Electrochemistry JEE Main previous year questions with solutions

5 solved JEE Main questions on Electrochemistry, free to read — no sign-in needed. The full chapter has 241 questions; sign in to attempt the remaining 236 in the exam simulator.

Answers checked against the official answer keys.

1. Q1JEE Main 2026 (08 Apr, Shift 2)Cells and Electrode Potential, Nernst Equation
   Given at $298$ K: $E^{\ominus}_{Fe^{2+}/Fe} = X$ Volt; $E^{\ominus}_{Fe^{3+}/Fe} = Y$ Volt. The $E^{\ominus}_{Fe^{3+}/Fe^{2+}}$ in Volt at $298$ K is given by:
   1. A.$2X-3Y$
   2. B.$3Y-2X$
   3. C.$3Y+2X$
   4. D.$Y+X$
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   Answer: (B)

   The half-reactions and their standard Gibbs free energy changes are given by: $Fe^{2+} + 2e^{-} \rightarrow Fe$ $\Delta G^{\ominus}_{1} = -2FX$ $Fe^{3+} + 3e^{-} \rightarrow Fe$ $\Delta G^{\ominus}_{2} = -3FY$ The required half-reaction is: $Fe^{3+} + e^{-} \rightarrow Fe^{2+}$ $\Delta G^{\ominus}_{3} = -1FE^{\ominus}_{Fe^{3+}/Fe^{2+}}$ This reaction is obtained by subtracting the first reaction from the second reaction: $\Delta G^{\ominus}_{3} = \Delta G^{\ominus}_{2} - \Delta G^{\ominus}_{1}$ $-FE^{\ominus}_{Fe^{3+}/Fe^{2+}} = -3FY - (-2FX)$ $-FE^{\ominus}_{Fe^{3+}/Fe^{2+}} = -3FY + 2FX$ $E^{\ominus}_{Fe^{3+}/Fe^{2+}} = 3Y - 2X$ Answer: $3Y-2X$
2. Q2JEE Main 2025 (04 Apr, Shift 1)Commercial Cells and Batteries
   On charging the lead storage battery, the oxidation state of lead changes from $x_1$ to $y_1$ at the anode and from $x_2$ to $y_2$ at the cathode. The values of $x_1, y_1, x_2, y_2$ are respectively :
   1. A.$+4,+2,0,+2$
   2. B.$+2,0,+2,+4$
   3. C.$0,+2,+4,+2$
   4. D.$+2,0,0,+4$
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   Answer: (B)

   For charging of lead storage battery cell reaction is $2 \mathrm{PbSO}_4(\mathrm{~s})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{Pb}(\mathrm{s})+\mathrm{PbO}_2(\mathrm{~s})+2 \mathrm{H}_2 \mathrm{SO}_4(\mathrm{aq})$ At anode $\mathrm{PbSO}_4$ reduced back to Pb and at cathode $\mathrm{PbSO}_4$ oxidised back to $\mathrm{PbO}_2$. $\begin{aligned} & \because \quad \mathrm{x}_1=+2, \mathrm{y}_1=0 \\ & \mathrm{x}_2=+2, \mathrm{y}_2=4 \end{aligned}$
3. Q3JEE Main 2024 (04 Apr, Shift 1)Concentration cell
   What pressure (bar) of $\mathrm{H}_2$ would be required to make emf of hydrogen electrode zero in pure water at $25^{\circ} \mathrm{C}$ ?
   1. A.$10^{-7}$
   2. B.0.5
   3. C.1
   4. D.$10^{-14}$
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   Answer: (D)

   $\begin{aligned} & 2 \mathrm{e}^{-}+2 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{H}_2(\mathrm{~g}) \\ & \mathrm{E}=\mathrm{E}^{\circ}-\frac{0.059}{\mathrm{n}} \log \frac{\mathrm{P}_{\mathrm{H}_2}}{\left[\mathrm{H}^{+}\right]^2} \\ & 0=0-\frac{0.059}{2} \log \frac{\mathrm{P}_{\mathrm{H}_2}}{\left(10^{-7}\right)^2} \\ & \log \frac{\mathrm{P}_{\mathrm{H}_2}}{\left(10^{-7}\right)^2}=0 \\ & \frac{\mathrm{P}_{\mathrm{H}_2}}{10^{-14}}=1 \\ & \mathrm{P}_{\mathrm{H}_2}=10^{-14} \text { bar }\end{aligned}$
4. Q4JEE Main 2023 (10 Apr, Shift 2)Conductance and Conductivity
   The specific conductance of $0.0025M$ acetic acid is $5\times {10}^{-5}S{cm}^{-1}$ at a certain temperature. The dissociation constant of acetic acid is ___________ $\times {10}^{-7}$.(Nearest integer) Consider limiting molar conductivity of ${CH}_{3}COOH$ as $400S{cm}^{2}{mol}^{-1}$
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   Answer: 66

   The relation between Molar conductance ${\lambda }_{m}$ and specific conductance $\kappa$ is ${\lambda }_{m}=\dfrac{k\times 1000}{M}$ M = molarity ${\lambda }_{m}=\dfrac{\left(5\times {10}^{-5}\right)\times \left({10}^{3}\right)}{\left(2.5\times {10}^{-3}\right)}=20S{cm}^{2}{mol}^{-1}$ Now, the degree of dissociation, $\alpha =\dfrac{{\lambda }_{m}}{{\lambda }^{\infty }}$ $\alpha =\dfrac{20}{400}=\dfrac{1}{20}$ The acid dissociation constant, ${K}_{a}=\dfrac{{C\alpha }^{2}}{(1-\alpha )}=\dfrac{\left(2.5\times {10}^{-3}\right)\left(\dfrac{1}{20}\times \dfrac{1}{20}\right)}{\left(\dfrac{19}{20}\right)}$ $=65.789\times {10}^{–7}$ $\approx 66\times {10}^{–7}$
5. Q5JEE Main 2022 (26 Jun, Shift 1)Types of half cells
   The ${\left(\dfrac{\partial E}{\partial T}\right)}_{P}$ of different types of half cells are as follows : $\begin{matrix}A & B & C & D \\ 1\times {10}^{-4} & 2\times {10}^{-4} & 0.1\times {10}^{-4} & 0.2\times {10}^{-4}\end{matrix}$ (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.$D$
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   Answer: (C)

   Metal Which have lower value of ${\left(\dfrac{dE}{dT}\right)}_{p}$ is best used for standard or reference half Electrode because if can be used for wider range of temperature Without much derivation from standard Value.