JEE Main 2026 Jan 22 Shift 2, Chemistry Q21: Electrochemical Cell And Nernst Equation

JEE Main2026Jan 22, Shift 2Chemistry

Consider the following electrochemical cell :

$P t ∣ O_{2} (g) (1 bar) ∣ H C l (a q) ∣∣ M^{2 +} (a q, 1.0 M) ∣ M (s)$

The pH above which, oxygen gas would start to evolve at anode is ______ (nearest integer).

Given : $E_{M^{2 +} / M}^{0} = 0.994 V$ , $E_{O_{2} / H_{2} O}^{0} = 1.23 V$ , and $\frac{R T}{F} (2.303) = 0.059 V$ at the given condition.

Your answer

Solution

For the cell to be spontaneous (oxygen to evolve at anode), $E_{ce l l} \geq 0$ .

Reaction at anode (oxidation): $H_{2} O \to 2 H^{+} + \frac{1}{2} O_{2} + 2 e^{-}$ , so $E_{an o d e}^{o x i} = - E_{O_{2} / H_{2} O}^{\circ} = - 1.23$ V (at standard).

Cathode: $M^{2 +} + 2 e^{-} \to M$ , $E_{c a t h o d e} = 0.994$ V (at 1 M $M^{2 +}$ ).

Using the Nernst correction for the anode at non-standard $[H^{+}]$ :

$E_{an o d e}^{o x i} = - 1.23 - \frac{0.059}{2} lo g \frac{[ H ^{+} ] ^{2} P _{O_{2}}^{1/2}}{1} = - 1.23 + 0.059 p H$ (since $P_{O_{2}} = 1$ ).

At the threshold $E_{ce l l} = E_{c a t h o d e} + E_{an o d e}^{o x i} = 0$ :

$0.994 + (- 1.23 + 0.059 p H) = 0 \Rightarrow 0.059 p H = 0.236 \Rightarrow p H \approx 4$ .

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JEE Main 2026 Jan 22 Shift 2, Chemistry Q21: Electrochemical Cell And Nernst Equation

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