Electrochemistry
366 Questions
Start JEE Mains Test
2021
Q201
JEE Mains
Numerical
14 Mar 2026
Assume a cell with the following reaction
$C{u_{(s)}} + 2A{g^ + }(1 \times {10^{ - 3}}M) \to C{u^{2 + }}(0.250M) + 2A{g_{(s)}}$
$E_{cell}^\Theta = 2.97$ V
Ecell for the above reaction is ______________ V. (Nearest integer)
[Given : log 2.5 = 0.3979, T = 298 K]
$C{u_{(s)}} + 2A{g^ + }(1 \times {10^{ - 3}}M) \to C{u^{2 + }}(0.250M) + 2A{g_{(s)}}$
$E_{cell}^\Theta = 2.97$ V
Ecell for the above reaction is ______________ V. (Nearest integer)
[Given : log 2.5 = 0.3979, T = 298 K]
Correct Answer: 3
Explanation:
$E = {E^o} - {{0.059} \over 2}\log {{[C{u^{ + 2}}]} \over {{{[A{g^ + }]}^2}}}$
$ = 2.97 - {{0.059} \over 2}\log {{0.25} \over {{{({{10}^{ - 3}})}^2}}} = 2.81V$
$ = 2.97 - {{0.059} \over 2}\log {{0.25} \over {{{({{10}^{ - 3}})}^2}}} = 2.81V$
2021
Q202
JEE Mains
Numerical
14 Mar 2026
Potassium chlorate is prepared by electrolysis of KCl in basic solution as shown by following equation.
6OH$-$ + Cl$-$ $\to$ ClO3$-$ + 3H2O + 6e$-$
A current of xA has to be passed for 10h to produce 10.0g of potassium chlorate. The value of x is ____________. (Nearest integer)
(Molar mass of KClO3 = 122.6 g mol$-$1, F = 96500 C)
6OH$-$ + Cl$-$ $\to$ ClO3$-$ + 3H2O + 6e$-$
A current of xA has to be passed for 10h to produce 10.0g of potassium chlorate. The value of x is ____________. (Nearest integer)
(Molar mass of KClO3 = 122.6 g mol$-$1, F = 96500 C)
Correct Answer: 1
Explanation:
$W = {E \over F} \times I \times t$
$10 = {{122.6} \over {96500 \times 6}} \times x \times 10 \times 3600$
$x = 1.311$
Ans. (1)
$10 = {{122.6} \over {96500 \times 6}} \times x \times 10 \times 3600$
$x = 1.311$
Ans. (1)
2021
Q203
JEE Mains
Numerical
14 Mar 2026
The molar conductivities at infinite dilution of barium chloride, sulphuric acid and hydrochloric acid are 280, 860 and 426 S cm2 mol$-$1 respectively. The molar conductivity at infinite dilution of barium sulphate is _________ S cm2 mol$-$1. (Round off to the Nearest Integer ).
Correct Answer: 288
Explanation:
From Kohlrausch's law
$\Lambda _m^\infty (BaS{O_4}) = \lambda _m^\infty (B{a^{2 + }}) + \lambda _m^\infty (SO_4^{2 - })$
$\Lambda _m^\infty (BaS{O_4}) = \Lambda _m^\infty (BaC{l_2}) + \Lambda _m^\infty ({H_2}S{O_4}) - 2\Lambda _m^\infty (HCl)$
$ = 280 + 860 - 2(426)$
$ = 288$ S cm2 mol$-$1
$\Lambda _m^\infty (BaS{O_4}) = \lambda _m^\infty (B{a^{2 + }}) + \lambda _m^\infty (SO_4^{2 - })$
$\Lambda _m^\infty (BaS{O_4}) = \Lambda _m^\infty (BaC{l_2}) + \Lambda _m^\infty ({H_2}S{O_4}) - 2\Lambda _m^\infty (HCl)$
$ = 280 + 860 - 2(426)$
$ = 288$ S cm2 mol$-$1
2021
Q204
JEE Mains
Numerical
14 Mar 2026
For the reaction
2Fe3+(aq) + 2I$-$(aq) $ \to $ 2Fe2+(aq) + I2(s)
the magnitude of the standard molar Gibbs free energy change, $\Delta$rG$_m^o$ = $-$ ___________ kJ (Round off to the Nearest Integer).
$\left[ {\matrix{ {E_{F{e^{2 + }}/Fe(s)}^o = - 0.440V;} & {E_{F{e^{3 + }}/Fe(s)}^o = - 0.036V} \cr {E_{{I_2}/2{I^ - }}^o = 0.539V;} & {F = 96500C} \cr } } \right]$
2Fe3+(aq) + 2I$-$(aq) $ \to $ 2Fe2+(aq) + I2(s)
the magnitude of the standard molar Gibbs free energy change, $\Delta$rG$_m^o$ = $-$ ___________ kJ (Round off to the Nearest Integer).
$\left[ {\matrix{ {E_{F{e^{2 + }}/Fe(s)}^o = - 0.440V;} & {E_{F{e^{3 + }}/Fe(s)}^o = - 0.036V} \cr {E_{{I_2}/2{I^ - }}^o = 0.539V;} & {F = 96500C} \cr } } \right]$
Correct Answer: 46
Explanation:
$E_{F{e^{3 + }}/Fe}^o = 0.036V$
$F{e^{3 + }} + 3{e^ - } \to Fe$
$ \therefore $ $\Delta G_1^o = - nF{E^o}$
$ = - 3F( - 0.036)$
$E_{F{e^{2 + }}/Fe}^o = 0.440V$
$F{e^{2 + }} + 2{e^ - } \to Fe;{E^o} = - 0.440V$
$Fe \to F{e^{2 + }} + 2{e^ - };{E^o} = 0.440V$
$ \therefore $ $\Delta G_2^o = - 2F(0.440)$
$E_{{I_2}/2{I^ - }}^o = 0.539V$
${I_2} + 2{e^ - } \to 2{I^ - };{E^o} = 0.539V$
$2{I^ - } \to {I_2} + 2{e^ - };{E^o} = - 0.539V$
$\Delta G_3^o = - 2F( - 0.539)$
$ \therefore $ $\Delta {G^o} = 2\left[ {\Delta G_1^o + \Delta G_2^o} \right] + \Delta G_3^o$
$ = 2\left[ {3F(0.036) - 2F(0.440)} \right] + 2F(0.539)$
$ = - 45934 = - 45.9KJ \simeq - 46KJ$
$F{e^{3 + }} + 3{e^ - } \to Fe$
$ \therefore $ $\Delta G_1^o = - nF{E^o}$
$ = - 3F( - 0.036)$
$E_{F{e^{2 + }}/Fe}^o = 0.440V$
$F{e^{2 + }} + 2{e^ - } \to Fe;{E^o} = - 0.440V$
$Fe \to F{e^{2 + }} + 2{e^ - };{E^o} = 0.440V$
$ \therefore $ $\Delta G_2^o = - 2F(0.440)$
$E_{{I_2}/2{I^ - }}^o = 0.539V$
${I_2} + 2{e^ - } \to 2{I^ - };{E^o} = 0.539V$
$2{I^ - } \to {I_2} + 2{e^ - };{E^o} = - 0.539V$
$\Delta G_3^o = - 2F( - 0.539)$
$ \therefore $ $\Delta {G^o} = 2\left[ {\Delta G_1^o + \Delta G_2^o} \right] + \Delta G_3^o$
$ = 2\left[ {3F(0.036) - 2F(0.440)} \right] + 2F(0.539)$
$ = - 45934 = - 45.9KJ \simeq - 46KJ$
2021
Q205
JEE Mains
Numerical
14 Mar 2026
A KCl solution of conductivity 0.14 S m$-$1 shows a resistance of 4.19$\Omega$ in a conductivity cell. If the same cell is filled with an HCl solution, the resistance drops to 1.03$\Omega$. The conductivity of the HCl solution is ____________ $\times$ 10$-$2 S m$-$1. (Round off to the Nearest Integer).
Correct Answer: 57
Explanation:
For KCl solution,
$R = \left( {{1 \over K}} \right)\left( {{l \over A}} \right) \Rightarrow {l \over A} = R \times K = 4.19 \times 0.14$
= 0.58
For HCl solution,
$R = \left( {{1 \over K}} \right)\left( {{l \over A}} \right)$
$ \Rightarrow K = {{(l/A)} \over R} = {{0.58} \over {1.03}} = 0.56 = 56 \times {10^{ - 2}}$ Sm$-$1
$R = \left( {{1 \over K}} \right)\left( {{l \over A}} \right) \Rightarrow {l \over A} = R \times K = 4.19 \times 0.14$
= 0.58
For HCl solution,
$R = \left( {{1 \over K}} \right)\left( {{l \over A}} \right)$
$ \Rightarrow K = {{(l/A)} \over R} = {{0.58} \over {1.03}} = 0.56 = 56 \times {10^{ - 2}}$ Sm$-$1
2021
Q206
JEE Mains
Numerical
14 Mar 2026
A 5.0 m mol dm$-$3 aqueous solution of KCl has a conductance of 0.55 mS when measured in a cell of cell constant 1.3 cm$-$1. The molar conductivity of this solution is ___________ mSm2 mol$-$1. (Round off to the Nearest Integer).
