Chemistry · 2026
JEE Main · 8 April 2026, Shift 2 · Q56
Given at 298 K: E_Fe^2+ / Fe^ = X Volt; E_Fe^3+ / Fe^ = Y Volt The E_Fe^3+ / Fe^2+^ in Volt at 298 K is given by:
Given at $\displaystyle 298$ K :
$$\begin{aligned}
& \mathrm{E}_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{\ominus}=\mathrm{X} \text { Volt } \\
& \mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}}^{\ominus}=\mathrm{Y} \text { Volt }
\end{aligned}
$$
The $\displaystyle \mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{\ominus}$ in Volt at $\displaystyle 298$ K is given by :
Official answer
From NTA’s final answer key for this paper.
(2)
$\displaystyle 3 \mathrm{Y}-2 \mathrm{X}$
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JEE Main 2026 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.