cathode in the cells commonly used in clocks involves:
  1. (A)reduction of Mn from +4 to +3
  2. (B)oxidation of Mn from +3 to +4
  3. (C)reduction of Mn from +7 to +2
  4. (D)oxidation of Mn from +7 to +2

Correct answer: (A)

Step-by-step solution →
Q73·ChemistrySingle correctJEE Main 2024
What pressure (bar) of H2H_2H2​ would be required to make emf of hydrogen electrode zero in pure water at 25∘C25^\circ C25∘C?
  1. (A)10−1410^{-14}10−14
  2. (B)10−710^{-7}10−7
  3. (C)111
  4. (D)0.50.50.5

Correct answer: (A)

Step-by-step solution →
Q74·ChemistrySingle correctJEE Main 2024
One of the commonly used electrode is calomel electrode. Under which of the following categories calomel electrode comes?
  1. (A)Metal – Insoluble Salt – Anion electrodes
  2. (B)Oxidation – Reduction electrodes
  3. (C)Gas – Ion electrodes
  4. (D)Metal ion – Metal electrodes

Correct answer: (A)

Step-by-step solution →
Q75·ChemistryNumericalJEE Main 2024
Only 2 mL2\,mL2mL of KMnO4KMnO_4KMnO4​ solution of unknown molarity is required to reach the end point of a titration of 20 mL20\,mL20mL of oxalic acid (2 M)(2\,M)(2M) in acidic medium. The molarity of KMnO4KMnO_4KMnO4​ solution should be ___ M.

Correct answer: 8

Step-by-step solution →
Q76·ChemistrySingle correctJEE Main 2024
Fuel cell, using hydrogen and oxygen as fuels, A. has been used in spaceship, B. has an efficiency of 40% to produce electricity, C. uses aluminium as catalysts, D. is eco-friendly, E. is actually a type of Galvanic cell only. Choose the correct answer from the options given below:
  1. (A)A, B, C only
  2. (B)A, B, D only
  3. (C)A, B, D, E only
  4. (D)A, D, E only

Correct answer: (D)

Step-by-step solution →
Q77·ChemistrySingle correctJEE Main 2024
For a strong electrolyte, a plot of molar conductivity against (concentration)1/2^{1/2}1/2 is a straight line, with a negative slope, the correct unit for the slope is
  1. (A)S cm2 mol−3/2 L1/2\text{S cm}^2\,\text{mol}^{-3/2}\,\text{L}^{1/2}S cm2mol−3/2L1/2
  2. (B)S cm2 mol−1 L1/2\text{S cm}^2\,\text{mol}^{-1}\,\text{L}^{1/2}S cm2mol−1L1/2
  3. (C)S cm2 mol−3/2 L\text{S cm}^2\,\text{mol}^{-3/2}\,\text{L}S cm2mol−3/2L
  4. (D)S cm2 mol−3/2 L−1/2\text{S cm}^2\,\text{mol}^{-3/2}\,\text{L}^{-1/2}S cm2mol−3/2L−1/2

Correct answer: (A)

Step-by-step solution →
Q78·ChemistrySingle correctJEE Main 2024
Which of the following reactions are disproportionation reactions? (A) Cu+→Cu2++CuCu^+\to Cu^{2+}+CuCu+→Cu2++Cu (B) 3MnO42−+4H+→2MnO4−+MnO2+2H2O3MnO_4^{2-}+4H^+\to 2MnO_4^-+MnO_2+2H_2O3MnO42−​+4H+→2MnO4−​+MnO2​+2H2​O (C) 2KMnO4→K2MnO4+MnO2+O22KMnO_4\to K_2MnO_4+MnO_2+O_22KMnO4​→K2​MnO4​+MnO2​+O2​ (D) 2MnO4−+3Mn2++2H2O→5MnO2+4H+2MnO_4^-+3Mn^{2+}+2H_2O\to 5MnO_2+4H^+2MnO4−​+3Mn2++2H2​O→5MnO2​+4H+. Choose the correct answer from the options given below:
  1. (A)(A), (B)
  2. (B)(B), (C), (D)
  3. (C)(A), (B), (C)
  4. (D)(A), (D)

Correct answer: (A)

Step-by-step solution →
Q79·ChemistryNumericalJEE Main 2024
Consider the following redox reaction: MnO4−+H++H2C2O4⇌Mn2++H2O+CO2\text{MnO}_4^-+\text{H}^++\text{H}_2\text{C}_2\text{O}_4\rightleftharpoons\text{Mn}^{2+}+\text{H}_2\text{O}+\text{CO}_2MnO4−​+H++H2​C2​O4​⇌Mn2++H2​O+CO2​. The standard reduction potentials are given as below (Ered∘E^{\circ}_{\text{red}}Ered∘​): EMnO4−/Mn2+∘=+1.51E^{\circ}_{\text{MnO}_4^-/\text{Mn}^{2+}}=+1.51EMnO4−​/Mn2+∘​=+1.51 V, ECO2/H2C2O4∘=−0.49E^{\circ}_{\text{CO}_2/\text{H}_2\text{C}_2\text{O}_4}=-0.49ECO2​/H2​C2​O4​∘​=−0.49 V. If the equilibrium constant of the above reaction is given as Keq=10xK_{\text{eq}}=10^xKeq​=10x, then the value of x=x=x= __________ (nearest integer).

Correct answer: 338

Step-by-step solution →
Q80·ChemistrySingle correctJEE Main 2024
In acidic medium, K2Cr2O7K_2Cr_2O_7K2​Cr2​O7​ shows oxidising action as represented in the half reaction Cr2O72−+XH++Ye−→2Cr3++ZH2OCr_2O_7^{2-}+XH^++Ye^-\to 2Cr^{3+}+ZH_2OCr2​O72−​+XH++Ye−→2Cr3++ZH2​O. XXX, YYY, ZZZ and AAA are respectively:
  1. (A)8,6,48,6,48,6,4 and Cr2O5Cr_2O_5Cr2​O5​
  2. (B)14,7,614,7,614,7,6 and Cr3+Cr^{3+}Cr3+
  3. (C)8,4,68,4,68,4,6 and Cr2O5Cr_2O_5Cr2​O5​
  4. (D)14,6,714,6,714,6,7 and Cr3+Cr^{3+}Cr3+

Correct answer: (D)

Step-by-step solution →
Q81·ChemistryNumericalJEE Main 2024
The amount of electricity in Coulomb required for the oxidation of 1 mol of H2O\text{H}_2\text{O}H2​O to O2\text{O}_2O2​ is __________ ×105\times 10^5×105 C.

Correct answer: 2

Step-by-step solution →
Q82·ChemistryNumericalJEE Main 2024
The potential for the given half cell at 298K298K298K is (−) ‾×10−2 V(-)\,\underline{\quad}\times10^{-2}\,V(−)​×10−2V. 2H(aq)++2e−→H2(g)2H^+_{(aq)}+2e^-\to H_2(g)2H(aq)+​+2e−→H2​(g); [H+]=1M[H^+]=1M[H+]=1M, PH2=2P_{H_2}=2PH2​​=2 atm. (Given: 2.303 RT/F=0.06 V2.303\,RT/F=0.06\,V2.303RT/F=0.06V, log⁡2=0.3\log2=0.3log2=0.3)

Correct answer: 0.9

Step-by-step solution →
Q83·ChemistrySingle correctJEE Main 2024
Given below are two statements: Statement-I: S8S_8S8​ solid undergoes disproportionation reaction under alkaline conditions to form S2−S^{2-}S2− and S2O32−S_2O_3^{2-}S2​O32−​. Statement-II: ClO4−ClO_4^-ClO4−​ can undergo disproportionation reaction under acidic condition. In the light of the above statements, choose the most appropriate answer from the options given below:
  1. (A)Statement I is correct but statement II is incorrect.
  2. (B)Statement I is incorrect but statement II is correct
  3. (C)Both Statement I and Statement II are incorrect
  4. (D)Both Statement I and statement II are correct

Correct answer: (A)

Step-by-step solution →
Q84·ChemistryNumericalJEE Main 2024
One Faraday of electricity liberates x×10−1x\times10^{-1}x×10−1 gram atom of copper from copper sulphate, x is ______.

Correct answer: 5

Step-by-step solution →
Q85·ChemistryNumericalJEE Main 2024
Number of moles of H+H^+H+ ions required by 1 mole of MnO4−MnO_4^-MnO4−​ to oxidise oxalate ion to CO2CO_2CO2​ is ______.

Correct answer: 8

Step-by-step solution →
Q86·ChemistrySingle correctJEE Main 2024
Identify the factor from the following that does not affect electrolytic conductance of a solution.
  1. (A)The nature of the electrolyte added.
  2. (B)The nature of the electrode used.
  3. (C)Concentration of the electrolyte.
  4. (D)The nature of solvent used.

Correct answer: (B)

Step-by-step solution →
Q87·ChemistrySingle correctJEE Main 2024
The metals that are employed in the battery industries are A. Fe B. Mn C. Ni D. Cr E. Cd Choose the correct answer from the options given below:
  1. (A)B, C and E only
  2. (B)A, B, C, D and E
  3. (C)A, B, C and D only
  4. (D)B, D and E only

Correct answer: (A)

Step-by-step solution →
Q88·ChemistrySingle correctJEE Main 2024
Reduction potential of ions are given below: EClO4−∘=1.19E^\circ_{ClO_4^-}=1.19EClO4−​∘​=1.19 V, EIO4−∘=1.65E^\circ_{IO_4^-}=1.65EIO4−​∘​=1.65 V, EBrO4−∘=1.74E^\circ_{BrO_4^-}=1.74EBrO4−​∘​=1.74 V. The correct order of their oxidising power is:
  1. (A)ClO4−>IO4−>BrO4−ClO_4^- > IO_4^- > BrO_4^-ClO4−​>IO4−​>BrO4−​
  2. (B)ClO4−>BrO4−>IO4−ClO_4^- > BrO_4^- > IO_4^-ClO4−​>BrO4−​>IO4−​
  3. (C)BrO4−>ClO4−>IO4−BrO_4^- > ClO_4^- > IO_4^-BrO4−​>ClO4−​>IO4−​
  4. (D)BrO4−>IO4−>ClO4−BrO_4^- > IO_4^- > ClO_4^-BrO4−​>IO4−​>ClO4−​

Correct answer: (D)

Step-by-step solution →
Q89·ChemistryNumericalJEE Main 2024
2MnO4−+bI−+cH2O→xI2+yMnO2+zOH−2MnO_4^- + bI^- + cH_2O \to xI_2 + yMnO_2 + zOH^-2MnO4−​+bI−+cH2​O→xI2​+yMnO2​+zOH−. If the above equation is balanced with integer coefficients, the value of zzz is ___

Correct answer: 8

Step-by-step solution →
Q90·ChemistryNumericalJEE Main 2024
Total number of species from the following which can undergo disproportionation reaction is ______. H2O2H_2O_2H2​O2​, ClO3−ClO_3^-ClO3−​, P4P_4P4​, Cl2Cl_2Cl2​, AgAgAg, Cu+1Cu^{+1}Cu+1, F2F_2F2​, NO2NO_2NO2​, K+K^+K+

Correct answer: 6

Step-by-step solution →
Q91·ChemistrySingle correctJEE Main 2024
In alkaline medium, MnO4−MnO_4^-MnO4−​ oxidises I−I^-I− to
  1. (A)IO4−IO_4^-IO4−​
  2. (B)IO−IO^-IO−
  3. (C)I2I_2I2​
  4. (D)IO3−IO_3^-IO3−​

Correct answer: (D)

Step-by-step solution →
Q92·ChemistryNumericalJEE Main 2024
The mass of zinc produced by the electrolysis of zinc sulphate solution with a steady current of 0.015 A for 15 minutes is ______ ×10−4\times10^{-4}×10−4 g. (Atomic mass of zinc = 65.4 amu)

Correct answer: 45.75

Step-by-step solution →
Q93·ChemistryNumericalJEE Main 2024
A constant current was passed through a solution of AuCl4−AuCl_4^-AuCl4−​ ion between gold electrodes. After a period of 10.0 minutes, the increase in mass of cathode was 1.314 g. The total charge passed through the solution is ___ ×10−2\times10^{-2}×10−2 F. (Given atomic mass of Au = 197)

Correct answer: 2

Step-by-step solution →
Q94·ChemistrySingle correctJEE Main 2024
Chlorine undergoes disproportionation in alkaline medium as shown below : a Cl2(g)+b OH−(aq)→c ClO−(aq)+d Cl−(aq)+e H2O(l)a\,Cl_2(g)+b\,OH^-(aq)\to c\,ClO^-(aq)+d\,Cl^-(aq)+e\,H_2O(l)aCl2​(g)+bOH−(aq)→cClO−(aq)+dCl−(aq)+eH2​O(l). The values of a, b, c and d in a balanced redox reaction are respectively :
  1. (A)1, 2, 1 and 1
  2. (B)2, 2, 1 and 3
  3. (C)3, 4, 4 and 2
  4. (D)2, 4, 1 and 3

Correct answer: (A)

Step-by-step solution →
Q95·ChemistryNumericalJEE Main 2024
The hydrogen electrode is dipped in a solution of pH = 3 at 25∘25^\circ25∘C. The potential of the electrode will be __________ ×10−2\times10^{-2}×10−2 V. (2.303RTF=0.059 V)\left(\dfrac{2.303RT}{F}=0.059\text{ V}\right)(F2.303RT​=0.059 V)

Correct answer: -18

Step-by-step solution →
Q96·ChemistryNumericalJEE Main 2024
The mass of silver (Molar mass of Ag : 108 g\,mol−1^{-1}−1) displaced by a quantity of electricity which displaces 5600 mL of O2\text{O}_2O2​ at S.T.P. will be __________ g.

Correct answer: 108

Step-by-step solution →
Q97·ChemistrySingle correctJEE Main 2024
Which of the following statements is not correct about rusting of iron ?
  1. (A)Coating of iron surface by tin prevents rusting, even if the tin coating is peeling off.
  2. (B)When pH lies above 9 or 10, rusting of iron does not take place.
  3. (C)Dissolved acidic oxides SO2SO_2SO2​, NO2NO_2NO2​ in water act as catalyst in the process of rusting.
  4. (D)Rusting of iron is envisaged as setting up of electrochemical cell on the surface of iron object.

Correct answer: (A)

Step-by-step solution →
Q98·ChemistryNumericalJEE Main 2024
1 mole of PbS is oxidised by "X" moles of O3O_3O3​ to get "Y" moles of O2O_2O2​. X+YX+YX+Y is __________.

Correct answer: 8

Step-by-step solution →
Q99·ChemistrySingle correctJEE Advanced 2023
Plotting 1/Λm1/\Lambda_\mathrm{m}1/Λm​ against cΛmc\Lambda_\mathrm{m}cΛm​ for aqueous solutions of a monobasic weak acid (HX) resulted in a straight line with y-axis intercept of P and slope of S. The ratio P/S is [Λm\Lambda_\mathrm{m}Λm​ = molar conductivity Λm0\Lambda_\mathrm{m}^{0}Λm0​ = limiting molar conductivity c = molar concentration Ka\mathrm{K_a}Ka​ = dissociation constant of HX]
  1. (A)KaΛm0\mathrm{K_a}\Lambda_\mathrm{m}^{0}Ka​Λm0​
  2. (B)KaΛm0/2\mathrm{K_a}\Lambda_\mathrm{m}^{0}/2Ka​Λm0​/2
  3. (C)2KaΛm02\mathrm{K_a}\Lambda_\mathrm{m}^{0}2Ka​Λm0​
  4. (D)1/(KaΛm0)1/\left(\mathrm{K_a}\Lambda_\mathrm{m}^{0}\right)1/(Ka​Λm0​)

Correct answer: (A)

Step-by-step solution →
Q100·ChemistryIntegerJEE Advanced 2023
Consider the following molecules: Br3O8\mathrm{Br_3O_8}Br3​O8​, F2O\mathrm{F_2O}F2​O, H2S4O6\mathrm{H_2S_4O_6}H2​S4​O6​, H2S5O6\mathrm{H_2S_5O_6}H2​S5​O6​, and C3O2\mathrm{C_3O_2}C3​O2​. Count the number of atoms existing in their zero oxidation state in each molecule. Their sum is____.

