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Calculate the $\mathrm{E}_{\text {cell }}^o$ for $\mathrm{Zn}_{(\mathrm{s})}\left|\mathrm{Zn}_{(\mathrm{IM})}^{+}\right|\left|\mathrm{Cd}_{(\mathrm{IM})}^{+}\right| \mathrm{Cd}_{(\mathrm{s})}$ at $25^{\circ} \mathrm{C}\left[\mathrm{E}_{\mathrm{Zn}}^{\circ}=-0.763 \mathrm{~V} ; \mathrm{E}_{\mathrm{Cd}}^{\circ}=-0.403 \mathrm{~V}\right]$
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Verified Answer
The correct answer is:
$0.36 \mathrm{~V}$
For the given cell reaction, anode is $\mathrm{Zn}$ and cathode is $\mathrm{Cd}$.
$\begin{aligned}
\mathrm{E}_{\text {cell }}^o & =\mathrm{E}_{\text {cathode }}^{\circ}-\mathrm{E}_{\text {anode }}^0 \\
& =-0.403-(-0.763) \\
& =0.36 \mathrm{~V}
\end{aligned}$
$\begin{aligned}
\mathrm{E}_{\text {cell }}^o & =\mathrm{E}_{\text {cathode }}^{\circ}-\mathrm{E}_{\text {anode }}^0 \\
& =-0.403-(-0.763) \\
& =0.36 \mathrm{~V}
\end{aligned}$
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