JEE Main 2026 Apr 6 Shift 2, Chemistry Q7: Balancing Of Redox Reactions

JEE Main2026Apr 6, Shift 2Chemistry

For a general redox reaction:

Anode: $Red_{1} \to Ox_{1}^{n_{1} +} + n_{1} e^{-}$

Cathode: $Ox_{2} + n_{2} e^{-} \to Red_{2}^{n_{2} -}$

Which of the following statement is incorrect?

A
The overall reaction can be written as $n_{2} Red_{1} + n_{1} Ox_{2} ⇌ n_{2} Ox_{1}^{n_{1} +} + n_{1} Red_{2}^{n_{2} -}$
B
The electrons do not appear in the overall reaction because electrons produced at the anode are consumed at the cathode.
C
Here $n$ is the number of electrons transferred in redox reaction.
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.

Solution

Options (1)-(3) are correct standard statements. Multiplying the anode half-reaction by $n_{2}$ and cathode by $n_{1}$ balances electrons, which then cancel in the sum. The Nernst equation $E = E^{\circ} - \frac{2.303 R T}{n F} lo g Q$ rearranges to a straight line with slope $\propto 1/ n$ .

Statement (4) is incorrect. Reversible electrical work equals the product (not ratio) of charge and potential difference: $w_{e l ec} = - n F E$ .

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JEE Main 2026 Apr 6 Shift 2, Chemistry Q7: Balancing Of Redox Reactions

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