Correct Answer: 14
Explanation:
Conductance = ${{Conductivity} \over {Cell\,cons\tan t}}$
$ \therefore $ Conductivity = 0.55 $\times$ 10$-$3 $\times$ 1.3 S cm$-$1
Molar conductivity = ${{Conductivity\,(S\,c{m^{ - 1}}) \times 1000} \over {Molarity\,(mol/L)}}$
$ = {{0.55 \times {{10}^{ - 3}} \times 1.3 \times 100} \over {5 \times {{10}^{ - 3}}}}$
= 143 S cm2 mol$-$1
= 14.3 mS m2 mol$-$1
$ \approx $ 14 mS m2 mol$-$1
$ \therefore $ Conductivity = 0.55 $\times$ 10$-$3 $\times$ 1.3 S cm$-$1
Molar conductivity = ${{Conductivity\,(S\,c{m^{ - 1}}) \times 1000} \over {Molarity\,(mol/L)}}$
$ = {{0.55 \times {{10}^{ - 3}} \times 1.3 \times 100} \over {5 \times {{10}^{ - 3}}}}$
= 143 S cm2 mol$-$1
= 14.3 mS m2 mol$-$1
$ \approx $ 14 mS m2 mol$-$1
2021
Q207
JEE Mains
Numerical
14 Mar 2026
Emf of the following cell at 298K in V is x $\times$ 10$-$2.
Zn|Zn2+(0.1 M)||Ag+ (0.01 M)|Ag
The value of x is _________. (Rounded off to the nearest integer)
[Given : $E_{Z{n^{2 + }}/Zn}^\theta = - 0.76V;E_{A{g^{2 + }}/Ag}^\theta = + 0.80V;{{2.303RT} \over F} = 0.059$]
Zn|Zn2+(0.1 M)||Ag+ (0.01 M)|Ag
The value of x is _________. (Rounded off to the nearest integer)
[Given : $E_{Z{n^{2 + }}/Zn}^\theta = - 0.76V;E_{A{g^{2 + }}/Ag}^\theta = + 0.80V;{{2.303RT} \over F} = 0.059$]
Correct Answer: 147
Explanation:
Zn | Zn2+(0.1 M) || Ag+ (0.01 M) | Ag
Zn(s) + 2Ag+ $ \rightleftharpoons $ 2Ag(s) + Zn+2
$E_{cell}^0 = E_{A{g^ + }/Ag}^0 - E_{Z{n^{2 + }}/Zn}^0$
$ = 0.80 - ( - 0.76)$
$ = 1.56V$
${E_{cell}} = 1.56 - {{ 0.059} \over 2}\log {{[Z{n^{2 + }}]} \over {{{[A{g^ + }]}^2}}}$
$ = 1.56 - {{0.059} \over 2}\log {{0.1} \over {{{(0.01)}^2}}}$
$ = 1.56 - {{0.059} \over 2} \times 3$
$ = 1.56 - 0.0885$
$ = 1.4715$
$ = 147.15 \times {10^{ - 2}}$
Zn(s) + 2Ag+ $ \rightleftharpoons $ 2Ag(s) + Zn+2
$E_{cell}^0 = E_{A{g^ + }/Ag}^0 - E_{Z{n^{2 + }}/Zn}^0$
$ = 0.80 - ( - 0.76)$
$ = 1.56V$
${E_{cell}} = 1.56 - {{ 0.059} \over 2}\log {{[Z{n^{2 + }}]} \over {{{[A{g^ + }]}^2}}}$
$ = 1.56 - {{0.059} \over 2}\log {{0.1} \over {{{(0.01)}^2}}}$
$ = 1.56 - {{0.059} \over 2} \times 3$
$ = 1.56 - 0.0885$
$ = 1.4715$
$ = 147.15 \times {10^{ - 2}}$
2021
Q208
JEE Mains
Numerical
14 Mar 2026
Consider the following reaction
$MnO_4^ - + 8{H^ + } + 5{e^ - } \to M{n^{ + 2}} + 4{H_2}O,{E^o} = 1.51V$.
The quantity of electricity required in Faraday to reduce five moles of $MnO_4^ - $ is ___________. (Integer answer)
$MnO_4^ - + 8{H^ + } + 5{e^ - } \to M{n^{ + 2}} + 4{H_2}O,{E^o} = 1.51V$.
The quantity of electricity required in Faraday to reduce five moles of $MnO_4^ - $ is ___________. (Integer answer)
Correct Answer: 25
Explanation:
$MnO_4^ - + 8{H^ + } + 5{e^ - } \to M{n^{ + 2}} + 4{H_2}O,{E^o} = 1.51V$
1 mole of MnO4- required 5 moles of electrons or 5 F electricity.
$ \therefore $ 5 moles of MnO4- required 25 F electricity.
1 mole of MnO4- required 5 moles of electrons or 5 F electricity.
$ \therefore $ 5 moles of MnO4- required 25 F electricity.
2021
Q209
JEE Mains
Numerical
14 Mar 2026
Copper reduces NO$_3^ - $ into NO and NO2 depending upon the concentration of HNO3 in solution. (Assuming fixed [Cu2+] and PNO = PNO2), the HNO3 concentration at which the thermodynamic tendency for reduction of NO$_3^ - $ into NO and NO2 by copper is same is 10x M. The value of 2x is _______. (Rounded off to the nearest integer)
[Given, $E_{C{u^{2 + }}/Cu}^o = 0.34$ V, $E_{NO_3^ - /NO}^o = 0.96$ V, $E_{NO_3^ - /N{O_2}}^o = 0.79$ V and at 298 K, ${{RT} \over F}$(2.303) = 0.059]
[Given, $E_{C{u^{2 + }}/Cu}^o = 0.34$ V, $E_{NO_3^ - /NO}^o = 0.96$ V, $E_{NO_3^ - /N{O_2}}^o = 0.79$ V and at 298 K, ${{RT} \over F}$(2.303) = 0.059]
Correct Answer: 1
Explanation:
Cell-I $(HN{O_3} \to NO)$
$3Cu + 2NO_3^ - + 8{H^ + } \to 3C{u^{2 + }} + 2NO + 4{H_2}O$
${Q_1} = {{{{[C{u^{2 + }}]}^3} \times {{({p_{NO}})}^2}} \over {{{[NO_3^ - ]}^2} \times {{[{H^ + }]}^8}}}$
$\because$ $E_1^o = 0.96 - ( - 0.34) = 1.3\,V$
${E_1} = 1.3 - {{0.059} \over 6}\log {Q_1}$
Cell-II $(HN{O_3} \to N{O_2})$
$Cu + 2NO_3^ - + 4{H^ + } \to C{u^{2 + }} + 2N{O_2} + 2{H_2}O$
${Q_2} = {{[C{u^2}] \times {{({p_{N{O_2}}})}^2}} \over {{{[NO_3^ - ]}^2} \times {{[{H^ + }]}^4}}}$
$\because$ $E_2^o = 0.79 - ( - 0.34)\,V = 1.13\,V$
${E_2} = 1.13 - {{0.059} \over 2}\log {Q_2}$
Now, ${E_1} = {E_2}$
$1.3 - {{0.059} \over 6}\log {Q_1} = 1.13 - {{0.059} \over 2}\log {Q_2}$
$0.17 = {{0.059} \over 6}[\log {Q_1} - 3\log {Q_2} - = {{0.059} \over 6}\log {{{Q_1}} \over {{Q_2}}}$
$ = {{0.059} \over 6}\log {{{{[C{u^{2 + }}]}^3} \times {{({p_{NO}})}^2}} \over {{{[NO_3^ - ]}^2} \times {{[{H^ + }]}^8}}} \times {{{{[NO_3^ - ]}^6} \times {{[{H^ + }]}^{12}}} \over {{{[C{u^{2 + }}]}^3} \times {{({p_{N{O_2}}})}^6}}}$
$ = {{0.059} \over 6}\log {{{{[{H^ + }]}^4} \times {{[NO_3^ - ]}^4}} \over {{{({p_{N{O_2}}})}^4}}}$ [$\because$ ${p_{NO}} = {p_{N{O_2}}}$]
$ = {{0.059} \over 6}\log {{{{[HN{O_3}]}^4}} \over {{{({p_{N{O_2}}})}^4}}}$
Now, ${p_{N{O_2}}} \equiv [HN{O_3}]$
So, $0.17 = {{0.059} \over 6}\log {[HN{O_3}]^8}$
$ = {{0.059} \over 6} \times 8\log [HN{O_3}]$
$\log [HN{O_3}] = 2.16$
$[HN{O_3}] = {10^{2.16}}M = {10^x}M$
$\therefore$ $x = 2.16$
$ \Rightarrow 2x = 2 \times 2.16 = 4.32 \simeq 4$
2021
Q210
JEE Mains
Numerical
14 Mar 2026
The magnitude of the change in oxidising power of the $MnO_4^ - /M{n^{2 + }}$ couple is x $\times$ 10$-$4 V, if the H+ concentration is decreased from 1M to 10$-$4 M at 25$^\circ$C. (Assume concentration of $MnO_4^ - $ and $M{n^{2 + }}$ to be same on change in H+ concentration). The value of x is ___________. $\left[ {Given\,:{{2.303RT} \over F} = 0.059} \right]$
