JEE Main 2025 — Electrochemistry Question with Solution
JEE Main 2025 (24 Jan Shift 1)
Question
For the given cell
The standard cell potential of the above reaction is Given:
$\begin{array}{lr}
\mathrm{Ag}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{Ag} & \mathrm{E}^\theta=\mathrm{xV} \\
\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Fe} & \mathrm{E}^\theta=\mathrm{yV} \\
\mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} & \mathrm{E}^\theta=\mathrm{zV}
\end{array}$
Choose an option
Show full solutionCorrect option: C
Correct answer
C
Step-by-step explanation
$\begin{aligned} & \Delta \mathrm{G}_3^0=\Delta \mathrm{G}_1^0+\Delta \mathrm{G}_2^0 \\ & -3 \mathrm{~F}(-\mathrm{z})=-2 \mathrm{~F}(-\mathrm{y})+\Delta \mathrm{G}_2{ }^0 \\ & \Delta \mathrm{G}_2^0=3 \mathrm{Fz}-2 \mathrm{Fy} \end{aligned}$ Also $\begin{aligned} & 3 \mathrm{Fz}-2 \mathrm{Fy}=-1 \mathrm{~F}\left(\mathrm{E}_{\mathrm{Fe}^{t^2} / \mathrm{Fe}^{43}}^0\right) \\ & \mathrm{E}_{\mathrm{Fe}^{2+2} / \mathrm{Fe}^{t^3}}^0=2 \mathrm{y}-3 \mathrm{z} \end{aligned}$ for reaction will be $\begin{aligned} & \mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^0+\mathrm{E}_{\mathrm{Fe}^{t^2 / \mathrm{Fe}}}^{03} \\ & =\mathrm{x}+2 \mathrm{y}-3 \mathrm{z} \end{aligned}$
Practice this on the real CBT interface
Solve this JEE Main question (and the rest of the Electrochemistry chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.
Solve interactively →About this question
This is a previous-year question from JEE Main 2025, covering the Electrochemistry chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.