Correct answer: 06

Step-by-step solution →
Q101·ChemistryIntegerJEE Advanced 2023
H2S\mathrm{H_2S}H2​S (5 moles) reacts completely with acidified aqueous potassium permanganate solution. In this reaction, the number of moles of water produced is x\mathbf{x}x, and the number of moles of electrons involved is y\mathbf{y}y. The value of (x+y)(\mathbf{x} + \mathbf{y})(x+y) is ____.

Correct answer: 18

Step-by-step solution →
Q102·ChemistryNumericalJEE Main 2023
The number of correct statements from the following is _______ (A) Conductivity always decreases with decrease in concentration for both strong and weak electrolytes. (B) The number of ions per unit volume that carry current in a solution increases on dilution. (C) Molar conductivity increases with decrease in concentration. (D) The variation in molar conductivity is different for strong and weak electrolytes. (E) For weak electrolytes, the change in molar conductivity with dilution is due to decrease in degree of dissociation.

Correct answer: 3

Step-by-step solution →
Q103·ChemistryNumericalJEE Main 2023
The total change in the oxidation state of manganese involved in the reaction of KMnO4KMnO_4KMnO4​ and potassium iodide in the acidic medium is _______

Correct answer: 5

Step-by-step solution →
Q104·ChemistryNumericalJEE Main 2023
See the following chemical reaction: Cr2O72−+XH++6Fe2+→YCr3++6Fe3++ZH2OCr_2O_7^{2-} + XH^+ + 6Fe^{2+} \rightarrow YCr^{3+} + 6Fe^{3+} + ZH_2OCr2​O72−​+XH++6Fe2+→YCr3++6Fe3++ZH2​O. The sum of X, Y and Z is ____.

Correct answer: 23

Step-by-step solution →
Q105·ChemistryNumericalJEE Main 2023
At 298 K, the standard reduction potential for Cu2+/CuCu^{2+}/CuCu2+/Cu electrode is 0.34 V. Given: KspK_{sp}Ksp​ of Cu(OH)2=1×10−20Cu(OH)_2 = 1 \times 10^{-20}Cu(OH)2​=1×10−20. Take 2.303RTF=0.059\dfrac{2.303RT}{F} = 0.059F2.303RT​=0.059 V. The reduction potential at pH = 14 for the above couple is (−)x×10−2(-)x \times 10^{-2}(−)x×10−2 V. The value of x is ____.

Correct answer: 25

Step-by-step solution →
Q106·ChemistryNumericalJEE Main 2023
KMnO4KMnO_{4}KMnO4​ is titrated with ferrous ammonium sulphate hexahydrate in presence of dilute H2SO4H_{2}SO_{4}H2​SO4​. Number of water molecules produced for 222 molecules of KMnO4KMnO_{4}KMnO4​ is _____.

Correct answer: 68

Step-by-step solution →
Q107·ChemistryNumericalJEE Main 2023
A metal surface of 100 cm2100\,cm^{2}100cm2 area has to be coated with nickel layer of thickness 0.001 mm0.001\,mm0.001mm. A current of 2 A2\,A2A was passed through a solution of Ni(NO3)2Ni(NO_{3})_{2}Ni(NO3​)2​ for ′x′'x'′x′ seconds to coat the desired layer. The value of xxx is _____ (Nearest integer). (ρNi(\rho_{Ni}(ρNi​ (density of Nickel) is 10 g mL−110\,g\,mL^{-1}10gmL−1. Molar mass of Nickel is 60 g mol−1, F=96500 C mol−1)60\,g\,mol^{-1},\,F=96500\,C\,mol^{-1})60gmol−1,F=96500Cmol−1)

Correct answer: 161

Step-by-step solution →
Q108·ChemistrySingle correctJEE Main 2023
For lead storage battery, pick the correct statement. A. During charging of battery, PbSO4_44​ on anode is converted into PbO2_22​. B. During discharging of battery, PbSO4_44​ on cathode is converted into PbO2_22​. C. Lead storage battery consists of grid of lead packed with PbO2_22​ as anode. D. Lead storage battery has ~38% solution of sulphuric acid as an electrolyte. Choose the correct answer from the options given below:
  1. (A)A, D only
  2. (B)B, C, D only
  3. (C)A, B, D only
  4. (D)B, C only

Correct answer: (A)

Step-by-step solution →
Q109·ChemistryNumericalJEE Main 2023
The number of correct statements from the following is ____. A. EcellE_{cell}Ecell​ is an intensive parameter. B. A negative E∘E^\circE∘ means that the redox couple is a stronger reducing agent than the H+/H2H^+/H_2H+/H2​ couple. C. The amount of electricity required for oxidation or reduction depends on the stoichiometry of the electrode reaction. D. The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte.

Correct answer: 4

Step-by-step solution →
Q110·ChemistryNumericalJEE Main 2023
KClO3+6FeSO4+3H2SO4→KCl+3Fe2(SO4)3+3H2OKClO_3 + 6FeSO_4 + 3H_2SO_4 \rightarrow KCl + 3Fe_2(SO_4)_3 + 3H_2OKClO3​+6FeSO4​+3H2​SO4​→KCl+3Fe2​(SO4​)3​+3H2​O The above reaction was studied at 300 K by monitoring the concentration of FeSO4FeSO_4FeSO4​ in which initial concentration was 10 M and after half an hour became 8.8 M. The rate of production of Fe2(SO4)3Fe_2(SO_4)_3Fe2​(SO4​)3​ is __________ ×10−6\times 10^{-6}×10−6 mol L−1L^{-1}L−1 s−1s^{-1}s−1. (Nearest integer)

Correct answer: 333

Step-by-step solution →
Q111·ChemistryNumericalJEE Main 2023
In an electrochemical reaction of lead, at standard temperature, if E(Pb2+/Pb)0=mE^0_{(Pb^{2+}/Pb)} = mE(Pb2+/Pb)0​=m Volt and E(Pb4+/Pb)0=nE^0_{(Pb^{4+}/Pb)} = nE(Pb4+/Pb)0​=n Volt, then the value of E(Pb2+/Pb4+)0E^0_{(Pb^{2+}/Pb^{4+})}E(Pb2+/Pb4+)0​ is given by m−xnm - xnm−xn. The value of x is __________. (Nearest integer)

Correct answer: 2

Step-by-step solution →
Q112·ChemistrySingle correctJEE Main 2023
Given below are two statements. Statement I: Aqueous solution of K2_22​Cr2_22​O7_77​ is preferred as a primary standard in volumetric analysis as compared to Na2_22​Cr2_22​O7_77​ aqueous solution. Statement II: K2_22​Cr2_22​O7_77​ has a higher solubility in water than Na2_22​Cr2_22​O7_77​. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)Both Statement I and Statement II are true
  2. (B)Both Statement I and Statement II are false
  3. (C)Statement I is true but Statement II is false
  4. (D)Statement I is false but Statement II is true

Correct answer: (C)

Step-by-step solution →
Q113·ChemistryNumericalJEE Main 2023
FeO42−→−2.2VFe3+→−0.70VFe2+→−0.45VFe0FeO_4^{2-}\xrightarrow{-2.2V}Fe^{3+}\xrightarrow{-0.70V}Fe^{2+}\xrightarrow{-0.45V}Fe^{0}FeO42−​−2.2V​Fe3+−0.70V​Fe2+−0.45V​Fe0. EFeO42−/Fe2+0E^0_{FeO_4^{2-}/Fe^{2+}}EFeO42−​/Fe2+0​ is x×10−1x\times10^{-1}x×10−1 V. The value of x is _________.

Correct answer: 1825

Step-by-step solution →
Q114·ChemistryNumericalJEE Main 2023
In alkaline medium, the reduction of the permanganate anion involves a gain of _______ electrons.

Correct answer: 3

Step-by-step solution →
Q115·ChemistryNumericalJEE Main 2023
The specific conductance of 0.00250.00250.0025 M acetic acid is 5×10−55\times10^{-5}5×10−5 S cm−1^{-1}−1 at a certain temperature. The dissociation constant of acetic acid is _______ ×10−7\times10^{-7}×10−7 (nearest integer). (Consider limiting molar conductivity of CH3COOH\text{CH}_3\text{COOH}CH3​COOH as 400400400 S cm2^22 mol−1^{-1}−1).

Correct answer: 66

Step-by-step solution →
Q116·ChemistrySingle correctJEE Main 2023
2IO3−+xI−+12H+→6I2+6H2O2IO_3^- + xI^- + 12H^+ \rightarrow 6I_2 + 6H_2O2IO3−​+xI−+12H+→6I2​+6H2​O What is the value of x?
  1. (A)12
  2. (B)2
  3. (C)6
  4. (D)10

Correct answer: (D)

Step-by-step solution →
Q117·ChemistrySingle correctJEE Main 2023
The reaction 12H2(g)+AgCl(s)⇌H+(aq)+Cl−(aq)+Ag(s)\frac{1}{2}H_2(g) + AgCl(s) \rightleftharpoons H^+(aq) + Cl^-(aq) + Ag(s)21​H2​(g)+AgCl(s)⇌H+(aq)+Cl−(aq)+Ag(s) occurs in which of the given galvanic cell.
  1. (A)Pt∣H2(g)∣KCl(soln)∣AgCl(s)∣AgPt | H_2(g) | KCl(sol^n) | AgCl(s) | AgPt∣H2​(g)∣KCl(soln)∣AgCl(s)∣Ag
  2. (B)Pt∣H2(g)∣HCl(soln)∣AgCl(s)∣AgPt | H_2(g) | HCl(sol^n) | AgCl(s) | AgPt∣H2​(g)∣HCl(soln)∣AgCl(s)∣Ag
  3. (C)Ag∣AgCl(s)∣KCl(soln)∣AgCl(s)∣AgAg | AgCl(s) | KCl(sol^n) | AgCl(s) | AgAg∣AgCl(s)∣KCl(soln)∣AgCl(s)∣Ag
  4. (D)Pt∣H2(g)∣HCl(soln)∣AgNO3(soln)∣AgPt | H_2(g) | HCl(sol^n) | AgNO_3(sol^n) | AgPt∣H2​(g)∣HCl(soln)∣AgNO3​(soln)∣Ag

Correct answer: (B)

Step-by-step solution →
Q118·ChemistryNumericalJEE Main 2023
The number of incorrect statements from the following is ____. A. The electrical work that a reaction can perform at constant pressure and temperature is equal to the reaction Gibbs energy. B. Ecell0E^0_{cell}Ecell0​ is dependent on the pressure. C. (dEcell0dT)=ΔrS0nF\left(\dfrac{dE^0_{cell}}{dT}\right)=\dfrac{\Delta_r S^0}{nF}(dTdEcell0​​)=nFΔr​S0​. D. A cell is operating reversibly if the cell potential is exactly balanced by an opposing source of potential difference.

Correct answer: 1

Step-by-step solution →
Q119·ChemistryNumericalJEE Main 2023
The sum of the oxidation states of the metals in Fe(CO)5Fe(CO)_5Fe(CO)5​, VO2+VO^{2+}VO2+ and WO3WO_3WO3​ is ____.

Correct answer: 10

Step-by-step solution →
Q120·ChemistrySingle correctJEE Main 2023
The product, which is not obtained during the electrolysis of brine solution is
  1. (A)NaOH
  2. (B)Cl2Cl_2Cl2​
  3. (C)H2H_2H2​
  4. (D)HCl

Correct answer: (D)

Step-by-step solution →
Q121·ChemistryNumericalJEE Main 2023
The standard reduction potentials at 298 K for the following half cells are given: NO3−+4H++3e−→NO(g)+2H2ONO_3^- + 4H^+ + 3e^- \rightarrow NO(g) + 2H_2ONO3−​+4H++3e−→NO(g)+2H2​O, E0=0.97E^0 = 0.97E0=0.97 V; V2+(aq)+2e−→VV^{2+}(aq) + 2e^- \rightarrow VV2+(aq)+2e−→V, E0=−1.19E^0 = -1.19E0=−1.19 V; Fe3+(aq)+3e−→FeFe^{3+}(aq) + 3e^- \rightarrow FeFe3+(aq)+3e−→Fe, E0=−0.04E^0 = -0.04E0=−0.04 V; Ag+(aq)+e−→Ag(s)Ag^+(aq) + e^- \rightarrow Ag(s)Ag+(aq)+e−→Ag(s), E0=0.80E^0 = 0.80E0=0.80 V; Au3+(aq)+3e−→Au(s)Au^{3+}(aq) + 3e^- \rightarrow Au(s)Au3+(aq)+3e−→Au(s), E0=1.40E^0 = 1.40E0=1.40 V. The number of metal(s) which will be oxidized by NO3−NO_3^-NO3−​ in aqueous solution is ____.

Correct answer: 3

Step-by-step solution →
Q122·ChemistrySingle correctJEE Main 2023
During the reaction of permanganate with thiosulphate, the change in oxidation number of manganese occurs by value of 3. Identify which of the below medium will favour the reaction.
  1. (A)aqueous acidic
  2. (B)aqueous neutral
  3. (C)both aqueous acidic and neutral
  4. (D)both aqueous acidic and faintly alkaline

Correct answer: (B)

Step-by-step solution →
Q123·ChemistrySingle correctJEE Main 2023
The standard electrode potential of M+/MM^{+}/MM+/M in aqueous solution does not depend on:
  1. (A)Ionisation of a solid metal atom
  2. (B)Sublimation of a solid metal
  3. (C)Ionisation of a gaseous metal atom
  4. (D)Hydration of a gaseous metal ion

Correct answer: (A)

Step-by-step solution →
Q124·ChemistryNumericalJEE Main 2023
Sum of oxidation states of bromine in bromic acid and perbromic acid is _______ .

Correct answer: 12

Step-by-step solution →
Q125·ChemistryNumericalJEE Main 2023
At what pH, the given half cell MnO4−(0.1 M) ∣ Mn2+(0.001 M)\text{MnO}_4^-(0.1\,\text{M})\,|\,\text{Mn}^{2+}(0.001\,\text{M})MnO4−​(0.1M)∣Mn2+(0.001M) will have electrode potential of 1.2821.2821.282 V? _______ (Nearest Integer). (Given EMnO4−∣Mn2+∘=1.54E^\circ_{\text{MnO}_4^-|\text{Mn}^{2+}}=1.54EMnO4−​∣Mn2+∘​=1.54 V, 2.303RTF=0.059\tfrac{2.303RT}{F}=0.059F2.303RT​=0.059 V).