Correct Answer: 3776
Explanation:
Reaction,
$MnO_4^ - + {H^ + } + 5{e^ - } \to M{n^{2 + }} + 4{H_2}O$
n = 5
Applying Nernst equation, ${E_{cell}} = E_{cell}^o - {{0.0591} \over n}\log {{[P]} \over {[R]}}$
or ${E_{cell}} = E_{cell}^o - {{0.0591} \over n}\log {{[M{n^{2 + }}]} \over {[MnO_4^ - ]}}{\left[ {{1 \over {{H^ + }}}} \right]^8}$
(I) Given, [H+] = 1 M
${E_1} = E^\circ - {{0.0591} \over 5}\log {{[M{n^{2 + }}]} \over {[MnO_4^ - ]}}$
(II) Now, [H+] = 10$-$4 M
${E_2} = E^\circ - {{0.0591} \over 5}\log {{[M{n^{2 + }}]} \over {[MnO_4^ - ]}} \times {1 \over {{{({{10}^{ - 4}})}^8}}}$
$\therefore$ $\left| {{E_1} - {E_2}} \right|$
$\left| {{E_1} - {E_2}} \right| = {{0.0591} \over 5} \times 32 = 0.3776\,V = 3776 \times {10^{ - 4}}$
x = 3776
$MnO_4^ - + {H^ + } + 5{e^ - } \to M{n^{2 + }} + 4{H_2}O$
n = 5
Applying Nernst equation, ${E_{cell}} = E_{cell}^o - {{0.0591} \over n}\log {{[P]} \over {[R]}}$
or ${E_{cell}} = E_{cell}^o - {{0.0591} \over n}\log {{[M{n^{2 + }}]} \over {[MnO_4^ - ]}}{\left[ {{1 \over {{H^ + }}}} \right]^8}$
(I) Given, [H+] = 1 M
${E_1} = E^\circ - {{0.0591} \over 5}\log {{[M{n^{2 + }}]} \over {[MnO_4^ - ]}}$
(II) Now, [H+] = 10$-$4 M
${E_2} = E^\circ - {{0.0591} \over 5}\log {{[M{n^{2 + }}]} \over {[MnO_4^ - ]}} \times {1 \over {{{({{10}^{ - 4}})}^8}}}$
$\therefore$ $\left| {{E_1} - {E_2}} \right|$
$\left| {{E_1} - {E_2}} \right| = {{0.0591} \over 5} \times 32 = 0.3776\,V = 3776 \times {10^{ - 4}}$
x = 3776
2021
Q211
JEE Mains
MCQ
14 Mar 2026
Match List - I with List - II
Choose the most appropriate answer from the options given below :
| List - I (Parmeter) |
List - II (Unit) |
||
|---|---|---|---|
| (a) | Cell constant | (i) | $S\,c{m^2}mo{l^{ - 1}}$ |
| (b) | Molar conductivity | (ii) | Dimensionless |
| (c) | Conductivity | (iii) | ${m^{ - 1}}$ |
| (d) | Degree of dissociation of electrolyte | (iv) | ${\Omega ^{ - 1}}{m^{ - 1}}$ |
Choose the most appropriate answer from the options given below :
A.
(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
B.
(a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)
C.
(a)-(i), (b)-(iv), (c)-(iii), (d)-(ii)
D.
(a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)
2021
Q212
JEE Mains
MCQ
14 Mar 2026
Given below are two statements :
Statement I : The limiting molar conductivity of KCl (strong electrolyte) is higher compared to that of CH3COOH (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 :
Statement I : The limiting molar conductivity of KCl (strong electrolyte) is higher compared to that of CH3COOH (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 :
A.
Statement I is true but Statement II is false.
B.
Statement I is false but Statement II is true.
C.
Both Statement I and Statement II are true.
D.
Both Statement I and Statement II are false.
2021
Q213
JEE Mains
MCQ
14 Mar 2026
Compound A used as a strong oxidizing agent is amphoteric in nature. It is the part of lead storage batteries. Compound A is :
A.
PbO
B.
PbO2
C.
PbSO4
D.
Pb3O4
2021
Q214
JEE Mains
MCQ
14 Mar 2026
The electrode potential of M2+/M of 3d-series elements shows positive value for :
A.
Zn
B.
Fe
C.
Cu
D.
Co
2021
Q215
JEE Advanced
Numerical
14 Mar 2026
At 298 K, the limiting molar conductivity of a weak monobasic acid is 4 $\times$ 102 S cm2 mol$-$1. At 298 K, for an aqueous solution of the acid the degree of dissociation is $\alpha$ and the molar conductivity is y $\times$ 102 S cm2 mol$-$1. At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes 3y $\times$ 102 S cm2 mol$-$1.
The value of $\alpha$ is __________.
The value of $\alpha$ is __________.
Correct Answer: 0.22
Explanation:
Degree of dissociation = $\alpha$
Limiting molar conductivity, $\Lambda _m^o$ = 4 $\times$ 102 S cm2 mol$-$1
Molar conductivity, $\Lambda _m$ = y $\times$ 102 S cm2 mol$-$1
Molar conductivity of dilution, $\Lambda _m^o$ = 3y $\times$ 102 S cm2 mol$-$1
Concentration before dilution = C
Concentration after dilution = ${C \over {20}}$
Using relation, $\alpha = {{\Lambda _m^{}} \over {\Lambda _m^o}}$ .... (i)
Dissociation constant, ${K_a} = {{C{\alpha ^2}} \over {(1 - \alpha )}}$
Putting Eq. (i),
${K_a} = {{C\Lambda _m^2} \over {\Lambda _m^{o2}\left( {1 - {{\Lambda _m^{}} \over {\Lambda _m^o}}} \right)}} = {{C\Lambda _m^2} \over {\Lambda _m^o(\Lambda _m^o - \Lambda _m^{})}}$
Dissociation constant before dillution,
${K_a} = {{C{{(y \times {{10}^2})}^2}} \over {(4 \times {{10}^2})(4 \times {{10}^2} - y \times {{10}^2})}}$ ..... (ii)
Dissociation constant after dilution,
${K_a} = {{{C \over {20}}{{(3y \times {{10}^2})}^2}} \over {(4 \times {{10}^2})(4 \times {{10}^2} - 3y \times {{10}^2})}}$ ..... (iii)
Comparing Eqs. (ii) and (iii),
${{C{{(y \times {{10}^2})}^2}} \over {(4 \times {{10}^2})(4 \times {{10}^2} - y \times {{10}^2})}} = {{C{{(3y \times {{10}^2})}^2}} \over {20(4 \times {{10}^2})(4 \times {{10}^2} - 3y \times {{10}^2})}}$
${{{y^2} \times {{10}^4}} \over {{{10}^2}(4 - y)}} = {{9{y^2} \times {{10}^4}} \over {{{10}^2} \times 20(4 - 3y)}}$
${1 \over {(4 - y)}} = {9 \over {20(4 - 3y)}}$
$80 - 60y = 36 - 9y$
$80 - 36 = 60y - 9y$
$44 = 51y$
$ \Rightarrow y = {{44} \over {51}}$ .... (iv)
Putting in Eq. (i),
$\alpha = {{\Lambda _m^{}} \over {\Lambda _m^o}} = {{y \times {{10}^2}} \over {4 \times {{10}^2}}}$
$\alpha = {{44} \over {51 \times 4}} = {{11} \over {51}}$
$\alpha = 0.22$
The value of $\alpha$ is 0.22.
