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If $\mathrm{E}^{\circ}\left(\mathrm{Zn}^{2+}, \mathrm{Zn}\right)=-0.763 \mathrm{~V}$ and $\mathrm{E}^{\circ}\left(\mathrm{Fe}^{2+}, \mathrm{Fe}\right)=-0.44 \mathrm{~V}$, then the emf of the cell $\mathrm{Zn}\left|\mathrm{Zn}^{2+}(\mathrm{a}=0.00 \mathrm{l}) \| \mathrm{Fe}^{2+}(\mathrm{a}=0.005)\right| \mathrm{Fe}$ is
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Verified Answer
The correct answer is:
greater than 0.323 V
The cell reaction is
$\mathrm{Zn}+\mathrm{Fe}^{2+} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{Fe}$
From Nernst equation
$\begin{aligned} & \mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\circ}-\frac{0.0591}{\mathrm{n}} \log \frac{\mathrm{a}_{\mathrm{Zn}^{2+}}}{\mathrm{a}_{\mathrm{Fe}^{2+}}} \\ & =(0.763-0.44)-\frac{0.0591}{1} \log \frac{0.001}{0.005} \\ & =0.364 \mathrm{~V} \\ & \end{aligned}$
$\mathrm{Zn}+\mathrm{Fe}^{2+} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{Fe}$
From Nernst equation
$\begin{aligned} & \mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\circ}-\frac{0.0591}{\mathrm{n}} \log \frac{\mathrm{a}_{\mathrm{Zn}^{2+}}}{\mathrm{a}_{\mathrm{Fe}^{2+}}} \\ & =(0.763-0.44)-\frac{0.0591}{1} \log \frac{0.001}{0.005} \\ & =0.364 \mathrm{~V} \\ & \end{aligned}$
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