Correct answer: 3

Step-by-step solution →
Q126·ChemistryNumericalJEE Main 2023
1×10−51\times10^{-5}1×10−5 M AgNO3AgNO_3AgNO3​ is added to 111 L of saturated solution of AgBr. The conductivity of this solution at 298298298 K is _________ ×10−9\times10^{-9}×10−9 S m−1^{-1}−1. [Given: KSP(AgBr)=4.9×10−13K_{SP}(AgBr)=4.9\times10^{-13}KSP​(AgBr)=4.9×10−13 at 298298298 K, λAg+0=6×10−3\lambda^0_{Ag^+}=6\times10^{-3}λAg+0​=6×10−3, λBr−0=8×10−3\lambda^0_{Br^-}=8\times10^{-3}λBr−0​=8×10−3, λNO3−0=7×10−3\lambda^0_{NO_3^-}=7\times10^{-3}λNO3−​0​=7×10−3 S m2^22 mol−1^{-1}−1]

Correct answer: 14

Step-by-step solution →
Q127·ChemistrySingle correctJEE Main 2023
Which one of the following statements is correct for electrolysis of brine solution?
  1. (A)Cl2Cl_2Cl2​ is formed at cathode
  2. (B)O2O_2O2​ is formed at cathode
  3. (C)H2H_2H2​ is formed at anode
  4. (D)OH−OH^-OH− is formed at cathode

Correct answer: (D)

Step-by-step solution →
Q128·ChemistryNumericalJEE Main 2023
The resistivity of a 0.8M solution of an electrolyte is 5×10−3 Ω5\times10^{-3}\,\Omega5×10−3Ω cm. Its molar conductivity is _________ ×104 Ω−1cm2mol−1\times10^4\,\Omega^{-1}cm^2mol^{-1}×104Ω−1cm2mol−1 (Nearest integer)

Correct answer: 25

Step-by-step solution →
Q129·ChemistrySingle correctJEE Main 2023
When Cu2+Cu^{2+}Cu2+ ion is treated with KI, a white precipitate, X appears in solution. The solution is titrated with sodium thiosulphate, the compound Y is formed. X and Y respectively are
  1. (A)X = CuI2CuI_2CuI2​, Y = Na2S4O6Na_2S_4O_6Na2​S4​O6​
  2. (B)X = CuI2CuI_2CuI2​, Y = Na2S2O3Na_2S_2O_3Na2​S2​O3​
  3. (C)X = Cu2I2Cu_2I_2Cu2​I2​, Y = Na2S4O5Na_2S_4O_5Na2​S4​O5​
  4. (D)X = Cu2I2Cu_2I_2Cu2​I2​, Y = Na2S4O6Na_2S_4O_6Na2​S4​O6​

Correct answer: (D)

Step-by-step solution →
Q130·ChemistryNumericalJEE Main 2023
The electrode potential of the following cell at 298 K, X ∣ X2+(0.001 M) ∥ Y2+(0.01 M) ∣ YX\,|\,X^{2+}(0.001\,M)\,\|\,Y^{2+}(0.01\,M)\,|\,YX∣X2+(0.001M)∥Y2+(0.01M)∣Y, is _______ ×10−2\times10^{-2}×10−2 V (nearest integer). (Given: EX2+∣X0=−2.36E^{0}_{X^{2+}|X}=-2.36EX2+∣X0​=−2.36 V; EY2+∣Y0=+0.36E^{0}_{Y^{2+}|Y}=+0.36EY2+∣Y0​=+0.36 V; 2.303RTF=0.06\dfrac{2.303RT}{F}=0.06F2.303RT​=0.06 V)

Correct answer: 275

Step-by-step solution →
Q131·ChemistryNumericalJEE Main 2023
Consider the cell Pt(s)_{(s)}(s)​|H2_22​(g, 1 atm)|H+^++(aq, 1M)||Fe3+^{3+}3+(aq), Fe2+^{2+}2+(aq)|Pt(s). When the potential of the cell is 0.7120.7120.712 V at 298298298 K, the ratio [Fe2+^{2+}2+]/[Fe3+^{3+}3+] is _________ (Nearest integer). Given: Fe3+^{3+}3+ + e−^-− = Fe2+^{2+}2+, E0^00 = 0.771 V, 2.303RTF=0.06\dfrac{2.303RT}{F}=0.06F2.303RT​=0.06 V

Correct answer: 10

Step-by-step solution →
Q132·ChemistryNumericalJEE Main 2023
The number of electrons involved in the reduction of permanganate to manganese dioxide in acidic medium is

Correct answer: 3

Step-by-step solution →
Q133·ChemistrySingle correctJEE Main 2023
KMnO4KMnO_4KMnO4​ oxidises I−I^-I− in acidic and neutral/faintly alkaline solution, respectively, to:
  1. (A)IO3− & IO3−IO_3^-\ \&\ IO_3^-IO3−​ & IO3−​
  2. (B)I2 & IO3−I_2\ \&\ IO_3^-I2​ & IO3−​
  3. (C)I2 & I2I_2\ \&\ I_2I2​ & I2​
  4. (D)IO3− & I2IO_3^-\ \&\ I_2IO3−​ & I2​

Correct answer: (B)

Step-by-step solution →
Q134·ChemistryNumericalJEE Main 2023
The equilibrium constant for the reaction Zn(s)+Sn2+(aq)⇌Zn2+(aq)+Sn(s)Zn(s)+Sn^{2+}(aq)\rightleftharpoons Zn^{2+}(aq)+Sn(s)Zn(s)+Sn2+(aq)⇌Zn2+(aq)+Sn(s) is 1×10201\times10^{20}1×1020 at 298 K298\,K298K. The magnitude of standard electrode potential of Sn/Sn2+Sn/Sn^{2+}Sn/Sn2+ if EZn2+/Zn∘=−0.76 VE^\circ_{Zn^{2+}/Zn}=-0.76\,VEZn2+/Zn∘​=−0.76V is _____ ×10−2 V\times10^{-2}\,V×10−2V (Nearest integer). Given: 2.303RTF=0.059 V\dfrac{2.303RT}{F}=0.059\,VF2.303RT​=0.059V.

Correct answer: 17

Step-by-step solution →
Q135·ChemistrySingle correctJEE Main 2023
The standard electrode potential (M3+/M2+M^{3+}/M^{2+}M3+/M2+) for V, Cr, Mn and Co are −0.26-0.26−0.26 V, −0.41-0.41−0.41 V, +1.57+1.57+1.57 V and +1.97+1.97+1.97 V, respectively. The metal ions which can liberate H2_22​ from a dilute acid are:
  1. (A)Mn2+^{2+}2+ and Co2+^{2+}2+
  2. (B)Cr2+^{2+}2+ and Co2+^{2+}2+
  3. (C)V2+^{2+}2+ and Cr2+^{2+}2+
  4. (D)V2+^{2+}2+ and Mn2+^{2+}2+

Correct answer: (C)

Step-by-step solution →
Q136·ChemistryNumericalJEE Main 2023
Following figure shows dependence of molar conductance of two electrolytes on concentration. Λm∘\Lambda_m^\circΛm∘​ is the limiting molar conductivity. The number of incorrect statement(s) from the following is ________ . (A) Λm∘\Lambda_m^\circΛm∘​ for electrolyte A is obtained by extrapolation (B) For electrolyte B, Λm\Lambda_mΛm​ vs c\sqrt{c}c​ graph is a straight line with intercept equal to Λm∘\Lambda_m^\circΛm∘​ (C) At infinite dilution, the value of degree of dissociation approaches zero for electrolyte B. (D) Λm∘\Lambda_m^\circΛm∘​ for any electrolyte A or B can be calculated using λ∘\lambda^\circλ∘ for individual ions.

Correct answer: 2

Step-by-step solution →
Q137·ChemistryNumericalJEE Main 2023
Pt(s)∣H2(g)(1 bar)∣∣H+(aq)(1 M) ∣∣ M3+(aq),M+(aq)∣Pt(s)Pt(s)|H_2(g)(1\,bar)||H^+(aq)(1\,M)\,||\,M^{3+}(aq), M^+(aq)|Pt(s)Pt(s)∣H2​(g)(1bar)∣∣H+(aq)(1M)∣∣M3+(aq),M+(aq)∣Pt(s) The EcellE_{cell}Ecell​ for the given cell is 0.1115 V at 298 K when [M+(aq)][M3+(aq)]=10a\frac{[M^+(aq)]}{[M^{3+}(aq)]} = 10^a[M3+(aq)][M+(aq)]​=10a. The value of aaa is Given : Eθ M3+/M+=0.2E^\theta\,M^{3+}/M^+ = 0.2EθM3+/M+=0.2 V 2.303RTF=0.059\frac{2.303RT}{F} = 0.059F2.303RT​=0.059 V

Correct answer: 3

Step-by-step solution →
Q138·ChemistrySingle correctJEE Main 2023
Which one among the following metals is the weakest reducing agent?
  1. (A)Li
  2. (B)K
  3. (C)Rb
  4. (D)Na

Correct answer: (D)

Step-by-step solution →
Q139·ChemistryNumericalJEE Main 2023
Consider the cell Pt(s) ∣ H2(g) (1 atm) ∣ H+(aq,[H+]=1) ∣∣ Fe3+(aq),Fe2+(aq) ∣ Pt(s)Pt(s)\,|\,H_2(g)\,(1\,atm)\,|\,H^+(aq, [H^+]=1)\,||\,Fe^{3+}(aq), Fe^{2+}(aq)\,|\,Pt(s)Pt(s)∣H2​(g)(1atm)∣H+(aq,[H+]=1)∣∣Fe3+(aq),Fe2+(aq)∣Pt(s). Given EFe3+/Fe2+∘=0.771E^{\circ}_{Fe^{3+}/Fe^{2+}}=0.771EFe3+/Fe2+∘​=0.771 V and EH+/H2∘=0E^{\circ}_{H^+/H_2}=0EH+/H2​∘​=0 V, T = 298 K. If the potential of the cell is 0.712 V, the ratio of concentration of Fe2+Fe^{2+}Fe2+ to Fe3+Fe^{3+}Fe3+ is _______ (nearest integer).

Correct answer: 10

Step-by-step solution →
Q140·ChemistrySingle correctJEE Main 2023
Potassium dichromate acts as a strong oxidizing agent in acidic solution. During this process, the oxidation state change from
  1. (A)+2 to +1
  2. (B)+3 to +1
  3. (C)+6 to +2
  4. (D)+6 to +3

Correct answer: (D)

Step-by-step solution →
Q141·ChemistryNumericalJEE Main 2023
At 298 K, a 1 litre solution containing 10 mmol of Cr2O72−\text{Cr}_2\text{O}_7^{2-}Cr2​O72−​ and 100 mmol of Cr3+\text{Cr}^{3+}Cr3+ shows a pH of 3.0. Given: Cr2O72−→Cr3+\text{Cr}_2\text{O}_7^{2-}\to\text{Cr}^{3+}Cr2​O72−​→Cr3+; E∘=1.330E^\circ=1.330E∘=1.330 V and 2.303RTF=0.059\tfrac{2.303RT}{F}=0.059F2.303RT​=0.059 V. The potential for the half cell reaction is x×10−3x\times10^{-3}x×10−3 V. The value of xxx is _______ .

Correct answer: 917

Step-by-step solution →
Q142·ChemistrySingle correctJEE Main 2023
Choose the correct representation of conductometric titration of benzoic acid vs sodium hydroxide.
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (A)

Step-by-step solution →
Q143·ChemistryNumericalJEE Advanced 2022
The reduction potential (E0, in V)\left(\mathrm{E}^{0}\text{, in V}\right)(E0, in V) of MnO4−(aq) / Mn(s)\mathrm{MnO_4^-(aq)\,/\,Mn(s)}MnO4−​(aq)/Mn(s) is [Given:E(MnO4−(aq)/MnO2((s)))0=1.68 V; E(MnO2(s)/Mn2+(aq))0=1.21 V; E(Mn2+(aq)/Mn((s)))0=−1.03 V]\left[\text{Given}: \mathrm{E}^{0}_{\left(\mathrm{MnO_4^-(aq)/MnO_2((s))}\right)} = 1.68\,\mathrm{V};\ \mathrm{E}^{0}_{\left(\mathrm{MnO_2(s)/Mn^{2+}(aq)}\right)} = 1.21\,\mathrm{V};\ \mathrm{E}^{0}_{\left(\mathrm{Mn^{2+}(aq)/Mn((s))}\right)} = -1.03\,\mathrm{V}\right][Given:E(MnO4−​(aq)/MnO2​((s)))0​=1.68V; E(MnO2​(s)/Mn2+(aq))0​=1.21V; E(Mn2+(aq)/Mn((s)))0​=−1.03V]

Correct answer: 0.77

Step-by-step solution →
Q144·ChemistryIntegerJEE Advanced 2022
Consider the strong electrolytes ZmXn\mathrm{Z_mX_n}Zm​Xn​, UmYp\mathrm{U_mY_p}Um​Yp​ and VmXn\mathrm{V_mX_n}Vm​Xn​. Limiting molar conductivity (Λ0)\left(\Lambda^0\right)(Λ0) of UmYp\mathrm{U_mY_p}Um​Yp​ and VmXn\mathrm{V_mX_n}Vm​Xn​ are 250 and 440 S cm2 mol−1\mathrm{S\ cm^2\ mol^{-1}}S cm2 mol−1, respectively. The value of (m + n + p) is____. The plot of molar conductivity (Λ)\left(\Lambda\right)(Λ) of ZmXn\mathrm{Z_mX_n}Zm​Xn​ vs c1/2\mathrm{c^{1/2}}c1/2 is given below.

Correct answer: 7

Step-by-step solution →
Q145·ChemistryNumericalJEE Advanced 2022
The treatment of an aqueous solution of 3.74 g of Cu(NO3)2\mathrm{Cu(NO_3)_2}Cu(NO3​)2​ with excess KI results in a brown solution along with the formation of a precipitate. Passing H2S\mathrm{H_2S}H2​S through this brown solution gives another precipitate X\mathbf{X}X. The amount of X\mathbf{X}X (in g) is ________. [Given: Atomic mass of H = 1, N = 14, O = 16, S = 32, K = 39, Cu = 63, I = 127]

Correct answer: 0.32

Step-by-step solution →
Q146·ChemistrySingle correctJEE Main 2022
In neutral or faintly alkaline medium, KMnO4KMnO_{4}KMnO4​ being a powerful oxidant can oxidize, thiosulphate almost quantitatively, to sulphate. In this reaction overall change in oxidation state of manganese will be :
  1. (A)5
  2. (B)1
  3. (C)0
  4. (D)3

Correct answer: (D)

Step-by-step solution →
Q147·ChemistryNumericalJEE Main 2022
Resistance of a conductivity cell (cell constant 129 m−1m^{-1}m−1) filled with 74.5 ppm solution of KCl is 100 Ω (labelled as solution 1). When the same cell is filled with KCl solution of 149 ppm, the resistance is 50 Ω (labelled as solution 2). The ratio of molar conductivity of solution 1 and solution 2 is i.e. Λ1Λ2=x×10−3\frac{\Lambda_{1}}{\Lambda_{2}} = x \times 10^{-3}Λ2​Λ1​​=x×10−3. The value of x is ______. (Nearest integer) Given, molar mass of KCl is 74.5 g mol−1mol^{-1}mol−1

Correct answer: 1000

Step-by-step solution →
Q148·ChemistryNumericalJEE Main 2022
For a cell, Cu(s) |Cu2+^{2+}2+(0.001M| |Ag+^{+}+(0.01M)| Ag(s) the cell potential is found to be 0.43 V at 298 K. The magnitude of standard electrode potential for Cu2+^{2+}2+/Cu is ___×10−2\times 10^{-2}×10−2 V. [Given : EAg+/Ag⊖E^{\ominus}_{Ag^{+}/Ag}EAg+/Ag⊖​ = 0.80V and 2.303RTF\frac{2.303RT}{F}F2.303RT​ = 0.06V]

Correct answer: 34

Step-by-step solution →
Q149·ChemistrySingle correctJEE Main 2022
Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R Assertion A : Permanganate titrations are not performed in presence of hydrochloric acid. Reason R : Chlorine is formed as a consequence of oxidation of hydrochloric acid. In the light of the above statements, choose the correct answer from the options given below
  1. (A)Both A and R are true and R is the correct explanation of A
  2. (B)Both A and R are true but R is NOT the correct explanation of A
  3. (C)A is true but R is false
  4. (D)A is false but R is true

Correct answer: (A)

Step-by-step solution →
Q150·ChemistrySingle correctJEE Main 2022
Match List-I with List-II. Choose the correct answer from the options given below :
List-IList-II
A.Cd(s) + 2Ni(OH)3_33​(s) → CdO(s) + 2Ni(OH)2_22​(s) + H2_22​O(lll)I.Primary battery
B.Zn(Hg) + HgO(s) → ZnO(s) + Hg(lll)II.Discharging of secondary battery
C.2PbSO4_44​(s) + 2H2_22​O(lll) → Pb(s) + PbO2_22​(s) + 2H2_22​SO4_44​(aq)III.Fuel cell
D.2H2_22​(g) + O2_22​(g) → 2H2_22​O(lll)IV.Charging of secondary battery
  1. (A)(A) – (I), (B) – (II), (C) – (III), (D) – (IV)
  2. (B)(A) – (IV), (B) – (I), (C) – (II), (D) – (III)
  3. (C)(A) – (II), (B) – (I), (C) – (IV), (D) – (III)
  4. (D)(A) – (II), (B) – (I), (C) – (III), (D) – (IV)

Correct answer: (C)

Step-by-step solution →
Q151·ChemistryNumericalJEE Main 2022
On reaction with stronger oxidizing agent like KIO4\mathrm{KIO_4}KIO4​, hydrogen peroxide oxidizes with the evolution of O2\mathrm{O_2}O2​. The oxidation number of I in KIO4\mathrm{KIO_4}KIO4​ changes to _____.

