JEE MAIN - Chemistry (2025 - 24th January Morning Shift - No. 18)
For the given cell
$$\mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Ag}_{(\mathrm{aq})}^{+} \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Ag}_{(\mathrm{s})}$$
The standard cell potential of the above reaction is Given:
$$\begin{array}{lr} \mathrm{Ag}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{Ag} & \mathrm{E}^\theta=\mathrm{xV} \\ \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Fe} & \mathrm{E}^\theta=\mathrm{yV} \\ \mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} & \mathrm{E}^\theta=\mathrm{zV} \end{array}$$
Explanation
$\mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s})$
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$$\begin{aligned} & \Delta \mathrm{G}_3^0=\Delta \mathrm{G}_1^0+\Delta \mathrm{G}_2^0 \\ & -3 \mathrm{~F}(-\mathrm{z})=-2 \mathrm{~F}(-\mathrm{y})+\Delta \mathrm{G}_2^0 \\ & \Delta \mathrm{G}_2^0=3 \mathrm{Fz}-2 \mathrm{Fy} \end{aligned}$$
Also $\Delta \mathrm{G}_2^0=-\mathrm{nFE}_{\mathrm{Fe}^{+2} / \mathrm{Fe}^{+3}}^0$
$$\begin{aligned} & 3 \mathrm{Fz}-2 \mathrm{Fy}=-1 \mathrm{~F}\left(\mathrm{E}_{\mathrm{Fe}^{+2} / \mathrm{Fe}^{+3}}^0\right) \\ & \mathrm{E}_{\mathrm{Fe}^{+2} / \mathrm{Fe}^{+3}}^0=2 \mathrm{y}-3 \mathrm{z} \end{aligned}$$
$\mathrm{E}_{\text {Cell }}^0$ for reaction will be
$$\begin{aligned} & \mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}^0}^0+\mathrm{E}_{\mathrm{Fe}^{+2} / \mathrm{Fe}^{+3}}^0 \\ & =\mathrm{x}+2 \mathrm{y}-3 \mathrm{z} \end{aligned}$$
Option (2)
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