Limiting molar conductivity, $\Lambda _m^o$ = 4 $\times$ 102 S cm2 mol$-$1
Molar conductivity, $\Lambda _m$ = y $\times$ 102 S cm2 mol$-$1
Molar conductivity of dilution, $\Lambda _m^o$ = 3y $\times$ 102 S cm2 mol$-$1
Concentration before dilution = C
Concentration after dilution = ${C \over {20}}$
Using relation, $\alpha = {{\Lambda _m^{}} \over {\Lambda _m^o}}$ .... (i)
Dissociation constant, ${K_a} = {{C{\alpha ^2}} \over {(1 - \alpha )}}$
Putting Eq. (i),
${K_a} = {{C\Lambda _m^2} \over {\Lambda _m^{o2}\left( {1 - {{\Lambda _m^{}} \over {\Lambda _m^o}}} \right)}} = {{C\Lambda _m^2} \over {\Lambda _m^o(\Lambda _m^o - \Lambda _m^{})}}$
Dissociation constant before dillution,
${K_a} = {{C{{(y \times {{10}^2})}^2}} \over {(4 \times {{10}^2})(4 \times {{10}^2} - y \times {{10}^2})}}$ ..... (ii)
Dissociation constant after dilution,
${K_a} = {{{C \over {20}}{{(3y \times {{10}^2})}^2}} \over {(4 \times {{10}^2})(4 \times {{10}^2} - 3y \times {{10}^2})}}$ ..... (iii)
Comparing Eqs. (ii) and (iii),
${{C{{(y \times {{10}^2})}^2}} \over {(4 \times {{10}^2})(4 \times {{10}^2} - y \times {{10}^2})}} = {{C{{(3y \times {{10}^2})}^2}} \over {20(4 \times {{10}^2})(4 \times {{10}^2} - 3y \times {{10}^2})}}$
${{{y^2} \times {{10}^4}} \over {{{10}^2}(4 - y)}} = {{9{y^2} \times {{10}^4}} \over {{{10}^2} \times 20(4 - 3y)}}$
${1 \over {(4 - y)}} = {9 \over {20(4 - 3y)}}$
$80 - 60y = 36 - 9y$
$80 - 36 = 60y - 9y$
$44 = 51y$
$ \Rightarrow y = {{44} \over {51}}$ .... (iv)
Putting in Eq. (i),
$\alpha = {{\Lambda _m^{}} \over {\Lambda _m^o}} = {{y \times {{10}^2}} \over {4 \times {{10}^2}}}$
$\alpha = {{44} \over {51 \times 4}} = {{11} \over {51}}$
$\alpha = 0.22$
The value of $\alpha$ is 0.22.
2021
Q216
JEE Advanced
Numerical
14 Mar 2026
At 298 K, the limiting molar conductivity of a weak monobasic acid is 4 $\times$ 102 S cm2 mol$-$1. At 298 K, for an aqueous solution of the acid the degree of dissociation is $\alpha$ and the molar conductivity is y $\times$ 102 S cm2 mol$-$1. At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes 3y $\times$ 102 S cm2 mol$-$1.
The value of y is __________.
The value of y is __________.
Correct Answer: 0.86
Explanation:
Degree of dissociation = $\alpha$
Limiting molar conductivity, $\Lambda _m^o$ = 4 $\times$ 102 S cm2 mol$-$1
Molar conductivity, $\Lambda _m$ = y $\times$ 102 S cm2 mol$-$1
Molar conductivity of dilution, $\Lambda _m^o$ = 3y $\times$ 102 S cm2 mol$-$1
Concentration before dilution = C
Concentration after dilution = ${C \over {20}}$
Using relation, $\alpha = {{\Lambda _m^{}} \over {\Lambda _m^o}}$ .... (i)
Dissociation constant, ${K_a} = {{C{\alpha ^2}} \over {(1 - \alpha )}}$
Putting Eq. (i),
${K_a} = {{C\Lambda _m^2} \over {\Lambda _m^{o2}\left( {1 - {{\Lambda _m^{}} \over {\Lambda _m^o}}} \right)}} = {{C\Lambda _m^2} \over {\Lambda _m^o(\Lambda _m^o - \Lambda _m^{})}}$
Dissociation constant before dillution,
${K_a} = {{C{{(y \times {{10}^2})}^2}} \over {(4 \times {{10}^2})(4 \times {{10}^2} - y \times {{10}^2})}}$ ..... (ii)
Dissociation constant after dilution,
${K_a} = {{{C \over {20}}{{(3y \times {{10}^2})}^2}} \over {(4 \times {{10}^2})(4 \times {{10}^2} - 3y \times {{10}^2})}}$ ..... (iii)
Comparing Eqs. (ii) and (iii),
${{C{{(y \times {{10}^2})}^2}} \over {(4 \times {{10}^2})(4 \times {{10}^2} - y \times {{10}^2})}} = {{C{{(3y \times {{10}^2})}^2}} \over {20(4 \times {{10}^2})(4 \times {{10}^2} - 3y \times {{10}^2})}}$
${{{y^2} \times {{10}^4}} \over {{{10}^2}(4 - y)}} = {{9{y^2} \times {{10}^4}} \over {{{10}^2} \times 20(4 - 3y)}}$
${1 \over {(4 - y)}} = {9 \over {20(4 - 3y)}}$
$80 - 60y = 36 - 9y$
$80 - 36 = 60y - 9y$
$44 = 51y$
$ \Rightarrow y = {{44} \over {51}}$ = 0.86
Limiting molar conductivity, $\Lambda _m^o$ = 4 $\times$ 102 S cm2 mol$-$1
Molar conductivity, $\Lambda _m$ = y $\times$ 102 S cm2 mol$-$1
Molar conductivity of dilution, $\Lambda _m^o$ = 3y $\times$ 102 S cm2 mol$-$1
Concentration before dilution = C
Concentration after dilution = ${C \over {20}}$
Using relation, $\alpha = {{\Lambda _m^{}} \over {\Lambda _m^o}}$ .... (i)
Dissociation constant, ${K_a} = {{C{\alpha ^2}} \over {(1 - \alpha )}}$
Putting Eq. (i),
${K_a} = {{C\Lambda _m^2} \over {\Lambda _m^{o2}\left( {1 - {{\Lambda _m^{}} \over {\Lambda _m^o}}} \right)}} = {{C\Lambda _m^2} \over {\Lambda _m^o(\Lambda _m^o - \Lambda _m^{})}}$
Dissociation constant before dillution,
${K_a} = {{C{{(y \times {{10}^2})}^2}} \over {(4 \times {{10}^2})(4 \times {{10}^2} - y \times {{10}^2})}}$ ..... (ii)
Dissociation constant after dilution,
${K_a} = {{{C \over {20}}{{(3y \times {{10}^2})}^2}} \over {(4 \times {{10}^2})(4 \times {{10}^2} - 3y \times {{10}^2})}}$ ..... (iii)
Comparing Eqs. (ii) and (iii),
${{C{{(y \times {{10}^2})}^2}} \over {(4 \times {{10}^2})(4 \times {{10}^2} - y \times {{10}^2})}} = {{C{{(3y \times {{10}^2})}^2}} \over {20(4 \times {{10}^2})(4 \times {{10}^2} - 3y \times {{10}^2})}}$
${{{y^2} \times {{10}^4}} \over {{{10}^2}(4 - y)}} = {{9{y^2} \times {{10}^4}} \over {{{10}^2} \times 20(4 - 3y)}}$
${1 \over {(4 - y)}} = {9 \over {20(4 - 3y)}}$
$80 - 60y = 36 - 9y$
$80 - 36 = 60y - 9y$
$44 = 51y$
$ \Rightarrow y = {{44} \over {51}}$ = 0.86
2021
Q217
JEE Advanced
MSQ
14 Mar 2026
Some standard electrode potentials at 298 K are given below :
Pb2+ /Pb = $- $0.13 V
Ni2+ /Ni = $-$ 0.24 V
Cd2+ /Cd = $-$ 0.40 V
Fe2+ /Fe = $-$ 0.44 V
To a solution containing 0.001 M of X2+ and 0.1 M of Y2+, the metal rods X and Y are inserted (at 298 K) and connected by a conducting wire. This resulted in dissolution of X. The correct combination(s) of X and Y, respectively, is(are)
(Given : Gas constant, R = 8.314 J K$-$ mol$-$1, Faraday constant, F = 96500 C mol$-$1)
Pb2+ /Pb = $- $0.13 V
Ni2+ /Ni = $-$ 0.24 V
Cd2+ /Cd = $-$ 0.40 V
Fe2+ /Fe = $-$ 0.44 V
To a solution containing 0.001 M of X2+ and 0.1 M of Y2+, the metal rods X and Y are inserted (at 298 K) and connected by a conducting wire. This resulted in dissolution of X. The correct combination(s) of X and Y, respectively, is(are)
(Given : Gas constant, R = 8.314 J K$-$ mol$-$1, Faraday constant, F = 96500 C mol$-$1)
A.