Correct answer: 5

Step-by-step solution →
Q152·ChemistryNumericalJEE Main 2022
20 mL of 0.02 M K2Cr2O7K_{2}Cr_{2}O_{7}K2​Cr2​O7​ solution is used for the titration of 10 mL of Fe2+Fe^{2+}Fe2+ solution in the acidic medium. The molarity of Fe2+Fe^{2+}Fe2+ solution is ______ ×10−2\times 10^{-2}×10−2 M. (Nearest Integer)

Correct answer: 24

Step-by-step solution →
Q153·ChemistrySingle correctJEE Main 2022
Given below are two statements: Statement I: For KI, molar conductivity increases steeply with dilution. Statement II: For carbonic acid, molar conductivity increases slowly with dilution. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)Both Statement I and Statement II are true
  2. (B)Both Statement I and Statement II are false
  3. (C)Statement I is true but Statement II is false
  4. (D)Statement I is false but Statement II is true

Correct answer: (B)

Step-by-step solution →
Q154·ChemistryNumericalJEE Main 2022
In the titration of KMnO4KMnO_{4}KMnO4​ and oxalic acid in acidic medium, the change in oxidation number of carbon at the end point is______

Correct answer: 1

Step-by-step solution →
Q155·ChemistryNumericalJEE Main 2022
The amount of charge in F (Faraday) required to obtain one mole of iron from Fe3_33​O4_44​ is _____. (Nearest Integer)

Correct answer: 3

Step-by-step solution →
Q156·ChemistrySingle correctJEE Main 2022
The dark purple colour of KMnO4_44​ disappears in the titration with oxalic acid in acidic medium. The overall change in the oxidation number of manganese in the reaction is :
  1. (A)5
  2. (B)1
  3. (C)7
  4. (D)2

Correct answer: (A)

Step-by-step solution →
Q157·ChemistryNumericalJEE Main 2022
20 mL of 0.02 M hypo solution is used for the titration of 10 mL of copper sulphate solution, in the presence of excess of KI using starch as an indicator. The molarity of Cu2+Cu^{2+}Cu2+ is found to be _____ × 10−210^{-2}10−2 M [nearest integer] Given : 2Cu2+2Cu^{2+}2Cu2+ + 4I−4I^{-}4I− → Cu2I2Cu_{2}I_{2}Cu2​I2​ + I2I_{2}I2​ I2I_{2}I2​ + 2S2O32−2S_{2}O_{3}^{2-}2S2​O32−​ → 2I−2I^{-}2I− + S4O62−S_{4}O_{6}^{2-}S4​O62−​

Correct answer: 4

Step-by-step solution →
Q158·ChemistrySingle correctJEE Main 2022
Which of the given reactions is not an example of disproportionation reaction ?
  1. (A)2H2_22​O2_22​ → 2H2_22​O + O2_22​
  2. (B)2NO2_22​ + H2_22​O → HNO3_33​ + HNO2_22​
  3. (C)MnO4−_4^-4−​ + 4H+^++ + 3e−^-− → MnO2_22​ + 2H2_22​O
  4. (D)3MnO42−_4^{2-}42−​ + 4H+^++ → 2MnO4−_4^-4−​ + MnO2_22​ + 2H2_22​O

Correct answer: (C)

Step-by-step solution →
Q159·ChemistrySingle correctJEE Main 2022
The reaction of H2O2H_{2}O_{2}H2​O2​ with potassium permanganate in acidic medium leads to the formation of mainly:
  1. (A)Mn2+Mn^{2+}Mn2+
  2. (B)Mn4+Mn^{4+}Mn4+
  3. (C)Mn3+Mn^{3+}Mn3+
  4. (D)Mn6+Mn^{6+}Mn6+

Correct answer: (A)

Step-by-step solution →
Q160·ChemistryNumericalJEE Main 2022
A dilute solution of sulphuric acid is electrolysed using a current of 0.10 A for 2 hours to produce hydrogen and oxygen gas. The total volume of gases produced at STP is _______ cm3cm^3cm3. (Nearest integer) [Given : Faraday constant F = 96500 C mol−1mol^{-1}mol−1 at STP, molar volume of an ideal gas is 22.7 L mol−1mol^{-1}mol−1]

Correct answer: 127

Step-by-step solution →
Q161·ChemistryNumericalJEE Main 2022
The cell potential for the given cell at 298 K Pt | H2_22​(g,1 bar) | H+^++(aq) || Cu2+^{2+}2+(aq) | Cu(s) is 0.31V. The pH of the acidic solution is found to be 3, whereas the concentration of Cu2+^{2+}2+ is 10−x^{-x}−x M. The value of x is ______. (Given: ECu2+/Cu⊖^{\ominus}_{Cu^{2+}/Cu}Cu2+/Cu⊖​ = 0.34 V and 2.303RTF\frac{2.303RT}{F}F2.303RT​ = 0.06V)

Correct answer: 7

Step-by-step solution →
Q162·ChemistryNumericalJEE Main 2022
A 2.0 g sample containing MnO2MnO_2MnO2​ is treated with HCl liberating Cl2Cl_2Cl2​. The Cl2Cl_2Cl2​ gas is passed into a solution of KI and 60.0 mL of 0.1 M Na2S2O3Na_2S_2O_3Na2​S2​O3​ is required to titrate the liberated iodine. The percentage of MnO2MnO_2MnO2​ in the sample is __________. (Nearest integer) [Atomic masses (in u) Mn = 55; Cl = 35.5; O = 16, I = 127, Na = 23, K = 39, S = 32]

Correct answer: 13

Step-by-step solution →
Q163·ChemistryNumericalJEE Main 2022
The solubility product of a sparingly soluble salt A2X3A_2X_3A2​X3​ is 1.1×10−231.1 \times 10^{-23}1.1×10−23. If specific conductance of the solution is 3×10−53 \times 10^{-5}3×10−5 S m−1m^{-1}m−1, the limiting molar conductivity of the solution is x ×10−3\times 10^{-3}×10−3 S m2m^{2}m2 mol−1mol^{-1}mol−1. The value of x is __________.

Correct answer: 3

Step-by-step solution →
Q164·ChemistryNumericalJEE Main 2022
The quantity of electricity in Faraday needed to reduce 1 mol of Cr2O72−Cr_2O_7^{2-}Cr2​O72−​ to Cr3+Cr^{3+}Cr3+ is __________.

Correct answer: 6

Step-by-step solution →
Q165·ChemistryNumericalJEE Main 2022
For the given reactions Sn2+^{2+}2+ + 2e−^-− → Sn Sn4+^{4+}4+ + 4e−^-− → Sn The electrode potentials are; ESn2+/SnoE^{o}_{Sn^{2+}/Sn}ESn2+/Sno​ = –0.140 V and ESn4+/SnoE^{o}_{Sn^{4+}/Sn}ESn4+/Sno​ = 0.010 V. The magnitude of standard electrode potential for Sn4+^{4+}4+/Sn2+^{2+}2+ i.e. ESn4+/Sn2+oE^{o}_{Sn^{4+}/Sn^{2+}}ESn4+/Sn2+o​ is ______ × 10−2^{-2}−2 V. (Nearest integer)

Correct answer: 16

Step-by-step solution →
Q166·ChemistryNumericalJEE Main 2022
0.01 M KMnO4_44​ solution was added to 20.0 mL of 0.05 M Mohr's salt solution through a burette. The initial reading of 50 mL burette is zero. The volume of KMnO4_44​ solution left in the burette after the end point is ______ mL. (nearest integer)

Correct answer: 30

Step-by-step solution →
Q167·ChemistryNumericalJEE Main 2022
For the reaction taking place in the cell: Pt(s) | H2_22​(g) | H+^++(aq) || Ag+^++(aq) | Ag(s) ECello^o_{Cell}Cello​ = +0.5332 V. The value of Δf\Delta_fΔf​G0^00 is _________ kJ mol−1^{-1}−1. (in nearest integer)

Correct answer: -51

Step-by-step solution →
Q168·ChemistryNumericalJEE Main 2022
The limiting molar conductivities of NaI, NaNO3NaNO_{3}NaNO3​ and AgNO3AgNO_{3}AgNO3​ are 12.7, 12.0 and 13.3 mS m2^{2}2 mol−1^{-1}−1, respectively (all at 25°C). The limiting molar conductivity of AgI at this temperature is _____ mS m2^{2}2 mol−1^{-1}−1

Correct answer: 14

Step-by-step solution →
Q169·ChemistrySingle correctJEE Main 2022
The (∂E∂T)P\left(\frac{\partial E}{\partial T}\right)_{P}(∂T∂E​)P​ of different types of half cells are as follows : A B C D 1×10−41 \times 10^{-4}1×10−4 2×10−42 \times 10^{-4}2×10−4 0.1×10−40.1 \times 10^{-4}0.1×10−4 0.2×10−40.2 \times 10^{-4}0.2×10−4 (Where E is the electromotive force) Which of the above half cells would be preferred to be used as reference electrode ?
  1. (A)A
  2. (B)B
  3. (C)C
  4. (D)D

Correct answer: (C)

Step-by-step solution →
Q170·ChemistrySingle correctJEE Main 2022
Which one of the following is an example of disproportionation reaction?
  1. (A)3MnO42−+4H+→2MnO4−+MnO2+2H2O3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O3MnO42−​+4H+→2MnO4−​+MnO2​+2H2​O
  2. (B)MnO42−+4H++4e−→MnO2+2H2OMnO_4^{2-} + 4H^+ + 4e^- \rightarrow MnO_2 + 2H_2OMnO42−​+4H++4e−→MnO2​+2H2​O
  3. (C)10I−+2MnO4−+16H+→2Mn2++8H2O+5I210I^- + 2MnO_4^- + 16H^+ \rightarrow 2Mn^{2+} + 8H_2O + 5I_210I−+2MnO4−​+16H+→2Mn2++8H2​O+5I2​
  4. (D)8MnO4−+3S2O32−+H2O→8MnO2+6SO42−+2OH−8MnO_4^- + 3S_2O_3^{2-} + H_2O \rightarrow 8MnO_2 + 6SO_4^{2-} + 2OH^-8MnO4−​+3S2​O32−​+H2​O→8MnO2​+6SO42−​+2OH−

Correct answer: (A)

Step-by-step solution →
Q171·ChemistryNumericalJEE Main 2022
Cu(s) + Sn2+Sn^{2+}Sn2+ (0.001M) → Cu2+Cu^{2+}Cu2+ (0.01M) + Sn(s) The Gibbs free energy change for the above reaction at 298 K is x × 10−110^{-1}10−1 kJ mol−1mol^{-1}mol−1; The value of x is_______. [nearest integer] [Given : ECu2+/Cu⊖E^{\ominus}_{Cu^{2+}/Cu}ECu2+/Cu⊖​ = 0.34 V; ESn2+/Sn⊖E^{\ominus}_{Sn^{2+}/Sn}ESn2+/Sn⊖​ = −0.14 V; F = 96500 C mol−1mol^{-1}mol−1]

Correct answer: 983

Step-by-step solution →
Q172·ChemistryNumericalJEE Main 2022
A solution of Fe2_22​(SO4_44​)3_33​ is electrolyzed for 'x' min with a current of 1.5 A to deposit 0.3482 g of Fe. The value of x is______. [nearest integer] Given : 1 F = 96500 C mol−1^{-1}−1 Atomic mass of Fe = 56 g mol−1^{-1}−1

Correct answer: 20

Step-by-step solution →
Q173·ChemistrySingle correctJEE Main 2022
The correct order of reduction potentials of the following pairs is A. Cl2_22​/Cl−^-− B. I2_22​/I−^-− C. Ag+^++/Ag D. Na+^++/Na E. Li+^++/Li Choose the correct answer from the options given below.
  1. (A)A > C > B > D > E
  2. (B)A > B > C > D > E
  3. (C)A > C > B > E > D
  4. (D)A > B > C > E > D

Correct answer: (A)

Step-by-step solution →
Q174·ChemistryNumericalJEE Main 2022
In a cell, the following reactions take place Fe2+→Fe3+e−Fe^{2+} \rightarrow Fe^{3+}e^{-}Fe2+→Fe3+e− EFe3+/Fe2+o=0.77E^{o}_{Fe^{3+}/Fe^{2+}} = 0.77EFe3+/Fe2+o​=0.77 V 2I−→I2+2e−2I^{-} \rightarrow I_2 + 2e^{-}2I−→I2​+2e− EI2/I−o=0.54E^{o}_{I_2/I^{-}} = 0.54EI2​/I−o​=0.54 V The standard electrode potential for the spontaneous reaction in the cell is x × 10−210^{-2}10−2V 298 K. The value of x is __________ (Nearest Integer)

Correct answer: 23

Step-by-step solution →
Q175·ChemistryNumericalJEE Main 2022
The cell potential for the following cell Pt ∣ H2(g) ∣ H+(aq) ∣∣ Cu2+(0.01 M) ∣ Cu(s)\mathrm{Pt}\,|\,\mathrm{H}_2(g)\,|\,\mathrm{H}^{+}(aq)\,||\,\mathrm{Cu}^{2+}(0.01\,\mathrm{M})\,|\,\mathrm{Cu}(s)Pt∣H2​(g)∣H+(aq)∣∣Cu2+(0.01M)∣Cu(s) is 0.576 V at 298 K. The pH of the solution is ___. (Nearest integer)

Correct answer: 5

Step-by-step solution →
Q176·ChemistryMultiple correctJEE Advanced 2021
Some standard electrode potentials at 298 K are given below: Pb2+^{2+}2+/Pb — −0.13 V Ni2+^{2+}2+/Ni — −0.24 V Cd2+^{2+}2+/Cd — −0.40 V Fe2+^{2+}2+/Fe — −0.44 V To a solution containing 0.001 M of X2+\mathbf{X}^{2+}X2+ and 0.1 M of Y2+\mathbf{Y}^{2+}Y2+, the metal rods X\mathbf{X}X and Y\mathbf{Y}Y are inserted (at 298 K) and connected by a conducting wire. This resulted in dissolution of X\mathbf{X}X. The correct combination(s) of X\mathbf{X}X and Y\mathbf{Y}Y, respectively, is (are) (Given: Gas constant, R = 8.314 J K−1^{-1}−1 mol−1^{-1}−1, Faraday constant, F = 96500 C mol−1^{-1}−1)
  1. (A)Cd and Ni
  2. (B)Cd and Fe
  3. (C)Ni and Pb
  4. (D)Ni and Fe

Correct answer: (A), (B), (C)

Step-by-step solution →
Q177·ChemistryNumericalJEE Advanced 2021
At 298 K, the limiting molar conductivity of a weak monobasic acid is 4 × 102^{2}2 S cm2^{2}2 mol−1^{-1}−1. At 298 K, for an aqueous solution of the acid the degree of dissociation of α and the molar conductivity is y × 102^{2}2 S cm2^{2}2 mol−1^{-1}−1. At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes 3y×102^{2}2 S cm2^{2}2 mol−1^{-1}−1. The value of α\alphaα is ______.

Correct answer: 0.21 or 0.22

Step-by-step solution →
Q178·ChemistryNumericalJEE Advanced 2021
At 298 K, the limiting molar conductivity of a weak monobasic acid is 4 × 102^{2}2 S cm2^{2}2 mol−1^{-1}−1. At 298 K, for an aqueous solution of the acid the degree of dissociation of α and the molar conductivity is y × 102^{2}2 S cm2^{2}2 mol−1^{-1}−1. At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes 3y×102^{2}2 S cm2^{2}2 mol−1^{-1}−1. The value of y\mathbf{y}y is ______.

