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 236 questions; sign in to attempt the remaining 231 in the exam simulator.

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.$Y+X$
   Show answer & solution

   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.$+2,0,0,+4$
   Show answer & solution

   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.$10^{-14}$
   Show answer & solution

   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 $400S{cm}^{2}{mol}^{-1}$
   Show answer & solution

   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.$D$
   Show answer & solution

   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.