Cd and Ni
B.
Cd and Fe
C.
Ni and Pb
D.
Ni and Fe
2021
Q218
AP-EAPCET
MCQ
20 May 2026
In the electrolysis of a CuSO$_4$ solution, how many grames of Cu are plated out on the cathode, in the time that is required to liberate 5.6 L of O$_2$(g), measured at 1 atm and 273 K, at the anode?
A.
31.75 g
B.
14.2 g
C.
4.32 g
D.
3.175 g
2021
Q219
AP-EAPCET
MCQ
20 May 2026
If hydrogen electrons dipped in two solutions of pH = 3 and pH = 6 are connected by a salt bridge, the emf of the resulting cell is
A.
0.177 V
B.
0.3 V
C.
0.052 V
D.
0.104 V
2021
Q220
AP-EAPCET
MCQ
20 May 2026
At $291 \mathrm{~K}$, saturated solution of $\mathrm{BaSO}_4$ was found to have a specific conductivity of $3.648 \times 10^{-6} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}$ and that of water being used is $1.25 \times 10^{-6} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}$. If the ionic conductances of $\mathrm{Ba}^{2+}$ and $\mathrm{SO}_4^{2-}$ are 110 and $136.6 \mathrm{ohm}^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ respectively. The solubility of $\mathrm{BaSO}_4$ at $291 \mathrm{~K}$ will be [Atomic masses of $\mathrm{Ba}=137, \mathrm{~S}=32, \mathrm{O}=16]$
A.
$1.435 \times 10^{-3} \mathrm{gL}^{-1}$
B.
$2.266 \times 10^{-3} \mathrm{gL}^{-1}$
C.
$2.843 \times 10^{-3} \mathrm{gL}^{-1}$
D.
$1.768 \times 10^{-3} \mathrm{gL}^{-1}$
2021
Q221
AP-EAPCET
MCQ
20 May 2026
Find the emf of the following cell reaction. Given, $E_{\mathrm{Cr}^{3+} / \mathrm{Cr}^{2+}}^{\Upsilon}=-0.72 \mathrm{~V}$ and $E_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{\Upsilon}= -0.42 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$ is $\mathrm{Cr}\left|\mathrm{Cr}^{3+}(0.1 \mathrm{M})\right| \mid \mathrm{Fe}^{2+} (0.1 \mathrm{M}) \mid \mathrm{Fe}$
A.
0.30 V
B.
0.25 V
C.
1.14 V
D.
1.56 V
2021
Q222
AP-EAPCET
MCQ
20 May 2026
For $\mathrm{C{r_2}O_7^{2 - } + 14{H^ + } + 6{e^ - }\buildrel {Yields} \over \longrightarrow 2C{r^{3 + }} + 7{H_2}O,{E^\Upsilon } = 1.33}$ V at $[C{r_2}O_7^{2 - }] = 4.5$ millimole, $[C{r^{3 + }}] = 1.5$ millimole and $E = 1.067$ V, then calculate the pH of the solution.
A.
2
B.
3
C.
2.5
D.
1.5
2021
Q223
AP-EAPCET
MCQ
20 May 2026
Assertion (A) Sodium acetate on Kolbe’s electrolysis gives ethane.
Reason (R) Methyl free radical is formed at cathode.
A.
Both A and R are true and R is a correct explanation of A.
B.
Both A and R are true but R is not a correct explanation of A.
C.
A is true but R is false.
D.
A is false but R is true.
2021
Q224
AP-EAPCET
MCQ
20 May 2026
When a current of 10 A is passes through molten AlCl$_3$ for 1.608 minutes. The mass of Al deposited will be
[Atomic mass of Al = 27 g]
A.
0.09 g
B.
0.81 g
C.
1.35 g
D.
0.27 g
2021
Q225
AP-EAPCET
MCQ
20 May 2026
The molar conductivities $\left(\lambda_{\mathrm{m}}^{\Upsilon}\right)$ at infinite dilution of $\mathrm{KBr}, \mathrm{HBr}$ and $\mathrm{KNH}_2$ are 120.5, 420.6 and $90.48 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ respectively. Find the value of $\lambda_{\mathrm{m}}^\Upsilon$ for $\mathrm{NH}_3$.
A.
$511.0 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
B.
$390.5 \mathrm{~s} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
C.
$256.2 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
D.
$240.9 \mathrm{~s} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
2021
Q226
BITSAT
MCQ
11 Jun 2026
The molar conductivities of KCl, NaCl and KNO3 are 152, 128 and 111 S cm2 mol$-$1 respectively. What is the molar conductivity of NaNO3?
A.
101 S cm2 mol$-$1
B.
87 S cm2 mol$-$1
C.
$-$ 101 S cm2 mol$-$1
D.
$-$ 391 S cm2 mol$-$1
2021
Q227
BITSAT
MCQ
11 Jun 2026
The approximate time duration in hours to electroplate 30 g of calcium from molten calcium chloride using a current of 5 A is (Atomic mass of Ca = 40)
A.
80
B.
10
C.
16
D.
8
2021
Q228
BITSAT
MCQ
11 Jun 2026
Given, the reduction potential of Na+, Mg2+, Al3+ and Ag+ as $E_{N{a^ + }/Na}^o$ = $-$ 2.17 V; $E_{M{g^{2 + }}/Mg}^o$ = $-$ 2.37 V; $E_{A{g^ + }/Ag}^o$ = $-$ 0.08 V.
The least stable oxide is
A.
Ag2O
B.
Al2O3
C.
MgO
D.
Na2O
2020
Q229
JEE Mains
Numerical
14 Mar 2026
Potassium chlorate is prepared by the
electrolysis of KCl in basic solution
6OH- + Cl- $ \to $ ClO3- + 3H2O + 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 KClO3 using a current of 2 A is_________.
(Given : F = 96,500 C mol–1; molar mass of KCIO3 = 122 g mol–1)
6OH- + Cl- $ \to $ ClO3- + 3H2O + 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 KClO3 using a current of 2 A is_________.
(Given : F = 96,500 C mol–1; molar mass of KCIO3 = 122 g mol–1)
Correct Answer: 11
Explanation:
For synthesis of 1 mole of ClO3- , 6F of charge
is required.
$ \therefore $ To synthesise ${{10} \over {122}}$ moles of KClO3,
Charge required = ${{10} \over {122}} \times 6$ F
${{10} \over {122}} \times 6 = {{2 \times t(hr) \times 3600} \over {96500}} \times {{60} \over {100}}$
$ \Rightarrow $ t(hr) = ${{965 \times 100} \over {122 \times 2 \times 36}}$ = 10.98 hr $ \simeq $ 11 Hr
$ \therefore $ To synthesise ${{10} \over {122}}$ moles of KClO3,
Charge required = ${{10} \over {122}} \times 6$ F
${{10} \over {122}} \times 6 = {{2 \times t(hr) \times 3600} \over {96500}} \times {{60} \over {100}}$
$ \Rightarrow $ t(hr) = ${{965 \times 100} \over {122 \times 2 \times 36}}$ = 10.98 hr $ \simeq $ 11 Hr
2020
Q230
JEE Mains
Numerical
14 Mar 2026
An oxidation-reduction reaction in which
3 electrons are transferred has a $\Delta $Gº of 17.37 kJ mol–1 at
25 oC. The value of Eo
cell (in V) is ______ × 10–2.
(1 F = 96,500 C mol–1)
3 electrons are transferred has a $\Delta $Gº of 17.37 kJ mol–1 at
25 oC. The value of Eo
cell (in V) is ______ × 10–2.
(1 F = 96,500 C mol–1)
Correct Answer: -6
Explanation:
$\Delta $Gº = 17.37 kJ ; n = 3
$\Delta $Gº = -nF$E_{cell}^o$
$ \Rightarrow $ $E_{cell}^o$ = $ - {{17.37 \times 1000} \over {3 \times 96500}}$
= -0.06 = -6.00 $ \times $ 10-2
$\Delta $Gº = -nF$E_{cell}^o$
$ \Rightarrow $ $E_{cell}^o$ = $ - {{17.37 \times 1000} \over {3 \times 96500}}$
= -0.06 = -6.00 $ \times $ 10-2
2020
Q231
JEE Mains
Numerical
14 Mar 2026
An acidic solution of dichromate is electrolyzed
for 8 minutes using 2A current. As per the
following equation
Cr2O72- + 14H+ + 6e– $ \to $ 2Cr3+ + 7H2O
The amount of Cr3+ obtained was 0.104 g. The efficiency of the process(in%) is (Take : F = 96000 C, At. mass of chromium = 52) ______.