Correct answer: 0.86

Step-by-step solution →
Q179·ChemistryNumericalJEE Advanced 2021
A sample (5.6 g) containing iron is completely dissolved in cold dilute HCl to prepare a 250 mL of solution. Titration of 25.0 mL of this solution requires 12.5 mL of 0.03 M KMnO4_{4}4​ solution to reach the end point. Number of moles of Fe2+^{2+}2+ present in 250 mL solution is x × 10−2^{-2}−2 (consider complete dissolution of FeCl2_{2}2​). The amount of iron present in the sample of y% by weight. (Assume : KMnO4_{4}4​ reacts only with Fe2+^{2+}2+ in the solution Use : Molar mass of iron as 56 g mol−1^{-1}−1) The value of x\mathbf{x}x is ______.

Correct answer: 1.87 or 1.88

Step-by-step solution →
Q180·ChemistryNumericalJEE Advanced 2021
A sample (5.6 g) containing iron is completely dissolved in cold dilute HCl to prepare a 250 mL of solution. Titration of 25.0 mL of this solution requires 12.5 mL of 0.03 M KMnO4_{4}4​ solution to reach the end point. Number of moles of Fe2+^{2+}2+ present in 250 mL solution is x × 10−2^{-2}−2 (consider complete dissolution of FeCl2_{2}2​). The amount of iron present in the sample of y% by weight. (Assume : KMnO4_{4}4​ reacts only with Fe2+^{2+}2+ in the solution Use : Molar mass of iron as 56 g mol−1^{-1}−1) The value of y\mathbf{y}y is ______.

Correct answer: 18.75

Step-by-step solution →
Q181·ChemistryNumericalJEE Main 2021
If the conductivity of mercury at 0°C is 1.07 × 106^{6}6 S m−1^{-1}−1 and the resistance of a cell containing mercury is 0.243 Ω, then the cell constant of the cell is x × 104^{4}4 m−1^{-1}−1. The value of x is ______ .(Nearest integer)

Correct answer: 26

Step-by-step solution →
Q182·ChemistrySingle correctJEE Main 2021
Match List-I with List-II List-I (Parameter) (a) Cell constant (b) Molar conductivity (c) Conductivity (d) Degree of dissociation of electrolyte List-II (Unit) (i) S cm2 mol−1\mathrm{S\ cm^2\ mol^{-1}}S cm2 mol−1 (ii) Dimensionless (iii) m−1\mathrm{m^{-1}}m−1 (iv) Ω−1 m−1\mathrm{\Omega^{-1}\ m^{-1}}Ω−1 m−1 Choose the most appropriate answer from the options given below :
  1. (A)(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
  2. (B)(a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)
  3. (C)(a)-(i), (b)-(iv), (c)-(iii), (d)-(ii)
  4. (D)(a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)

Correct answer: (A)

Step-by-step solution →
Q183·ChemistryNumericalJEE Main 2021
Consider the following cell reaction : Cd(s)+Hg2SO4(s)+95H2O(l)⇌CdSO4.95H2O(s)+2Hg(l)\mathrm{Cd_{(s)}+Hg_2SO_{4(s)}+\frac{9}{5}H_2O_{(l)}\rightleftharpoons CdSO_4.\frac{9}{5}H_2O_{(s)}+2Hg_{(l)}}Cd(s)​+Hg2​SO4(s)​+59​H2​O(l)​⇌CdSO4​.59​H2​O(s)​+2Hg(l)​ The value of Ecell0\mathrm{E^{0}_{cell}}Ecell0​ is 4.315 V at 25°C. If Δ\DeltaΔH° = −-−825.2 kJ mol−1^{-1}−1, the standard entropy change Δ\DeltaΔS° in J K−1^{-1}−1 is __. (Nearest integer) [Given : Faraday constant = 96487 C mol−1^{-1}−1]

Correct answer: 25

Step-by-step solution →
Q184·ChemistryNumericalJEE Main 2021
When 10 mL of an aqueous solution of KMnO4KMnO_4KMnO4​ was titrated in acidic medium, equal volume of 0.1 M of an aqueous solution of ferrous sulphate was required for complete discharge of colour. The strength of KMnO4KMnO_4KMnO4​ in grams per litre is ________ ×10−2\times 10^{-2}×10−2. (Nearest integer) [Atomic mass of K = 39, Mn = 55, O = 16]

Correct answer: 316

Step-by-step solution →
Q185·ChemistryNumericalJEE Main 2021
The resistance of a conductivity cell with cell constant 1.14 cm−1^{-1}−1, containing 0.001 M KCl at 298 K is 1500 Ω. The molar conductivity of 0.001 M KCl solution at 298 K in S cm2^{2}2 mol−1^{-1}−1 is _____ . (Integer answer)

Correct answer: 760

Step-by-step solution →
Q186·ChemistrySingle correctJEE Main 2021
In polythionic acid, H2SxO6\mathrm{H_2S_xO_6}H2​Sx​O6​ (x = 3 to 5) the oxidation state(s) of sulphur is/are :
  1. (A)+ 5 only
  2. (B)+ 6 only
  3. (C)+ 3 and + 5 only
  4. (D)0 and + 5 only

Correct answer: (D)

Step-by-step solution →
Q187·ChemistrySingle correctJEE Main 2021
Given below are two statements : Statement I : The limiting molar conductivity of KCl (strong electrolyte) is higher compared to that of CH3COOH\mathrm{CH_3COOH}CH3​COOH (weak electrolyte). Statement II : Molar conductivity decreases with decrease in concentration of electrolyte. In the light of the above statements, choose the most appropriate answer from the options given below :
  1. (A)Statement I is true but Statement II is false.
  2. (B)Statement I is false but Statement II is true.
  3. (C)Both Statement I and Statement II are true.
  4. (D)Both Statement I and Statement II are false.

Correct answer: (D)

Step-by-step solution →
Q188·ChemistryNumericalJEE Main 2021
These are physical properties of an element (A) Sublimation enthalpy (B) Ionisation enthalpy (C) Hydration enthalpy (D) Electron gain enthalpy The total number of above properties that affect the reduction potential is _________ (Integer answer)

Correct answer: 3

Step-by-step solution →
Q189·ChemistryNumericalJEE Main 2021
For the galvanic cell, Zn(s)+Cu2+(0.02 M)→Zn2+(0.04 M)+Cu(s)\mathrm{Zn(s)} + \mathrm{Cu}^{2+} (0.02\ \mathrm{M}) \rightarrow \mathrm{Zn}^{2+} (0.04\ \mathrm{M}) + \mathrm{Cu(s)}Zn(s)+Cu2+(0.02 M)→Zn2+(0.04 M)+Cu(s), Ecell=E_{cell} =Ecell​= ______ ×10−2\times 10^{-2}×10−2 V. (Nearest integer) [Use : ECu/Cu2+0=−0.34E^{0}_{\mathrm{Cu/Cu}^{2+}} = -0.34ECu/Cu2+0​=−0.34 V, EZn/Zn2+0=+0.76E^{0}_{\mathrm{Zn/Zn}^{2+}} = +0.76EZn/Zn2+0​=+0.76 V, 2.303 RTF=0.059\frac{2.303\ RT}{F} = 0.059F2.303 RT​=0.059 V]

Correct answer: 109

Step-by-step solution →
Q190·ChemistryNumericalJEE Main 2021
For the cell Cu(s) | Cu2+^{2+}2+ (aq)(0.1M) ‖ Ag+^++ (aq)(0.01M) | Ag(s) The cell potential E1_11​ = 0.3095V For the cell Cu(s) | Cu2+^{2+}2+ (aq)(0.01M) ‖ Ag+^++ (aq)(0.001M) | Ag(s) The cell potential = __________×10−2^{-2}−2 V. (Round off to the Nearest Integer). [Use : 2.303RTF\frac{2.303RT}{F}F2.303RT​ = 0.059 ]

Correct answer: 28

Step-by-step solution →
Q191·ChemistrySingle correctJEE Main 2021
The product obtained from the electrolytic oxidation of acidified sulphate solutions, is:
  1. (A)HO3SOSO3HHO_3SOSO_3HHO3​SOSO3​H
  2. (B)HSO4−HSO_4^-HSO4−​
  3. (C)HO2SOSO2HHO_2SOSO_2HHO2​SOSO2​H
  4. (D)HO3SOOSO3HHO_3SOOSO_3HHO3​SOOSO3​H

Correct answer: (D)

Step-by-step solution →
Q192·ChemistryNumericalJEE Main 2021
When 10 ml of an aqueous solution of Fe2+^{2+}2+ ions was titrated in the presence of dil H2_22​SO4_44​ using diphenylamine indicator, 15 mL of 0.02 M solution of K2_22​Cr2_22​O7_77​ was required to get the end point. The molarity of the solution containing Fe2+^{2+}2+ ions is x ×10−2\times 10^{-2}×10−2M. The value of x is___. (Nearest integer)

Correct answer: 18

Step-by-step solution →
Q193·ChemistrySingle correctJEE Main 2021
The correct order of following 3d metal oxides, according to their oxidation numbers is: (a) CrO3CrO_3CrO3​ (b) Fe2O3Fe_2O_3Fe2​O3​ (c) MnO2MnO_2MnO2​ (d) V2O5V_2O_5V2​O5​ (e) Cu2OCu_2OCu2​O
  1. (A)(a) > (d) > (c) > (b) > (e)
  2. (B)(c) > (a) > (d) > (e) > (b)
  3. (C)(d) > (a) > (b) > (c) > (e)
  4. (D)(a) > (c) > (d) > (b) > (e)

Correct answer: (A)

Step-by-step solution →
Q194·ChemistryNumericalJEE Main 2021
Consider the cell at 25°C Zn | Zn2+^{2+}2+(aq), (1M) || Fe3+^{3+}3+(aq), Fe2+^{2+}2+(aq) | Pt(s) The fraction of total iron present as Fe3+^{3+}3+ ion at the cell potential of 1.500V is x ×10−2\times 10^{-2}×10−2. The value of x is ______. (Nearest integer) (Given: EFe3+/Fe2+0E^{0}_{Fe^{3+}/Fe^{2+}}EFe3+/Fe2+0​ = 0.77V, EZn2+/Zn0E^{0}_{Zn^{2+}/Zn}EZn2+/Zn0​ = −0.76V )

Correct answer: 24

Step-by-step solution →
Q195·ChemistrySingle correctJEE Main 2021
Identify the process in which change in the oxidation state is five:
  1. (A)CrO42−CrO_4^{2-}CrO42−​ → Cr3+Cr^{3+}Cr3+
  2. (B)MnO4−MnO_4^-MnO4−​ → Mn2+Mn^{2+}Mn2+
  3. (C)Cr2O72−Cr_2O_7^{2-}Cr2​O72−​ → 2Cr3+2Cr^{3+}2Cr3+
  4. (D)C2O42−C_2O_4^{2-}C2​O42−​ → 2CO22CO_22CO2​

Correct answer: (B)

Step-by-step solution →
Q196·ChemistryNumericalJEE Main 2021
Assume a cell with the following reaction Cu(s)+2Ag+(1×10−3 M)→Cu2+(0.250 M)+2Ag(s)\mathrm{Cu_{(s)} + 2Ag^+ \left(1 \times 10^{-3}\,M\right) \rightarrow Cu^{2+} \left(0.250\,M\right) + 2Ag_{(s)}}Cu(s)​+2Ag+(1×10−3M)→Cu2+(0.250M)+2Ag(s)​ Ecell⊖E_{\mathrm{cell}}^{\ominus}Ecell⊖​ = 2.97 V EcellE_{\mathrm{cell}}Ecell​ for the above reaction is _____________ V. (Nearest integer) [ Given : log 2.5 = 0.3979, T = 298K]

Correct answer: 3

Step-by-step solution →
Q197·ChemistryNumericalJEE Main 2021
Potassium chlorate is prepared by electrolysis of KCl\mathrm{KCl}KCl in basic solution as shown by following equation. 6 OH−+Cl−→ClO3−+3 H2O+6 e−6\,\mathrm{OH}^- + \mathrm{Cl}^- \rightarrow \mathrm{ClO_3^-} + 3\,\mathrm{H_2O} + 6\,\mathrm{e}^-6OH−+Cl−→ClO3−​+3H2​O+6e− A current of xA has to be passed for 10 h to produce 10.0 g of potassium chlorate. The value of xxx is ________. (Nearest Integer) (Molar mass of KClO3=122.6 g mol−1, F=96500 C\mathrm{KClO_3} = 122.6\ \mathrm{g\ mol^{-1}},\ \mathrm{F} = 96500\ \mathrm{C}KClO3​=122.6 g mol−1, F=96500 C)

Correct answer: 1

Step-by-step solution →
Q198·ChemistrySingle correctJEE Main 2021
The species given below that does NOT show disproportionation reaction is:
  1. (A)BrO−BrO^{-}BrO−
  2. (B)BrO3−BrO_{3}^{-}BrO3−​
  3. (C)BrO2−BrO_{2}^{-}BrO2−​
  4. (D)BrO4−BrO_{4}^{-}BrO4−​

Correct answer: (D)

Step-by-step solution →
Q199·ChemistryNumericalJEE Main 2021
For the reaction 2Fe3+^{3+}3+(aq) + 2I−^{-}−(aq) → 2Fe2+^{2+}2+(aq) + I2_{2}2​(s) the magnitude of the standard molar free energy change, ΔrGm∘\Delta_{r}G^{\circ}_{m}Δr​Gm∘​ = – ______ kJ (Round off to the Nearest Integer). [EFe2+/Fe(s)∘E^{\circ}_{Fe^{2+}/Fe(s)}EFe2+/Fe(s)∘​ = −0.440 V; EFe3+/Fe(s)∘E^{\circ}_{Fe^{3+}/Fe(s)}EFe3+/Fe(s)∘​ = −0.036 V; EI2/2I−∘E^{\circ}_{I_{2}/2I^{-}}EI2​/2I−∘​ = 0.539 V; F = 96500 C]

Correct answer: 45

Step-by-step solution →
Q200·ChemistrySingle correctJEE Main 2021
The oxidation states of nitrogen in NO, NO2_{2}2​, N2_{2}2​O and NO3−_{3}^{-}3−​ are in the order of :
  1. (A)NO3−_{3}^{-}3−​ > NO2_{2}2​ > NO > N2_{2}2​O
  2. (B)NO2_{2}2​ > NO3−_{3}^{-}3−​ > NO > N2_{2}2​O
  3. (C)N2_{2}2​O > NO2_{2}2​ > NO > NO3−_{3}^{-}3−​
  4. (D)NO > NO2_{2}2​ > N2_{2}2​O > NO3−_{3}^{-}3−​

Correct answer: (A)

Step-by-step solution →
Q201·ChemistryNumericalJEE Main 2021
The molar conductivities at infinite dilution of barium chloride, sulphuric arid and hydrochloric acid are 280, 860 and 426 Scm2^{2}2 mol−1^{-1}−1 respectively. The molar conductivity at infinite dilution of barium sulphate is ________ S cm2^{2}2 mol−1^{-1}−1 (Round off to the Nearest Integer).

Correct answer: 288

Step-by-step solution →
Q202·ChemistryNumericalJEE Main 2021
15 mL of aqueous solution of Fe2+\mathrm{Fe^{2+}}Fe2+ in acidic medium completely reacted with 20 mL of 0.03 M aqueous Cr2O72−\mathrm{Cr_2O_7^{2-}}Cr2​O72−​. The molarity of the Fe2+\mathrm{Fe^{2+}}Fe2+ solution is ________ ×10−2\times 10^{-2}×10−2 M (Round off to the Nearest Integer).

Correct answer: 24

Step-by-step solution →
Q203·ChemistryNumericalJEE Main 2021
A KCl solution of conductivity 0.14 S m−1^{-1}−1 shows a resistance of 4.19 Ω\OmegaΩ in a conductivity cell. If the same cell is filled with an HCl solution, the resistance drops to 1.03 Ω\OmegaΩ. The conductivity of the HCl solution is ____ ×\times× 10−2^{-2}−2 S m−1^{-1}−1. (Round off to the Nearest Integer).

