Given at 298 K : E Fe 2+ / Fe ominus = X Volt E Fe 3+ / Fe ominus = Y Volt The E Fe 3+ / Fe 2+ ominus in Volt at 298 K is given by :

Chemistry · Electrochemistry

JEE Main 2026 — 8 April, Evening Shift — Question 48

Given at $298 \mathrm{~K}: \mathrm{E}_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{\ominus}=\mathrm{X}$ Volt

\mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}}^{\ominus}=\mathrm{Y} \text { Volt }

The $\mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{\ominus}$ in Volt at 298 K is given by :

Option A:
$2 X-3 Y$
Option B:
$3 Y-2 X$
Correct
Option C:
$3 Y+2 X$
Option D:
$Y+X$

Answer: B

Step-by-step solution

(i) $\mathrm{Fe}^{+2}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Fe}(\mathrm{s}) \ldots \ldots . . \Delta \mathrm{G}_{1}{ }^{\circ}$ (ii) $\mathrm{Fe}^{+3}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe}(\mathrm{s}) \ldots \ldots . . \Delta \mathrm{G}_{2}^{\circ}$ (iii) $\mathrm{Fe}^{+3}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{+2}(\mathrm{aq}) \ldots \ldots \Delta \mathrm{G}_{3}{ }^{\circ}$ iii $=$ ii -i $-1 \times \mathrm{F} \times \mathrm{E}^{\circ}{ }_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}=-3 \times \mathrm{F} \times \mathrm{y}-(-2 \times \mathrm{F} \times \mathrm{x})$ $\mathrm{E}^{\circ}{ }_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}=3 \mathrm{y}-2 \mathrm{x}$ Option is correct

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 8 April, Evening Shift
Subject: Chemistry
Chapter: Electrochemistry
Topic: Nernst Equation and Electrochemical Series