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Chemistry · Electrochemistry

JEE Main 2026 — 6 April, Evening Shift — Question 51

For a general redox reaction

AnodeRedX1→OxX1Xn1++nX1eX−CathodeOxX2+nX2eX−→RedX2Xn2−\begin{array}{l l} \text{Anode} & \ce{Red1 -> Ox1^{n1+} + n1e^-} \\ \text{Cathode} & \ce{Ox2 + n2e^- -> Red2^{n2-}} \end{array}

Which of the following statement is incorrect?

  1. Option A:

    The overall reaction can be written as n2Red⁡1+n1Ox2⇌n2Ox1n1++n1Red⁡2n2−\mathrm{n}_{2} \operatorname{Red}_{1}+\mathrm{n}_{1} \mathrm{Ox}_{2} \rightleftharpoons \mathrm{n}_{2} \mathrm{Ox}_{1}^{\mathrm{n}_{1}^{+}}+\mathrm{n}_{1} \operatorname{Red}_{2}{ }^{\mathrm{n}_{2}-}

  2. Option B:

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

  3. Option C:

    Here nn is the number of electrons transferred in redox reaction.

  4. Option 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

Anode: [Red1→Ox1n1++n1e−]×n2\left[\mathrm{Red}_1\rightarrow\mathrm{Ox}_1^{n_1+}+n_1e^-\right]\times n_2

Cathode: [Ox2+n2e−→Red2n2−]×n1\left[\mathrm{Ox}_2+n_2e^-\rightarrow\mathrm{Red}_2^{n_2-}\right]\times n_1

Overall reaction: n2Red1+n1Ox2→n2Ox1n1++n1Red2n2−n_2\mathrm{Red}_1+n_1\mathrm{Ox}_2 \rightarrow n_2\mathrm{Ox}_1^{n_1+}+n_1\mathrm{Red}_2^{n_2-}

The electrons do not appear in overall reaction.

E=E∘−2.303RTnFlog⁡QE=E^\circ-2.303\frac{RT}{nF}\log Q

E−E∘RT/F=−2.303nlog⁡Q\frac{E-E^\circ}{RT/F}=-\frac{2.303}{n}\log Q

Slope =−2.303n=-\frac{2.303}{n}

Straight line passing through origin.

Electrical work == Charge ×\times Potential difference.

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
6 April, Evening Shift
Subject
Chemistry
Chapter
Electrochemistry
Topic
Nernst Equation and Electrochemical Series
For a general redox reaction begin array l l Anode & ce Red1 - Ox1… | JEE Main 2026 PYQ with Solution · DhiX AI