JEE Main 2026 · 6 Apr · Shift 2 · Chemistry Q57

← JEE Main 2026 · 6 Apr · Shift 2

ChemistrySingle correct+4 / -1Question 57 of 75

For a general redox reaction

Anode: $\mathrm{Red_1}\to\mathrm{Ox_1}^{n_1^+}+n_1e^-$

Cathode: $\mathrm{Ox_2}+n_2e^-\to\mathrm{Red_2}^{n_2^-}$

Which of the following statement is incorrect ?

The overall reaction can be written as

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^-}

The electrons do not appear in the overall reaction because electrons produced at the anode are consumed at the cathode.

The graph of

\dfrac{E-E^\circ}{RT/F}

against

\log_{10}Q

shown. Here n is the number of electrons transferred in redox reaction.

Option C: (E − E°)/(RT/F) against log₁₀Q from −4 to 4: straight lines for n = 1, 2, 3 all pass through (0, 0) with negative slopes, steepest for n = 1, labelled 'Slope ∝ 1/n'.

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.

Answer and worked solution

Official answer: D

Each statement.

A: Balancing electrons gives $n_2\mathrm{Red_1}+n_1\mathrm{Ox_2}\rightleftharpoons n_2\mathrm{Ox_1}+n_1\mathrm{Red_2}$ . Correct.
B: The electrons cancel in the overall reaction. Correct.
C: By the Nernst equation, $\dfrac{E-E^\circ}{RT/F}=-\dfrac{\ln Q}n$ , a straight line through the origin with slope proportional to $\tfrac1n$ . Correct in form.
D: Electrical work is charge times potential difference ( $W=nFE$ ), not their ratio. Incorrect.

(D)

Topics:Nernst equation Gibbs energy and EMF

← Previous question Next question →

JEE Main 2026 · 6 Apr · Shift 2, Chemistry question 57