Cr2O72- + 14H+ + 6e– $ \to $ 2Cr3+ + 7H2O
The amount of Cr3+ obtained was 0.104 g. The efficiency of the process(in%) is (Take : F = 96000 C, At. mass of chromium = 52) ______.
Correct Answer: 60
Explanation:
Cr2O72-
+ 14H+ + 6e– $ \to $ 2Cr3+ + 7H2O
I = 2 A, t = 8 min
From 1st law of faraday,
wCr+3 = z $ \times $ i $ \times $ t
$ \Rightarrow $ wCr+3 = ${{52} \over {96000 \times 3}} \times 2 \times 8 \times 60$
= ${{52} \over {300}}$
Mass of Cr3+ ions actually obtained = 0.104 gm
% efficiency =
= ${{0.104} \over {{{52} \over {300}}}}$ $ \times $ 100
= 60 %
I = 2 A, t = 8 min
From 1st law of faraday,
wCr+3 = z $ \times $ i $ \times $ t
$ \Rightarrow $ wCr+3 = ${{52} \over {96000 \times 3}} \times 2 \times 8 \times 60$
= ${{52} \over {300}}$
Mass of Cr3+ ions actually obtained = 0.104 gm
% efficiency =
Actual obtained Amt
Theo. obtained Amt
$ \times $ 100
= ${{0.104} \over {{{52} \over {300}}}}$ $ \times $ 100
= 60 %
2020
Q232
JEE Mains
Numerical
14 Mar 2026
The photoelectric current from Na (Work function, w0
= 2.3 eV) is stopped by the output voltage of
the cell
Pt(s) | H2 (g, 1 Bar) | HCl (aq., pH =1) | AgCl(s) | Ag(s).
The pH of aq. HCl required to stop the photoelectric current form K(w0 = 2.25 eV), all other conditions remaining the same, is _______ $ \times $ 10-2 (to the nearest integer).
Given, 2.303${{RT} \over F}$ = 0.06 V;
$E_{AgCl|Ag|C{l^ - }}^0$ = 0.22 V
Pt(s) | H2 (g, 1 Bar) | HCl (aq., pH =1) | AgCl(s) | Ag(s).
The pH of aq. HCl required to stop the photoelectric current form K(w0 = 2.25 eV), all other conditions remaining the same, is _______ $ \times $ 10-2 (to the nearest integer).
Given, 2.303${{RT} \over F}$ = 0.06 V;
$E_{AgCl|Ag|C{l^ - }}^0$ = 0.22 V
Correct Answer: 142
Explanation:
${1 \over 2}$H2 + AgCl $ \to $ H+ + Ag + Cl-
From Nernst Equation,
Ecell = $E_{cell}^0$ - ${{0.06} \over 1}\log \left[ {{H^ + }} \right]\left[ {C{l^ - }} \right]$
= 0.22 – 0.06 log 10–2 = 0.34 V
Work function of Na metal = 2.3 eV
KE of photoelectron = 0.34 eV
Energy of incident radiation = 2.3 + 0.34 = 2.64 eV
For K atom,
Energy of incident radiation for K metal = 2.64 eV
Work function of K metal = 2.25 eV
KE of photoelectrons = 2.64 – 2.25 = 0.39 eV
$ \therefore $ Ecell = 0.39 V
Using Nernst Equation,
0.39 = 0.22 - ${{0.06} \over 1}\log \left[ {{H^ + }} \right]\left[ {C{l^ - }} \right]$
As $\left[ {{H^ + }} \right] = \left[ {C{l^ - }} \right]$
0.39 = 0.22 - ${{0.06} \over 1}\log {\left[ {{H^ + }} \right]^2}$
$ \Rightarrow $ 0.39 = 0.22 - $0.12 \times \log \left[ {{H^ + }} \right]$
$ \Rightarrow $ 0.39 = 0.22 + 0.12 $ \times $ pH
$ \Rightarrow $ pH = 1.42 = 142 $ \times $ 10-2
From Nernst Equation,
Ecell = $E_{cell}^0$ - ${{0.06} \over 1}\log \left[ {{H^ + }} \right]\left[ {C{l^ - }} \right]$
= 0.22 – 0.06 log 10–2 = 0.34 V
Work function of Na metal = 2.3 eV
KE of photoelectron = 0.34 eV
Energy of incident radiation = 2.3 + 0.34 = 2.64 eV
For K atom,
Energy of incident radiation for K metal = 2.64 eV
Work function of K metal = 2.25 eV
KE of photoelectrons = 2.64 – 2.25 = 0.39 eV
$ \therefore $ Ecell = 0.39 V
Using Nernst Equation,
0.39 = 0.22 - ${{0.06} \over 1}\log \left[ {{H^ + }} \right]\left[ {C{l^ - }} \right]$
As $\left[ {{H^ + }} \right] = \left[ {C{l^ - }} \right]$
0.39 = 0.22 - ${{0.06} \over 1}\log {\left[ {{H^ + }} \right]^2}$
$ \Rightarrow $ 0.39 = 0.22 - $0.12 \times \log \left[ {{H^ + }} \right]$
$ \Rightarrow $ 0.39 = 0.22 + 0.12 $ \times $ pH
$ \Rightarrow $ pH = 1.42 = 142 $ \times $ 10-2
2020
Q233
JEE Mains
Numerical
14 Mar 2026
For the disproportionation reaction
2Cu+(aq) ⇌ Cu(s) + Cu2+(aq) at 298 K. ln K
(where K is the equilibrium constant) is
___________ × 10–1.
Given :
($E_{C{u^{2 + }}/C{u^ + }}^0 = 0.16V$
$E_{C{u^ + }/Cu}^0 = 0.52V$
${{RT} \over F} = 0.025$)
2Cu+(aq) ⇌ Cu(s) + Cu2+(aq) at 298 K. ln K
(where K is the equilibrium constant) is
___________ × 10–1.
Given :
($E_{C{u^{2 + }}/C{u^ + }}^0 = 0.16V$
$E_{C{u^ + }/Cu}^0 = 0.52V$
${{RT} \over F} = 0.025$)
Correct Answer: 144
Explanation:
$E_{cell}^0$ = $E_{C{u^ + }/Cu}^0$ - $E_{C{u^{2 + }}/C{u^ + }}^0$
= 0.52 – 0.16
= 0.36 V
At equilibrium, Ecell = 0
$E_{cell}^0$ = ${{RT} \over {nF}}$ln K
$ \Rightarrow $ ln K = ${{E_{cell}^0 \times nF} \over {RT}}$
= ${{0.36 \times 1} \over {0.025}}$ = 14.4 = 144 $ \times $ 10-1
= 0.52 – 0.16
= 0.36 V
At equilibrium, Ecell = 0
$E_{cell}^0$ = ${{RT} \over {nF}}$ln K
$ \Rightarrow $ ln K = ${{E_{cell}^0 \times nF} \over {RT}}$
= ${{0.36 \times 1} \over {0.025}}$ = 14.4 = 144 $ \times $ 10-1
2020
Q234
JEE Mains
Numerical
14 Mar 2026
The Gibbs change (in J) for the given reaction at
[Cu2+] = [Sn2+] = 1 M and 298K is :
Cu(s) + Sn2+(aq.) $ \to $ Cu2+(aq.) + Sn(s);
($E_{S{n^{2 + }}|Sn}^0 = - 0.16\,V$,
$E_{C{u^{2 + }}|Cu}^0 = 0.34\,V$)
Take F = 96500 C mol–1)
[Cu2+] = [Sn2+] = 1 M and 298K is :
Cu(s) + Sn2+(aq.) $ \to $ Cu2+(aq.) + Sn(s);
($E_{S{n^{2 + }}|Sn}^0 = - 0.16\,V$,
$E_{C{u^{2 + }}|Cu}^0 = 0.34\,V$)
Take F = 96500 C mol–1)
Correct Answer: 96500
Explanation:
$\Delta $G = $\Delta $Go + RTln $\left[ {{{S{n^{ + 2}}} \over {C{u^{ + 2}}}}} \right]$
= –2 × 96500 [(–0.16) – 0.34] + RT$\left[ {{1 \over 1}} \right]$
= 96500 J
= –2 × 96500 [(–0.16) – 0.34] + RT$\left[ {{1 \over 1}} \right]$
= 96500 J
2020
Q235
JEE Mains
Numerical
14 Mar 2026
108 g of silver (molar mass 108 g mol–1) is
deposited at cathode from AgNO3(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: 5.66to5.68
Explanation:
Cathode : Ag+(aq) + e- $ \to $ Ag(s)
Moles of Ag deposited = ${{108} \over {108}}$ = 1 mole
Anode : 2H2O $ \to $ O2 + 4H+ + 4e-
Here we have to find volume of O2 evolved.