Correct answer: 57

Step-by-step solution →
Q204·ChemistryNumericalJEE Main 2021
2 MnO4−_{4}^{-}4−​ + b C2_{2}2​O42−_{4}^{2-}42−​ + c H+^{+}+ → x Mn2+^{2+}2+ + y CO2_{2}2​ + z H2_{2}2​O If the above equation is balanced with integer coefficients, the value of c is _______. (Round off to the Nearest Integer).

Correct answer: 16

Step-by-step solution →
Q205·ChemistryNumericalJEE Main 2021
A 5.0 m mol dm−3^{-3}−3 aqueous solution of KCl has a conductance of 0.55 mS when measured in a cell constant 1.3 cm−1^{-1}−1. The molar conductivity of this solution is _______ mSm2^{2}2 mol−1^{-1}−1. (Round off to the Nearest Integer)

Correct answer: 14

Step-by-step solution →
Q206·ChemistryNumericalJEE Main 2021
Emf of the following cell at 298 K in V is x ×10−2\times10^{-2}×10−2. Zn∣Zn2+Zn|Zn^{2+}Zn∣Zn2+ (0.1 M)∣∣Ag+||Ag^+∣∣Ag+(0.01 M)∣|∣ Ag The value of x is ___________. (Rounded off to the nearest integer) [Given: EZn2+/Zn0E^0_{Zn^{2+}/Zn}EZn2+/Zn0​ = −0.76V; EAg+/Ag0E^0_{Ag^+/Ag}EAg+/Ag0​ = +0.80V; 2.303RTF\frac{2.303RT}{F}F2.303RT​ = 0.059]

Correct answer: 147

Step-by-step solution →
Q207·ChemistryNumericalJEE Main 2021
In mildly alkaline medium, thiosulphate ion is oxidized by MnO4−MnO_4^-MnO4−​ to "A". The oxidation state of sulphur in "A" is_________.

Correct answer: 6

Step-by-step solution →
Q208·ChemistryNumericalJEE Main 2021
Consider the following reaction MnO4−_4^{-}4−​ + 8H+^++ + 5e−^-− → Mn+2^{+2}+2 + 4H2_22​O, E° = 1.51 V. The quantity of electricity required in Faraday to reduce five moles of MnO4−_4^{-}4−​ is_____.

Correct answer: 25

Step-by-step solution →
Q209·ChemistryNumericalJEE Main 2021
In basic medium CrO42−_{4}^{2-}42−​ oxidizes S2_{2}2​O32−_{3}^{2-}32−​ to form SO42−_{4}^{2-}42−​ and itself changes into Cr(OH)4−_{4}^{-}4−​. The volume of 0.154 M CrO42−_{4}^{2-}42−​ required to react with 40 mL of 0.25 M S2_{2}2​O32−_{3}^{2-}32−​ is _______ mL. (Rounded-off to the nearest integer)

Correct answer: 173

Step-by-step solution →
Q210·ChemistryNumericalJEE Main 2021
Copper reduces NO3_{3}3​ −into NO and NO2 depending upon the concentration of HNO3 in solution. (Assuming fixed [Cu2+^{2+}2+] and PNO=PNO2), the HNO3 concentration at which the thermodynamic tendency for reduction of NO3_{3}3​ − into NO and NO2 by copper is same is 10x^{x}x M. The value of 2x is ______. (Rounded-off to the nearest integer) [Given:_{[Given :}[Given:​ECu0_{Cu}^{0}Cu0​2+_{/}_{Cu}=0.34V ,E_{NO}^{0}_{3}−_{/}_{NO}=0.96V ,E_{NO}^{0}_{3}−_{/}_{NO}_{2}=0.79V and at 298 K, RTF(2.303)=0.059]]_{]}]​

Correct answer: 4

Step-by-step solution →
Q211·ChemistryNumericalJEE Main 2021
The reaction of sulphur in alkaline medium is given below: S8(s)+a OH−(aq)⟶b S2−(aq)+c S2O32−(aq)+d H2O(ℓ)\mathrm{S_{8(s)}} + a\,\mathrm{OH^-}_{(aq)} \longrightarrow b\,\mathrm{S^{2-}}_{(aq)} + c\,\mathrm{S_2O_3^{2-}}_{(aq)} + d\,\mathrm{H_2O}_{(\ell)}S8(s)​+aOH−(aq)​⟶bS2−(aq)​+cS2​O32−​(aq)​+dH2​O(ℓ)​ The values of 'a' is __________. (Integer answer)

Correct answer: 12

Step-by-step solution →
Q212·ChemistryNumericalJEE Main 2021
The magnitude of the change in oxidising power of the MnO4−MnO_{4}^{-}MnO4−​ / Mn2+Mn^{2+}Mn2+ couple is x × 10−410^{-4}10−4 V, if the H+H^{+}H+ concentration is decreased from 1M to 10−410^{-4}10−4 M at 25ºC. (Assume concentration of MnO4−MnO_{4}^{-}MnO4−​ and Mn2+Mn^{2+}Mn2+ to be same on change in H+H^{+}H+ concentration). The value of x is ______. (Rounded off to the nearest integer) [Given:2303RTF=0.059]\left[ Given : \frac{2303RT}{F} = 0.059 \right][Given:F2303RT​=0.059]

Correct answer: 3776

Step-by-step solution →
Q213·ChemistryIntegerJEE Advanced 2020
In the chemical reaction between stoichiometric quantities of KMnO4KMnO_{4}KMnO4​ and KI in weakly basic solution, what is the number of moles of I2I_{2}I2​ released for 4 moles of KMnO4KMnO_{4}KMnO4​ consumed ?

Correct answer: 6

Step-by-step solution →
Q214·ChemistryNumericalJEE Advanced 2020
Consider a 70% efficient hydrogen-oxygen fuel cell working under standard conditions at 1 bar and 298 K. Its cell reaction is H2(g)+12O2(g)→H2O(ℓ)H_2 (g) + \frac{1}{2} O_2(g) \rightarrow H_2O(\ell)H2​(g)+21​O2​(g)→H2​O(ℓ). The work derived from the cell on the consumption of 1.0×10−31.0 \times 10^{-3}1.0×10−3 mol of H2(g)H_2(g)H2​(g) is used to compress 1.00 mol of a monoatomic ideal gas in a thermally insulted container. What is the change in the temperature (in K) of the ideal gas ? The standard reduction potentials for the two half-cells are given below. O2(g)+4 H+(aq.)+4e−→2H2O (ℓ)O_2(g) + 4\,H^+ (aq.) + 4e^- \rightarrow 2H_2O\,(\ell)O2​(g)+4H+(aq.)+4e−→2H2​O(ℓ) , E∘=1.23E^\circ = 1.23E∘=1.23 V, 2H+(aq.)+2e−→H2(g)2H^+ (aq.) + 2e^- \rightarrow H_2(g)2H+(aq.)+2e−→H2​(g), E∘=0.00E^\circ = 0.00E∘=0.00 V. Use F=96500F = 96500F=96500 C mol−1mol^{-1}mol−1, R=8.314R = 8.314R=8.314 J mol−1K−1mol^{-1}K^{-1}mol−1K−1

Correct answer: 13.00 - 13.60

Step-by-step solution →
Q215·ChemistrySingle correctJEE Main 2020
For the given cell; Cu(s) | Cu2+^{2+}2+ (C1_11​M) || Cu2+^{2+}2+ (C2_22​M) | Cu(s) Change in Gibbs energy (ΔG) is negative, if
  1. (A)C1_11​ = C2_22​
  2. (B)C2=C12\frac{C_2 = C_1}{\sqrt{2}}2​C2​=C1​​
  3. (C)C1_11​ = 2C2_22​
  4. (D)C2_22​ = 2\sqrt{2}2​C1_11​

Correct answer: (D)

Step-by-step solution →
Q216·ChemistryNumericalJEE Main 2020
Potassium chlorate is prepared by the electrolysis of KCl in basic solution 6HO−+Cl−→ClO3−+3H2O+6e−6HO^- + Cl^- \rightarrow ClO_3^- + 3H_2O + 6e^-6HO−+Cl−→ClO3−​+3H2​O+6e− If only 60% of the current is utilized in the reaction, the time (rounded to the nearest hour) required to produce 10 g of KClO3KClO_3KClO3​ using a current of 2A is ________________. (Given : F = 96,500 C mol−1^{-1}−1; molar mass of KClO3KClO_3KClO3​ = 122 g mol−1^{-1}−1)

Correct answer: 11

Step-by-step solution →
Q217·ChemistryNumericalJEE Main 2020
The volume, in mL, of 0.02 M K2_22​Cr2_22​O7_77​ solution required to react with 0.288 g of ferrous oxalate in acidic medium is __________. (Molar mass of Fe = 56 mol−1^{-1}−1)

Correct answer: 50.00

Step-by-step solution →
Q218·ChemistryNumericalJEE Main 2020
An oxidation – reduction in which 3 electrons are transferred has a ΔG0\Delta G^0ΔG0 of 17.37 kJ mol−1mol^{-1}mol−1 at 25025^0250 C. The value of Ecell0E^0_{cell}Ecell0​ (in V is _____ × 10−210^{-2}10−2) (1 F = 96,500 C mol−1mol^{-1}mol−1)

Correct answer: -6

Step-by-step solution →
Q219·ChemistrySingle correctJEE Main 2020
The variation of molar conductivity with concentration of an electrolyte (X) in aqueous solution is shown in the given figure.
  1. (A)KNO3KNO_3KNO3​
  2. (B)CH3COOHCH_3COOHCH3​COOH
  3. (C)HClHClHCl
  4. (D)NaClNaClNaCl

Correct answer: (B)

Step-by-step solution →
Q220·ChemistryNumericalJEE Main 2020
At 20.0 mL solution containing 0.2 g impure H2_22​O2_22​ reacts completely with 0.316 g of KMnO4_44​ in acid solution. The purity of H2_22​O2_22​ (in %) is _______________. (mol. wt. of H2_22​O2_22​ = 34; mol. wt. of KMnO4_44​ = 158)

Correct answer: 85

Step-by-step solution →
Q221·ChemistrySingle correctJEE Main 2020
Identify the incorrect statement from the options below for the above cell:
  1. (A)If Eext<1.1E_{ext} < 1.1Eext​<1.1 V, Zn dissolves at anode and Cu deposits at cathode
  2. (B)If Eext=1.1E_{ext} = 1.1Eext​=1.1 V, no flow of e−^-− or current occurs
  3. (C)If Eext>1.1E_{ext} > 1.1Eext​>1.1 V, Zn dissolves at Zn electrode and Cu deposits at Cu electrode
  4. (D)If Eext>1.1E_{ext} > 1.1Eext​>1.1 V, e−^-− flows from Cu to Zn

Correct answer: (C)

Step-by-step solution →
Q222·ChemistrySingle correctJEE Main 2020
250 mL of a waste solution obtained from the workshop of a goldsmith contains 0.1 M AgNO3AgNO_3AgNO3​ and 0.1 M AuClAuClAuCl. The solution was electrolyzed at 2 V by passing a current of 1 A for 15 minutes. The metal/metals electrodeposited will be : (EAg+/Ago=0.8 V,EAu+/Auo=1.69 V)\left(E^{o}_{Ag^{+}/Ag} = 0.8\ V, E^{o}_{Au^{+}/Au} = 1.69\ V\right)(EAg+/Ago​=0.8 V,EAu+/Auo​=1.69 V)
  1. (A)silver and gold in equal mass proportion
  2. (B)only silver
  3. (C)only gold
  4. (D)silver and gold in proportion to their atomic weights

Correct answer: (C)

Step-by-step solution →
Q223·ChemistryNumericalJEE Main 2020
Consider the following equations : 2Fe2++H2O2→x A+y B2\mathrm{Fe}^{2+} + \mathrm{H}_2\mathrm{O}_2 \rightarrow x\,\mathrm{A} + y\,\mathrm{B}2Fe2++H2​O2​→xA+yB (in basic medium) (in acidic medium) 2MnO4−+6H++5H2O2→x′C+y′D+z′E2\mathrm{MnO}_4^- + 6\mathrm{H}^+ + 5\mathrm{H}_2\mathrm{O}_2 \rightarrow x'\mathrm{C} + y'\mathrm{D} + z'\mathrm{E}2MnO4−​+6H++5H2​O2​→x′C+y′D+z′E (in acidic medium) The sum of the stoichiometric coefficients x, y, x', y' and z' for products A, B, C, D and E, respectively, is ………..

Correct answer: 19

Step-by-step solution →
Q224·ChemistryNumericalJEE Main 2020
An acidic solution of dichromate is electrolyzed for 8 minutes using 2A current. As per the following equation Cr2_{2}2​O72−_{7}^{2-}72−​ + 14H+^{+}+ + 6e−^{-}− → 2Cr3+^{3+}3+ + 7H2_{2}2​O. The amount of Cr3+^{3+}3+ obtained was 0.104 g. The efficiency of the process (in %) is (Take: F = 96000 C, At. Mass of chromium = 52) ______.

Correct answer: 60

Step-by-step solution →
Q225·ChemistryNumericalJEE Main 2020
The photoelectric current from Na (work function, w0_00​ = 2.3 eV) is stopped by the output voltage of the cell Pt(s)/H2_22​(g, 1 bar) | HCl (aq., pH = 1) | AgCl(s) | Ag(s). The pH of aq. HCl required to stop the photoelectric current from K(w0_00​ = 2.25 eV), all other conditions remaining the same, is __________ ×\times× 10−2^{-2}−2 (to the nearest integer). Given, 2.303 RTF\frac{RT}{F}FRT​ = 0.06 V ; EAgCl ∣ Ag ∣ Cl−0^0_{AgCl\,|\,Ag\,|\,Cl^-}AgCl∣Ag∣Cl−0​ = 0.22 V

Correct answer: 142

Step-by-step solution →
Q226·ChemistryNumericalJEE Main 2020
108 g of silver (molar mass 108 g mol−1^{-1}−1) is deposited at cathode from AgNO3_{3}3​ (aq) solution by a certain quantity of electricity. The volume (in L) of oxygen gas produced at 273 K and 1 bar pressure from water by the same quantity of electricity is_.

