Q6 / 75 · JEE Main 2026 (08 April Shift 2) · Chemistry
Given at $298$ K: $E^{\ominus}_{Fe^{2+}/Fe} = X$ Volt; $E^{\ominus}_{Fe^{3+}/Fe} = Y$ Volt. The $E^{\ominus}_{Fe^{3+}/Fe^{2+}}$ in Volt at $298$ K is given by:
- A$2X-3Y$
- Correct answerB$3Y-2X$
- C$3Y+2X$
- D$Y+X$
Solution
The half-reactions and their standard Gibbs free energy changes are given by:
$Fe^{2+} + 2e^{-} \rightarrow Fe$
$\Delta G^{\ominus}_{1} = -2FX$
$Fe^{3+} + 3e^{-} \rightarrow Fe$
$\Delta G^{\ominus}_{2} = -3FY$
The required half-reaction is:
$Fe^{3+} + e^{-} \rightarrow Fe^{2+}$
$\Delta G^{\ominus}_{3} = -1FE^{\ominus}_{Fe^{3+}/Fe^{2+}}$
This reaction is obtained by subtracting the first reaction from the second reaction:
$\Delta G^{\ominus}_{3} = \Delta G^{\ominus}_{2} - \Delta G^{\ominus}_{1}$
$-FE^{\ominus}_{Fe^{3+}/Fe^{2+}} = -3FY - (-2FX)$
$-FE^{\ominus}_{Fe^{3+}/Fe^{2+}} = -3FY + 2FX$
$E^{\ominus}_{Fe^{3+}/Fe^{2+}} = 3Y - 2X$
Answer: $3Y-2X$
$Fe^{2+} + 2e^{-} \rightarrow Fe$
$\Delta G^{\ominus}_{1} = -2FX$
$Fe^{3+} + 3e^{-} \rightarrow Fe$
$\Delta G^{\ominus}_{2} = -3FY$
The required half-reaction is:
$Fe^{3+} + e^{-} \rightarrow Fe^{2+}$
$\Delta G^{\ominus}_{3} = -1FE^{\ominus}_{Fe^{3+}/Fe^{2+}}$
This reaction is obtained by subtracting the first reaction from the second reaction:
$\Delta G^{\ominus}_{3} = \Delta G^{\ominus}_{2} - \Delta G^{\ominus}_{1}$
$-FE^{\ominus}_{Fe^{3+}/Fe^{2+}} = -3FY - (-2FX)$
$-FE^{\ominus}_{Fe^{3+}/Fe^{2+}} = -3FY + 2FX$
$E^{\ominus}_{Fe^{3+}/Fe^{2+}} = 3Y - 2X$
Answer: $3Y-2X$
Hence, the correct answer is option: 'B'.