Equivalance of Ag = Equivalance of O2
$ \Rightarrow $ 1 $ \times $ 1 = nO2 $ \times $ 4
$ \Rightarrow $ nO2 = ${1 \over 4}$ mol
$ \therefore $ Volume of O2 evolved
= ${1 \over 4}$ $ \times $ 22.4
= 5.6 lit
Moles of Ag deposited = ${{108} \over {108}}$ = 1 mole
Anode : 2H2O $ \to $ O2 + 4H+ + 4e-
Here we have to find volume of O2 evolved.
Equivalance of Ag = Equivalance of O2
$ \Rightarrow $ 1 $ \times $ 1 = nO2 $ \times $ 4
$ \Rightarrow $ nO2 = ${1 \over 4}$ mol
$ \therefore $ Volume of O2 evolved
= ${1 \over 4}$ $ \times $ 22.4
= 5.6 lit
2020
Q236
JEE Mains
Numerical
14 Mar 2026
For an electrochemical cell
Sn(s) | Sn2+ (aq,1M)||Pb2+ (aq,1M)|Pb(s)
the ratio ${{\left[ {S{n^{2 + }}} \right]} \over {\left[ {P{b^{2 + }}} \right]}}$ when this cell attains equilibrium is _________.
(Given $E_{S{n^{2 + }}|Sn}^0 = - 0.14V$,
$E_{P{b^{2 + }}|Pb}^0 = - 0.13V$, ${{2.303RT} \over F} = 0.06$)
Sn(s) | Sn2+ (aq,1M)||Pb2+ (aq,1M)|Pb(s)
the ratio ${{\left[ {S{n^{2 + }}} \right]} \over {\left[ {P{b^{2 + }}} \right]}}$ when this cell attains equilibrium is _________.
(Given $E_{S{n^{2 + }}|Sn}^0 = - 0.14V$,
$E_{P{b^{2 + }}|Pb}^0 = - 0.13V$, ${{2.303RT} \over F} = 0.06$)
Correct Answer: 2.13TO2.16
Explanation:
Cell reaction is :
Sn(s) + Pb+2(aq) $ \to $ Sn+2(aq) + Pb(s)
Apply Nernst equation :
Ecell = $E_{cell}^0$ - ${{0.06} \over 2}\log {{\left[ {S{n^{ + 2}}} \right]} \over {\left[ {P{b^{ + 2}}} \right]}}$ ....(1)
$ \Rightarrow $ $E_{cell}^0$ = -0.13 + 0.14 = 0.01 V
At equilibrium : Ecell = 0
Substituting in (1), we get
0 = 0.01 - ${{0.06} \over 2}\log {{\left[ {S{n^{ + 2}}} \right]} \over {\left[ {P{b^{ + 2}}} \right]}}$
$ \Rightarrow $ $\log {{\left[ {S{n^{ + 2}}} \right]} \over {\left[ {P{b^{ + 2}}} \right]}}$ = ${1 \over 3}$
$ \Rightarrow $ ${{\left[ {S{n^{2 + }}} \right]} \over {\left[ {P{b^{2 + }}} \right]}}$ = 2.15
Sn(s) + Pb+2(aq) $ \to $ Sn+2(aq) + Pb(s)
Apply Nernst equation :
Ecell = $E_{cell}^0$ - ${{0.06} \over 2}\log {{\left[ {S{n^{ + 2}}} \right]} \over {\left[ {P{b^{ + 2}}} \right]}}$ ....(1)
$ \Rightarrow $ $E_{cell}^0$ = -0.13 + 0.14 = 0.01 V
At equilibrium : Ecell = 0
Substituting in (1), we get
0 = 0.01 - ${{0.06} \over 2}\log {{\left[ {S{n^{ + 2}}} \right]} \over {\left[ {P{b^{ + 2}}} \right]}}$
$ \Rightarrow $ $\log {{\left[ {S{n^{ + 2}}} \right]} \over {\left[ {P{b^{ + 2}}} \right]}}$ = ${1 \over 3}$
$ \Rightarrow $ ${{\left[ {S{n^{2 + }}} \right]} \over {\left[ {P{b^{2 + }}} \right]}}$ = 2.15
2020
Q237
JEE Mains
Numerical
14 Mar 2026
What would be the electrode potential for the given half cell reaction at pH = 5?
______.
2H2O $ \to $ O2 + 4H$ \oplus $ + 4e– ; $E_{red}^0$ = 1.23 V
(R = 8.314 J mol–1 K–1 ; Temp = 298 k;
oxygen under std. atm. pressure of 1 bar)
2H2O $ \to $ O2 + 4H$ \oplus $ + 4e– ; $E_{red}^0$ = 1.23 V
(R = 8.314 J mol–1 K–1 ; Temp = 298 k;
oxygen under std. atm. pressure of 1 bar)
Correct Answer: 1.52TO1.53
Explanation:
E = E0 - ${{0.0591} \over 4}\log {\left[ {{H^ + }} \right]^4}$
$ \Rightarrow $ E = 1.23 + 0.0591 × pH
$ \Rightarrow $ E = 1.23 + 0.0591 × (5)
$ \Rightarrow $ E = 1.52
$ \Rightarrow $ E = 1.23 + 0.0591 × pH
$ \Rightarrow $ E = 1.23 + 0.0591 × (5)
$ \Rightarrow $ E = 1.52
2020
Q238
JEE Mains
MCQ
14 Mar 2026
For the given cell :
Cu(s) | Cu2+(C1M) || Cu2+(C2M) | Cu(s)
change in Gibbs energy ($\Delta $G) is negative, if :
Cu(s) | Cu2+(C1M) || Cu2+(C2M) | Cu(s)
change in Gibbs energy ($\Delta $G) is negative, if :
A.
C2 = $\sqrt 2 $C1
B.
C2 = ${{{C_1}} \over {\sqrt 2 }}$
C.
C1 = 2C2
D.
C1 = C2
2020
Q239
JEE Mains
MCQ
14 Mar 2026
The variation of molar conductivity with concentration of an electrolyte (X) in aqueous solution
is shown in the given figure.
The electrolyte X is :
The electrolyte X is :
A.
HCl
B.
CH3COOH
C.
NaCl
D.
KNO3
2020
Q240
JEE Mains
MCQ
14 Mar 2026
250 mL of a waste solution obtained from the
workshop of a goldsmith contains 0.1 M AgNO3
and 0.1 M AuCl. The solution was electrolyzed
at 2V by passing a current of 1A for 15
minutes. The metal/metals electrodeposited will
be
[ $E_{A{g^ + }/Ag}^0$ = 0.80 V, $E_{A{u^ + }/Au}^0$ = 1.69 V ]
[ $E_{A{g^ + }/Ag}^0$ = 0.80 V, $E_{A{u^ + }/Au}^0$ = 1.69 V ]
A.
Silver and gold in equal mass proportion
B.
Silver and gold in proportion to their atomic
weights
C.
Only gold
D.
Only silver
2020
Q241
JEE Mains
MCQ
14 Mar 2026
$E_{C{u^{2 + }}|Cu}^0$ = +0.34 V
$E_{Z{n^{2 + }}|Zn}^0$ = -0.76 V
Identify the incorrect statement from the option below for the above cell :
A.
If Eext < 1.1 V, Zn dissolves at anode and Cu
deposits at cathode
B.
If Eext = 1.1 V, no flow of e– or current
occurs
C.
If Eext > 1.1 V, e– flows from Cu to Zn
D.
If Eext > 1.1 V, Zn dissolves at Zn electrode
and Cu deposits at Cu electrode
2020
Q242
JEE Mains
MCQ
14 Mar 2026
Let CNaCl
and CBaSO4 be the conductances (in S) measured for saturated aqueous solutions of NaCl
and BaSO4, respectively, at a temperature T.
Which of the following is false?
A.
Ionic mobilities of ions from both salts increase with T.
B.
CNaCl(T2) > CNaCl(T1) for T2 > T1
C.
CBaSO4(T2) > CBaSO4(T1) for T2 > T1
D.