Correct answer: 5.67

Step-by-step solution →
Q227·ChemistrySingle correctJEE Main 2020
Amongst the following, the form of water with the lowest ionic conductance at 298 K is :
  1. (A)sea water
  2. (B)distilled water
  3. (C)saline water used for intravenous injection
  4. (D)water from a well

Correct answer: (B)

Step-by-step solution →
Q228·ChemistrySingle correctJEE Main 2020
The compound that cannot act both as oxidising and reducing agent is:
  1. (A)H2_22​O2_22​
  2. (B)H2_22​SO3_33​
  3. (C)HNO2_22​
  4. (D)H3_33​PO4_44​

Correct answer: (D)

Step-by-step solution →
Q229·ChemistryNumericalJEE Main 2020
For an electrochemical cell Sn(s)∣Sn2+(aq,1M)∣∣Pb2+(aq,1M)∣Pb(s)Sn(s)|Sn^{2+}(aq, 1M)||Pb^{2+}(aq, 1M)|Pb(s)Sn(s)∣Sn2+(aq,1M)∣∣Pb2+(aq,1M)∣Pb(s) the ratio [Sn2+][Pb2+]\dfrac{[Sn^{2+}]}{[Pb^{2+}]}[Pb2+][Sn2+]​ when this cell attains equilibrium is ____. (Given: ESn2+∣Sn0=−0.14V,EPb2+∣Pb0=−0.13V,2.303RTF=0.06E^0_{Sn^{2+}|Sn} = -0.14V, E^0_{Pb^{2+}|Pb} = -0.13V, \dfrac{2.303RT}{F} = 0.06ESn2+∣Sn0​=−0.14V,EPb2+∣Pb0​=−0.13V,F2.303RT​=0.06)

Correct answer: 2.15

Step-by-step solution →
Q230·ChemistrySingle correctJEE Main 2020
The redox reaction among the following is
  1. (A)reaction of [Co(H2O)6]Cl3\mathrm{[Co(H_2O)_6]Cl_3}[Co(H2​O)6​]Cl3​ with AgNO3\mathrm{AgNO_3}AgNO3​
  2. (B)combination of dinitrogen with dioxygen at 2000 K
  3. (C)reaction of H2SO4\mathrm{H_2SO_4}H2​SO4​ with NaOH
  4. (D)formation of ozone from atmospheric oxygen in the presence of sunlight

Correct answer: (B)

Step-by-step solution →
Q231·ChemistrySingle correctJEE Main 2020
Oxidation number of potassium in K2_22​O, K2_22​O2_22​ and KO2_22​ respectively is
  1. (A)+1, +2 and +4
  2. (B)+2, +1 and +½
  3. (C)+1, +1 and +1
  4. (D)+1, +4 and +2

Correct answer: (C)

Step-by-step solution →
Q232·ChemistrySingle correctJEE Main 2020
Given that the standard potential (E°) of Cu2+^{2+}2+/Cu and Cu+^{+}+/Cu are 0.34 V and 0.522 V respectively, the E° of Cu2+^{2+}2+/Cu+^{+}+ is
  1. (A)-0.158 V
  2. (B)0.182 V
  3. (C)+0.158 V
  4. (D)-0.182 V

Correct answer: (C)

Step-by-step solution →
Q233·ChemistrySingle correctJEE Main 2020
The equation that is incorrect is
  1. (A)(Λm0)NaBr−(Λm0)NaI=(Λm0)KBr−(Λm0)NaBr(\Lambda_m^0)_{NaBr} - (\Lambda_m^0)_{NaI} = (\Lambda_m^0)_{KBr} - (\Lambda_m^0)_{NaBr}(Λm0​)NaBr​−(Λm0​)NaI​=(Λm0​)KBr​−(Λm0​)NaBr​
  2. (B)(Λm0)NaBr−(Λm0)NaCl=(Λm0)KBr−(Λm0)KCl(\Lambda_m^0)_{NaBr} - (\Lambda_m^0)_{NaCl} = (\Lambda_m^0)_{KBr} - (\Lambda_m^0)_{KCl}(Λm0​)NaBr​−(Λm0​)NaCl​=(Λm0​)KBr​−(Λm0​)KCl​
  3. (C)(Λm0)H2O=(Λm0)HCl+(Λm0)NaOH−(Λm0)NaCl(\Lambda_m^0)_{H_2O} = (\Lambda_m^0)_{HCl} + (\Lambda_m^0)_{NaOH} - (\Lambda_m^0)_{NaCl}(Λm0​)H2​O​=(Λm0​)HCl​+(Λm0​)NaOH​−(Λm0​)NaCl​
  4. (D)(Λm0)KCl−(Λm0)NaCl=(Λm0)KBr−(Λm0)NaBr(\Lambda_m^0)_{KCl} - (\Lambda_m^0)_{NaCl} = (\Lambda_m^0)_{KBr} - (\Lambda_m^0)_{NaBr}(Λm0​)KCl​−(Λm0​)NaCl​=(Λm0​)KBr​−(Λm0​)NaBr​

Correct answer: (A)

Step-by-step solution →
Q234·ChemistryNumericalJEE Advanced 2019
The amount of water produced (in g) in the oxidation of 1 mole of rhombic sulphur by conc. HNO3_{3}3​ to a compound with the highest oxidation state of sulphur is______ (Given data : Molar mass of water = 18 g mol−1^{-1}−1)

Correct answer: 288.00

Step-by-step solution →
Q235·ChemistrySingle correctJEE Main 2019
The decreasing order of electrical conductivity of the following aqueous solutions is: 0.1 M Formic acid (a) 0.1 M Acetic acid (b) 0.1 M Benzoic acid (c)
  1. (A)a > c > b
  2. (B)c > a > b
  3. (C)c > b > a
  4. (D)a > b > c

Correct answer: (A)

Step-by-step solution →
Q236·ChemistrySingle correctJEE Main 2019
Given : Co3+^{3+}3+e−^{-}− ⟶\longrightarrow⟶ Co2+^{2+}2+; E° = 1.81 V Pb4^{4}4 +2e−^{-}− ⟶\longrightarrow⟶ Pb2+^{2+}2+; E° = +1.67 V Ce4+^{4+}4+ + e−^{-}− ⟶\longrightarrow⟶ Ce3+^{3+}3+; E° +1.61 V Bi3+^{3+}3+ + 3e−^{-}− ⟶\longrightarrow⟶ Bi ; E° = +0.20 V Oxidizing power of the species will increase in the order
  1. (A)Ce4+^{4+}4+ < Pb4+^{4+}4+ < Bi3+^{3+}3+ < Co3+^{3+}3+
  2. (B)Co3+^{3+}3+ < Pb4+^{4+}4+ < Ce4+^{4+}4+ < Bi3+^{3+}3+
  3. (C)Bi3+^{3+}3+ < Ce4+^{4+}4+ < Pb4+^{4+}4+ < Co3+^{3+}3+
  4. (D)Co3+^{3+}3+ < Ce4+^{4+}4+ < Bi3+^{3+}3+ < Pb4+^{4+}4+

Correct answer: (C)

Step-by-step solution →
Q237·ChemistrySingle correctJEE Main 2019
An example of a disproportionation reaction is :
  1. (A)2KMnO4→K2MnO4+MnO2+O22KMnO_4 \rightarrow K_2MnO_4 + MnO_2 + O_22KMnO4​→K2​MnO4​+MnO2​+O2​
  2. (B)2NaBr+Cl2→2naCl+Br22NaBr + Cl_2 \rightarrow 2naCl + Br_22NaBr+Cl2​→2naCl+Br2​
  3. (C)2CuBr→CuBr2+Cu2CuBr \rightarrow CuBr_2 + Cu2CuBr→CuBr2​+Cu
  4. (D)2MnO4−+10I−+16H+→2Mn2++5I2+8H2O2MnO_4^- + 10I^- + 16H^+ \rightarrow 2Mn^{2+} + 5I_2 + 8H_2O2MnO4−​+10I−+16H+→2Mn2++5I2​+8H2​O

Correct answer: (C)

Step-by-step solution →
Q238·ChemistrySingle correctJEE Main 2019
Consider the statements S1 and S2: S1: Conductivity always increases with decrease in the concentration of electrolyte. S2: Molar conductivity always increases with decrease in the concentration of electrolyte. The correct option among the following is:
  1. (A)S1 is wrong and S2 is correct
  2. (B)Both S1 and S2 are wrong
  3. (C)S1 is correct and S2 is wrong
  4. (D)Both S1 and S2 are correct

Correct answer: (A)

Step-by-step solution →
Q239·ChemistrySingle correctJEE Main 2019
Which one of the following graphs between molar conductivity (Λm_mm​) versus C\sqrt{C}C​ is correct?
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (B)

Step-by-step solution →
Q240·ChemistrySingle correctJEE Main 2019
A solution of Ni(NO3_33​)2_22​ is electrolysed between platinum electrodes using 0.1 Faraday electricity. How many mole of Ni will be deposited at the cathode?
  1. (A)0.20
  2. (B)0.05
  3. (C)0.10
  4. (D)0.15

Correct answer: (B)

Step-by-step solution →
Q241·ChemistrySingle correctJEE Main 2019
The standard Gibbs energy for the given cell reaction in kJ mol−1mol^{-1}mol−1 at 298 K is: Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s),E∘=2VZn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s), E^\circ = 2 VZn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s),E∘=2V at 298K [Faraday's constant F = 96500 C mol−1mol^{-1}mol−1]
  1. (A)-192
  2. (B)384
  3. (C)-384
  4. (D)192

Correct answer: (C)

Step-by-step solution →
Q242·ChemistrySingle correctJEE Main 2019
In order to oxidize a mixture of one mole of each of FeC2O4FeC_2O_4FeC2​O4​, Fe2(C2O4)3Fe_2(C_2O_4)_3Fe2​(C2​O4​)3​, FeSO4FeSO_4FeSO4​ and Fe2(SO4)3Fe_2(SO_4)_3Fe2​(SO4​)3​ in acidic medium, the number of moles of KMnO4KMnO_4KMnO4​ required is:
  1. (A)1
  2. (B)1.5
  3. (C)2
  4. (D)3

Correct answer: (C)

Step-by-step solution →
Q243·ChemistrySingle correctJEE Main 2019
Calculate the standard cell potential (in V) of the cell in which following reaction takes place: Fe2+(aq)+Ag+(aq)→Fe3+(aq)+Ag(s)Fe^{2}+(aq)+Ag^{+}(aq)\rightarrow Fe^{3+}(aq)+Ag(s)Fe2+(aq)+Ag+(aq)→Fe3+(aq)+Ag(s) Given that: EAg′/Ag0=xVE^{0}_{Ag'/Ag}=xVEAg′/Ag0​=xV EFe2+/Fe0=yVE^{0}_{Fe^{2+}/Fe}=yVEFe2+/Fe0​=yV EFe3+/Fe0=zVE^{0}_{Fe^{3+}/Fe}=zVEFe3+/Fe0​=zV
  1. (A)x \u2212 z
  2. (B)x + y \u2212 z
  3. (C)x \u2212 y
  4. (D)x + 2y \u2212 3z

Correct answer: (D)

Step-by-step solution →
Q244·ChemistrySingle correctJEE Main 2019
Given that EO2/H2O0=+1.23VE^0_{O_2/H_2O} = +1.23VEO2​/H2​O0​=+1.23V; ES2O82−/SO42−0=2.05VE^0_{S_2O_8^{2-}/SO_4^{2-}} = 2.05VES2​O82−​/SO42−​0​=2.05V; EBr2/Br−0=+1.09VE^0_{Br_2/Br^-} = +1.09VEBr2​/Br−0​=+1.09V; EAu3+/Au0=1.4VE^0_{Au^{3+}/Au} = 1.4VEAu3+/Au0​=1.4V. The strongest oxidizing agent is:
  1. (A)O2O_2O2​
  2. (B)S2O82−S_2O_8^{2-}S2​O82−​
  3. (C)Au3+Au^{3+}Au3+
  4. (D)Br2Br_2Br2​

Correct answer: (B)

Step-by-step solution →
Q245·ChemistrySingle correctJEE Main 2019
The standard electrode potential E⊖E^{\ominus}E⊖ and its temperature coefficient (dE⊖dT)\left(\dfrac{dE^{\ominus}}{dT}\right)(dTdE⊖​) for a cell are 2 V and −5×10−4-5\times10^{-4}−5×10−4 VK−1K^{-1}K−1 at 300 K respectively. The cell reaction is : Zn(s)+Cu2+(aq)⇌Zn2+(aq)+Cu(s)Zn(s)+Cu^{2+}(aq)\rightleftharpoons Zn^{2+}(aq)+Cu(s)Zn(s)+Cu2+(aq)⇌Zn2+(aq)+Cu(s) Standard reaction enthalpy (ΔrH⊖\Delta_rH^{\ominus}Δr​H⊖)........
  1. (A)-412.8
  2. (B)-384.0
  3. (C)1920
  4. (D)206.4 kJ

Correct answer: (A)

Step-by-step solution →
Q246·ChemistrySingle correctJEE Main 2019
∧m∘\wedge^{\circ}_{m}∧m∘​ for NaCl, HCl and NaA are 126.4, 425.9 and 100.5 S cm2^{2}2mol−1^{-1}−1, respectively. If the conductivity of 0.001 M HA is 5×10−55\times10^{-5}5×10−5 S cm−1^{-1}−1, degree of dissociation of HA is :
  1. (A)0.50
  2. (B)0.25
  3. (C)0.125
  4. (D)0.75

Correct answer: (C)

Step-by-step solution →
Q247·ChemistrySingle correctJEE Main 2019
For the cell Zn(s)∣Zn2+(aq)∣∣Mx+(aq)∣M(s)Zn(s)|Zn^{2+}(aq)||M^{x+}(aq)|M(s)Zn(s)∣Zn2+(aq)∣∣Mx+(aq)∣M(s), different half cells and their standard electrode potentials are given below: If EZn2+/Zn0=−0.76E^{0}_{Zn^{2+}/Zn} = -0.76EZn2+/Zn0​=−0.76 V, which cathode will given a maximum value of Ecell0E^{0}_{cell}Ecell0​ per electron transferred?
Mx+(aq)/M(s)M^{x+}(aq)/M(s)Mx+(aq)/M(s)Au3+(aq)/Au(s)Au^{3+}(aq)/Au(s)Au3+(aq)/Au(s)Ag+(aq)/Ag(s)Ag^{+}(aq)/Ag(s)Ag+(aq)/Ag(s)Fe3+(aq)/Fe2+(aq)Fe^{3+}(aq)/Fe^{2+}(aq)Fe3+(aq)/Fe2+(aq)Fe2+(aq)/Fe(s)Fe^{2+}(aq)/Fe(s)Fe2+(aq)/Fe(s)
EMx+/M∘E^{\circ}_{M^{x+}/M}EMx+/M∘​/(V)1.400.800.77-0.44
  1. (A)Ag+/AgAg^{+}/AgAg+/Ag
  2. (B)Fe3+/Fe2+Fe^{3+}/Fe^{2+}Fe3+/Fe2+
  3. (C)Au3+/AuAu^{3+}/AuAu3+/Au
  4. (D)Fe2+/FeFe^{2+}/FeFe2+/Fe

Correct answer: (A)

Step-by-step solution →
Q248·ChemistrySingle correctJEE Main 2019
Given the equilibrium constant: KCK_{C}KC​ of the reaction: Cu(s)+2Ag+(aq)→Cu2+(aq)+2Ag(s)Cu(s) + 2Ag^{+}(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)Cu(s)+2Ag+(aq)→Cu2+(aq)+2Ag(s) is 10×101510 \times 10^{15}10×1015, calculate the EcellθE^{\theta}_{cell}Ecellθ​ of the reaction of 298 K [2.303RTF at 298K=0.059V]\left[2.303\dfrac{RT}{F} \text{ at } 298K = 0.059V\right][2.303FRT​ at 298K=0.059V]
  1. (A)0.04736 mV
  2. (B)0.4736 mV
  3. (C)0.4736 V
  4. (D)0.04736 V

Correct answer: (C)

Step-by-step solution →
Q249·ChemistrySingle correctJEE Main 2019
Consider the following Zn2+^{2+}2+ + 2e−^{-}− ⟶\longrightarrow⟶ Zn(s); Eo^{o}o = −0.76-0.76−0.76 V Ca2+^{2+}2+ + 2e−^{-}− ⟶\longrightarrow⟶ Ca(s); Eo^{o}o = −2.87-2.87−2.87 V Mg2+^{2+}2+ + 2e−^{-}− ⟶\longrightarrow⟶ Mg(s); Eo^{o}o = −2.36-2.36−2.36 V Ni2+^{2+}2+ + 2e−^{-}− ⟶\longrightarrow⟶ Ni(s); Eo^{o}o = −0.25-0.25−0.25 V The reducing power of the metals increases in the order
  1. (A)Ca < Zn < Mg < Ni
  2. (B)Ni < Zn < Mg < Ca
  3. (C)Zn < Mg < Ni < Ca
  4. (D)Ca < Mg < Zn < Ni

Correct answer: (B)

Step-by-step solution →
Q250·ChemistrySingle correctJEE Main 2019
The anodic half-cell of lead-acid battery is recharged using electricity of 0.05 Faraday. The amount of PbSO4_{4}4​ electrolyzed in g during the process is: (Molar mass of PbSO4_{4}4​ = 303 g mol−1^{-1}−1)
  1. (A)22.8
  2. (B)15.2
  3. (C)7.6
  4. (D)11.4

Correct answer: (C)

Step-by-step solution →
Q251·ChemistrySingle correctJEE Main 2019
If the standard electrode potential for a cell is 2 V at 300 K, the equilibrium constant (K) for the reaction Zn(s)+Cu2+(aq)⇌Zn2+(aq)+Cu(s)Zn(s) + Cu^{2+}(aq) \rightleftharpoons Zn^{2+}(aq) + Cu(s)Zn(s)+Cu2+(aq)⇌Zn2+(aq)+Cu(s) At 300 K is approximately (R = 8 JK−1^{-1}−1 mol−1^{-1}−1, F = 96000 C mol−1^{-1}−1)
  1. (A)e−80e^{-80}e−80
  2. (B)e−160e^{-160}e−160
  3. (C)e320e^{320}e320
  4. (D)e160e^{160}e160

Correct answer: (D)

Step-by-step solution →
Q252·ChemistryNumericalJEE Advanced 2018
Consider an electrochemical cell: A(s)∣An+(aq,2M)∣∣B2n+(aq,1M)∣B(s)A(s) | A^{n+} (aq, 2 M) || B^{2n+} (aq, 1 M) | B(s)A(s)∣An+(aq,2M)∣∣B2n+(aq,1M)∣B(s). The value of ΔH⊖\Delta H^{\ominus}ΔH⊖ for the cell reaction is twice that of ΔG⊖\Delta G^{\ominus}ΔG⊖ at 300 K. If the emf of the cell is zero, the ΔS⊖\Delta S^{\ominus}ΔS⊖ (in J K−1K^{-1}K−1 mol−1mol^{-1}mol−1) of the cell reaction per mole of B formed at 300 K is ____. (Given: ln⁡(2)=0.7\ln(2) = 0.7ln(2)=0.7, R (universal gas constant) = 8.3 J K−1K^{-1}K−1 mol−1mol^{-1}mol−1. HHH, SSS and GGG are enthalpy, entropy and Gibbs energy, respectively.)