CNaCl >> CBaSO4 at a given T
2020
Q243
JEE Mains
MCQ
14 Mar 2026
The equation that is incorrect is :
A.
${\left( {\Lambda _m^0} \right)_{KCl}} - {\left( {\Lambda _m^0} \right)_{NaCl}} = {\left( {\Lambda _m^0} \right)_{KBr}} - {\left( {\Lambda _m^0} \right)_{NaBr}}$
B.
${\left( {\Lambda _m^0} \right)_{NaBr}} - {\left( {\Lambda _m^0} \right)_{NaI}} = {\left( {\Lambda _m^0} \right)_{KBr}} - {\left( {\Lambda _m^0} \right)_{NaBr}}$
C.
${\left( {\Lambda _m^0} \right)_{NaBr}} - {\left( {\Lambda _m^0} \right)_{NaCl}} = {\left( {\Lambda _m^0} \right)_{KBr}} - {\left( {\Lambda _m^0} \right)_{KCl}}$
D.
${\left( {\Lambda _m^0} \right)_{{H_2}O}} = {\left( {\Lambda _m^0} \right)_{HCl}} + {\left( {\Lambda _m^0} \right)_{NaOH}} - {\left( {\Lambda _m^0} \right)_{NaCl}}$
2020
Q244
JEE Mains
MCQ
14 Mar 2026
Given that the standard potentials (Eo) of Cu2+/Cu and Cu+/Cu are 0.34 V and 0.522 V respectively, the Eo of Cu2+/Cu+ :
A.
- 0.182 V
B.
- 0.158 V
C.
0.182 V
D.
+0.158 V
2020
Q245
JEE Advanced
Numerical
14 Mar 2026
Consider a 70% efficient hydrogen-oxygen fuel cell working under standard conditions at 1 bar and 298 K. Its cell reaction is
${H_2}(g) + {1 \over 2}{O_2}(g)\buildrel {} \over \longrightarrow {H_2}O(l)$
The work derived from the cell on the consumption of 1.0 $ \times $ 10$-$3 mole of H2(g) is used to compress 1.00 mole of a monoatomic ideal gas in a thermally insulated 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 :
${O_2}(g) + 4{H^ + }(aq) + 4{e^ - }\buildrel {} \over \longrightarrow 2{H_2}O(l),$
${E^o} = 1.23V$
$2{H^ + }(aq) + 2{e^ - }\buildrel {} \over \longrightarrow {H_2}(g),$
${E^o} = 0.00\,V$
Use, $F = 96500\,C\,mo{l^{ - 1}}$, $R = 8.314\,J\,mo{l^{ - 1}}\,{K^{ - 1}}$.
${H_2}(g) + {1 \over 2}{O_2}(g)\buildrel {} \over \longrightarrow {H_2}O(l)$
The work derived from the cell on the consumption of 1.0 $ \times $ 10$-$3 mole of H2(g) is used to compress 1.00 mole of a monoatomic ideal gas in a thermally insulated 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 :
${O_2}(g) + 4{H^ + }(aq) + 4{e^ - }\buildrel {} \over \longrightarrow 2{H_2}O(l),$
${E^o} = 1.23V$
$2{H^ + }(aq) + 2{e^ - }\buildrel {} \over \longrightarrow {H_2}(g),$
${E^o} = 0.00\,V$
Use, $F = 96500\,C\,mo{l^{ - 1}}$, $R = 8.314\,J\,mo{l^{ - 1}}\,{K^{ - 1}}$.
Correct Answer: 13.32
Explanation:
Vessel is insulated, thus q = 0
For the given reaction : $\eqalign{ & {H_2}(g) + 1/2{O_2}(g)\buildrel {} \over \longrightarrow {H_2}O(l); \cr & {E^o} = 1.23 - 0.00 = 1.23V \cr} $
$\Delta $Go = $-$nFEo
= $-$2 $ \times $ 96500 $ \times $ 1.23 J/mol
Therefore, work derived from this fuel cell using 70% efficiency and on consumption of 1.0 $ \times $ 10$-$3 mol of H2(g)
= 2 $ \times $ 96500 $ \times $ 1.23 $ \times $ 0.7 $ \times $ 1 $ \times $ 10$-$3
= 166.17 J
This work done = change in internal energy (for monoatomic gas,
${C_{V'm}} = 3R/2$),
$166.17 = n{C_V}{,_m}\Delta T$
$ \Rightarrow \Delta T = {{166.17 \times 2} \over {1 \times 3 \times 8.314}}$
$ \Rightarrow 13.32\,K$
For the given reaction : $\eqalign{ & {H_2}(g) + 1/2{O_2}(g)\buildrel {} \over \longrightarrow {H_2}O(l); \cr & {E^o} = 1.23 - 0.00 = 1.23V \cr} $
$\Delta $Go = $-$nFEo
= $-$2 $ \times $ 96500 $ \times $ 1.23 J/mol
Therefore, work derived from this fuel cell using 70% efficiency and on consumption of 1.0 $ \times $ 10$-$3 mol of H2(g)
= 2 $ \times $ 96500 $ \times $ 1.23 $ \times $ 0.7 $ \times $ 1 $ \times $ 10$-$3
= 166.17 J
This work done = change in internal energy (for monoatomic gas,
${C_{V'm}} = 3R/2$),
$166.17 = n{C_V}{,_m}\Delta T$
$ \Rightarrow \Delta T = {{166.17 \times 2} \over {1 \times 3 \times 8.314}}$
$ \Rightarrow 13.32\,K$
2020
Q246
TS-EAMCET
MCQ
20 May 2026
A solution of $\mathrm{Fe}^{2+}$ is titrated potentiometrically using $\mathrm{Ce}^{4+}$ solution. When $80 \% \mathrm{Fe}^{2+}$ is titrated, the EMF of the system in $V$ is
(Given, $E^{\circ} \mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}=0.77 \mathrm{~V}$ and $\left.\mathrm{Fe}^{2+}+\mathrm{Ce}^{4+} \longrightarrow \mathrm{Fe}^{3+}+\mathrm{Ce}^{3+}\right) (\log 2=0.3, \log 3=0.5, \log 4=0.6)$
A.
0.806
B.
0.532
C.
0.734
D.
0.756
2020
Q247
TS-EAMCET
MCQ
20 May 2026
$\mathrm{Mg}^{2+}$ displaces hydrogen from acids but copper does not. A galvanic cell prepared by combining $\mathrm{Cu} / \mathrm{Cu}^{2+}$ and $\mathrm{Mg} / \mathrm{Mg}^{2+}$ has an EMF of 2.71 V at 298 K . If the potential of copper electrode is 0.34 V , what is the reduction potential of Mg electrode?
A.
+3.05 V
B.
-2.37 V
C.
+2.37 V
D.
2 V
2020
Q248
TS-EAMCET
MCQ
20 May 2026
The maximum work that can be obtained from the following cells is
$ X\left|X^{2+}(a q) \| Y^{+}(a q)\right| Y $
Given, $E_{X^{2+} / X}^{\circ}=-1.7 \mathrm{~V}, E_{Y^{2+} / Y}^{\circ}=0.8 \mathrm{~V}$
A.
$579 \mathrm{~kJ} / \mathrm{mol}$
B.
$482.5 \mathrm{~kJ} / \mathrm{mol}$
C.
$289.5 \mathrm{~kJ} / \mathrm{mol}$
D.
$301.8 \mathrm{~kJ} / \mathrm{mol}$
2020
Q249
TS-EAMCET
MCQ
20 May 2026
The standard electrode potentials of $\mathrm{Ag}^{+} / \mathrm{Ag}$ is +0.80 V and $\mathrm{Cu}^{+} / \mathrm{Cu}$ is +0.34 V . If these electrodes are connected through a salt-bridge, which of the following statements is correct?
A.
Silver electrode acts as anode and $E_{\text {cell }}^{\circ}$ is -0.34 V .
B.
Copper electrode acts as anode and $E_{\text {cell }}^{\circ}$ is +0.46 V .
C.
Silver electrode acts as a cathode and $E_{\text {cell }}^{\circ}$ is -0.34 V .
D.
Copper electrode acts as cathode and $E_{\text {cell }}^{\circ}$ is +0.46 V .
2020
Q250
BITSAT
MCQ
11 Jun 2026
A solution of copper sulphate is electrolyzed between copper electrodes by a current of 10.0A passing for one hour. Which of the following statements is correct regarding the changes occur at the electrodes and in the solution?
A.
11.84 g of copper will deposit on the cathode
B.
11.84 g of copper will deposit on the anode
C.
11.84 g of copper will deposit on the anode as well as on the cathode
D.
copper will not deposit on any of the electrode