Correct answer: -11.62

Step-by-step solution →
Q253·ChemistryNumericalJEE Advanced 2018
To measure the quantity of MnCl2MnCl_{2}MnCl2​ dissolved in an aqueous solution, it was completely converted to KMnO4KMnO_{4}KMnO4​ using the reaction, MnCl2+K2S2O8+H2O→KMnO4+H2SO4+HClMnCl_{2} + K_{2}S_{2}O_{8} + H_{2}O \rightarrow KMnO_{4} + H_{2}SO_{4} + HClMnCl2​+K2​S2​O8​+H2​O→KMnO4​+H2​SO4​+HCl (equation not balanced). Few drops of concentrated HCl were added to this solution and gently warmed. Further, oxalic acid (225 mg) was added in portions till the colour of the permanganate ion disappeared. The quantity of MnCl2MnCl_{2}MnCl2​ (in mg) present in the initial solution is ____. (Atomic weights in g mol−1mol^{-1}mol−1: Mn = 55, Cl = 35.5)

Correct answer: 126

Step-by-step solution →
Q254·ChemistryNumericalJEE Advanced 2018
For the electrochemical cell, Mg(s)∣Mg2+(aq,1 M)∥Cu2+(aq,1 M)∣Cu(s)Mg(s) \mid Mg^{2+} (aq, 1\ M) \parallel Cu^{2+} (aq, 1\ M) \mid Cu(s)Mg(s)∣Mg2+(aq,1 M)∥Cu2+(aq,1 M)∣Cu(s) the standard emf of the cell is 2.70 V at 300 K. When the concentration of Mg2+Mg^{2+}Mg2+ is changed to xxx M, the cell potential changes to 2.67 V at 300 K. The value of xxx is ___. (given, FR=11500\dfrac{F}{R} = 11500RF​=11500 K V−1V^{-1}V−1, where FFF is the Faraday constant and RRR is the gas constant, ln⁡(10)=2.30\ln(10) = 2.30ln(10)=2.30)

Correct answer: 10

Step-by-step solution →
Q255·ChemistryIntegerJEE Advanced 2017
The conductance of a 0.0015 M aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinized Pt electrodes. The distance between the electrodes is 120 cm with an area of cross section of 1 cm2cm^{2}cm2. The conductance of this solution was found to be 5×10−75 \times 10^{-7}5×10−7 S. The pH of the solution is 4. The value of limiting molar conductivity (Λmo)\left(\Lambda_{m}^{o}\right)(Λmo​) of this weak monobasic acid in aqueous solution is Z×102Z \times 10^{2}Z×102 S cm−1cm^{-1}cm−1 mol−1mol^{-1}mol−1. The value of Z is

Correct answer: 6

Step-by-step solution →
Q256·ChemistrySingle correctJEE Advanced 2017
For the following cell, Zn(s) ∣ ZnSO4(aq) ∥ CuSO4(aq) ∣ Cu(s)\mathrm{Zn}(s)\,|\,\mathrm{ZnSO_{4}}(aq)\,\|\,\mathrm{CuSO_{4}}(aq)\,|\,\mathrm{Cu}(s)Zn(s)∣ZnSO4​(aq)∥CuSO4​(aq)∣Cu(s) When the concentration of Zn2+\mathrm{Zn^{2+}}Zn2+ is 10 times the concentration of Cu2+\mathrm{Cu^{2+}}Cu2+, the expression for ΔG\Delta GΔG (in J mol−1^{-1}−1) is [F is Faraday constant; R is gas constant; T is temperature; E0(cell)E^{0}(cell)E0(cell) = 1.1 V]
  1. (A)1.1 F1.1\,F1.1F
  2. (B)2.303 RT−2.2 F2.303\,RT - 2.2\,F2.303RT−2.2F
  3. (C)2.303 RT+1.1 F2.303\,RT + 1.1\,F2.303RT+1.1F
  4. (D)−2.2 F-2.2\,F−2.2F

Correct answer: (B)

Step-by-step solution →
Q257·ChemistryIntegerJEE Advanced 2016
In neutral or faintly alkaline solution, 8 moles of permanganate anion quantitatively oxidize thiosulphate anions to produce X\mathbf{X}X moles of a sulphur containing product. The magnitude of X\mathbf{X}X is

Correct answer: 6

Step-by-step solution →
Q258·ChemistrySingle correctJEE Advanced 2016
For the following electrochemical cell at 298 K, Pt(s) ∣ H2(g,1 bar) ∣ H+(aq,1 M) ∣∣ M4+(aq),M2+(aq) ∣ Pt(s)Pt(s)\,|\,H_{2}(g, 1\ bar)\,|\,H^{+}(aq, 1\ M)\,||\,M^{4+}(aq), M^{2+}(aq)\,|\,Pt(s)Pt(s)∣H2​(g,1 bar)∣H+(aq,1 M)∣∣M4+(aq),M2+(aq)∣Pt(s) Ecell=0.092E_{cell} = 0.092Ecell​=0.092 V when [M2+(aq)][M4+(aq)]=10x\dfrac{[M^{2+}(aq)]}{[M^{4+}(aq)]} = 10^{x}[M4+(aq)][M2+(aq)]​=10x Given: EM4+/M2+0=0.151E^{0}_{M^{4+}/M^{2+}} = 0.151EM4+/M2+0​=0.151 V; 2.303RTF=0.0592.303\dfrac{RT}{F} = 0.0592.303FRT​=0.059 V The value of xxx is
  1. (A)−2-2−2
  2. (B)−1-1−1
  3. (C)1
  4. (D)2

Correct answer: (D)

Step-by-step solution →
Q259·ChemistryIntegerJEE Advanced 2015
All the energy released from the reaction X→Y\mathbf{X} \to \mathbf{Y}X→Y, ΔrG0=−193\Delta_{r}G^{0} = -193Δr​G0=−193 kJ mol−1^{-1}−1 is used for oxidizing M+\mathbf{M^{+}}M+ as M+→M3++2e−\mathbf{M^{+}} \to \mathbf{M^{3+}} + 2e^{-}M+→M3++2e−, E0=−0.25E^{0} = -0.25E0=−0.25 V. Under standard conditions, the number of moles of M+\mathbf{M^{+}}M+ oxidized when one\mathbf{one}one mole of X\mathbf{X}X is converted to Y\mathbf{Y}Y is [F = 96500 C mol−1^{-1}−1]

Correct answer: 4

Step-by-step solution →
Q260·ChemistryIntegerJEE Advanced 2015
The molar conductivity of a solution of a weak acid HX (0.01 M) is 10 times smaller than the molar conductivity of a solution of a weak acid HY (0.10 M). If λX−0≈λY−0\lambda^{0}_{\mathrm{X^{-}}} \approx \lambda^{0}_{\mathrm{Y^{-}}}λX−0​≈λY−0​, the difference in their pKa\mathrm{p}K_{a}pKa​ values, pKa(HX)−pKa(HY)\mathrm{p}K_{a}(\mathrm{HX}) - \mathrm{p}K_{a}(\mathrm{HY})pKa​(HX)−pKa​(HY), is (consider degree of ionization of both acids to be ≪1\ll 1≪1)

Correct answer: 3

Step-by-step solution →
Q261·ChemistryMultiple correctJEE Advanced 2014
In a galvanic cell, the salt bridge
  1. (A)does not participate chemically in the cell reaction.
  2. (B)stops the diffusion of ions from one electrode to another.
  3. (C)is necessary for the occurrence of the cell reaction.
  4. (D)ensures mixing of the two electrolytic solutions.

Correct answer: (A)

Step-by-step solution →
Q262·ChemistryMultiple correctJEE Advanced 2014
For the reaction: I−+ClO3−+H2SO4⟶Cl−+HSO4−+I2I^- + ClO_3^- + H_2SO_4 \longrightarrow Cl^- + HSO_4^- + I_2I−+ClO3−​+H2​SO4​⟶Cl−+HSO4−​+I2​ The correct statement(s) in the balanced equation is/are:
  1. (A)Stoichiometric coefficient of HSO4−HSO_4^-HSO4−​ is 6.
  2. (B)Iodide is oxidized.
  3. (C)Sulphur is reduced.
  4. (D)H2OH_2OH2​O is one of the products.

Correct answer: (A), (B), (D)

Step-by-step solution →
Q263·ChemistryIntegerJEE Advanced 2014
Consider the following list of reagents: Acidified K2Cr2O7K_2Cr_2O_7K2​Cr2​O7​, alkaline KMnO4KMnO_4KMnO4​, CuSO4CuSO_4CuSO4​, H2O2H_2O_2H2​O2​, Cl2Cl_2Cl2​, O3O_3O3​, FeCl3FeCl_3FeCl3​, HNO3HNO_3HNO3​ and Na2S2O3Na_2S_2O_3Na2​S2​O3​. The total number of reagents that can oxidise aqueous iodide to iodine is

Correct answer: 7

Step-by-step solution →
Q264·ChemistrySingle correctJEE Advanced 2013
The standard reduction potential data at 25°C is given below: E∘(Fe3+,Fe2+)=+0.77 V;E^{\circ}(\mathrm{Fe^{3+}}, \mathrm{Fe^{2+}}) = +0.77\ \mathrm{V};E∘(Fe3+,Fe2+)=+0.77 V; E∘(Fe2+,Fe)=−0.44 VE^{\circ}(\mathrm{Fe^{2+}}, \mathrm{Fe}) = -0.44\ \mathrm{V}E∘(Fe2+,Fe)=−0.44 V E∘(Cu2+,Cu)=+0.34 V;E^{\circ}(\mathrm{Cu^{2+}}, \mathrm{Cu}) = +0.34\ \mathrm{V};E∘(Cu2+,Cu)=+0.34 V; E∘(Cu+,Cu)=+0.52 VE^{\circ}(\mathrm{Cu^{+}}, \mathrm{Cu}) = +0.52\ \mathrm{V}E∘(Cu+,Cu)=+0.52 V E∘[O2(g)+4H++4e−→2H2O]=+1.23 V;E^{\circ}[\mathrm{O_2(g)} + 4\mathrm{H^{+}} + 4e^{-} \rightarrow 2\mathrm{H_2O}] = +1.23\ \mathrm{V};E∘[O2​(g)+4H++4e−→2H2​O]=+1.23 V; E∘[O2(g)+2H2O+4e−→4OH−]=+0.40 VE^{\circ}[\mathrm{O_2(g)} + 2\mathrm{H_2O} + 4e^{-} \rightarrow 4\mathrm{OH^{-}}] = +0.40\ \mathrm{V}E∘[O2​(g)+2H2​O+4e−→4OH−]=+0.40 V E∘(Cr3+,Cr)=−0.74 V;E^{\circ}(\mathrm{Cr^{3+}}, \mathrm{Cr}) = -0.74\ \mathrm{V};E∘(Cr3+,Cr)=−0.74 V; E∘(Cr2+,Cr)=−0.91 VE^{\circ}(\mathrm{Cr^{2+}}, \mathrm{Cr}) = -0.91\ \mathrm{V}E∘(Cr2+,Cr)=−0.91 V Match E0E^{0}E0 of the redox pair in List – I with the values given in List – II and select the correct answer using the code given below the lists:
List – IList – II
P.E∘(Fe3+,Fe)E^{\circ}(\mathrm{Fe^{3+}}, \mathrm{Fe})E∘(Fe3+,Fe)1.−0.18 V-0.18\ \mathrm{V}−0.18 V
Q.E∘(4H2O⇌4H++4OH−)E^{\circ}(4\mathrm{H_2O} \rightleftharpoons 4\mathrm{H^{+}} + 4\mathrm{OH^{-}})E∘(4H2​O⇌4H++4OH−)2.−0.4 V-0.4\ \mathrm{V}−0.4 V
R.E∘(Cu2++Cu⟶2Cu+)E^{\circ}(\mathrm{Cu^{2+}} + \mathrm{Cu} \longrightarrow 2\mathrm{Cu^{+}})E∘(Cu2++Cu⟶2Cu+)3.−0.04 V-0.04\ \mathrm{V}−0.04 V
S.E∘(Cr3+,Cr2+)E^{\circ}(\mathrm{Cr^{3+}}, \mathrm{Cr^{2+}})E∘(Cr3+,Cr2+)4.−0.83 V-0.83\ \mathrm{V}−0.83 V
  1. (A)P-4, Q-1, R-2, S-3
  2. (B)P-2, Q-3, R-4, S-1
  3. (C)P-1, Q-2, R-3, S-4
  4. (D)P-3, Q-4, R-1, S-2

Correct answer: (D)

Step-by-step solution →
Q265·ChemistrySingle correctJEE Advanced 2013
An aqueous solution of X is added slowly to an aqueous solution of Y as shown in List – I. The variation in conductivity of these reactions in List – II. Match List – I with List – II and select the correct answer using the code given below the lists:
List – IList – II
P.(C2H5)3N(C_2H_5)_3N(C2​H5​)3​N (X) + CH3COOHCH_3COOHCH3​COOH (Y)1.Conductivity decreases and then increases
Q.KI(0.1M)\mathrm{KI(0.1M)}KI(0.1M) (X) + AgNO3(0.01M)\mathrm{AgNO_3(0.01M)}AgNO3​(0.01M) (Y)2.Conductivity decreases and then does not change much
R.CH3COOHCH_3COOHCH3​COOH (X) + KOHKOHKOH (Y)3.Conductivity increases and then does not change much
S.NaOHNaOHNaOH (X) + HIHIHI (Y)4.Conductivity does not change much and then increases
  1. (A)P-3, Q-4, R-2, S-1
  2. (B)P-4, Q-3, R-2, S-1
  3. (C)P-2, Q-3, R-4, S-1
  4. (D)P-1, Q-4, R-3, S-2

Correct answer: (A)

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Redox Reactions and Electrochemistry — frequently asked

How many questions from Redox Reactions and Electrochemistry appear in JEE?

Redox Reactions and Electrochemistry has appeared in 177 of the last 186 JEE Main and JEE Advanced papers — about 95% of them — contributing 265 questions in total across those papers.

Is Redox Reactions and Electrochemistry an important chapter for JEE?

Judged by how often it is actually tested, it appears in roughly 95% of papers. Chapters above about 50% are effectively guaranteed to show up every session, so they repay thorough preparation; lower-frequency chapters are better treated as targeted revision.

Where do these Redox Reactions and Electrochemistry questions come from?

Every question is from an official JEE Main or JEE Advanced paper, transcribed from the original paper and tagged to this chapter. Answers follow the official answer key.

Other Chemistry chapters

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  • Equilibrium 199
  • Solutions 198
  • Chemical Thermodynamics 195

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