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Calculate standard potential of a cell having electrode reaction as
$\begin{aligned}
& \mathrm{Cd}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cd}_{\mathrm{s}} \mathrm{E}^{\circ}=-0.403 \mathrm{~V} \\
& \mathrm{Zn}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Zn}_{(\mathrm{s})} \mathrm{E}^{\circ}=-0.763 \mathrm{~V}
\end{aligned}$
Options:
$\begin{aligned}
& \mathrm{Cd}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cd}_{\mathrm{s}} \mathrm{E}^{\circ}=-0.403 \mathrm{~V} \\
& \mathrm{Zn}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Zn}_{(\mathrm{s})} \mathrm{E}^{\circ}=-0.763 \mathrm{~V}
\end{aligned}$
Solution:
2985 Upvotes
Verified Answer
The correct answer is:
0.360 V
$\begin{aligned} & \mathrm{E}_{\text {cell }}^{\circ}=\mathrm{E}_{\text {Cathode }}^{\circ}-\mathrm{E}_{\text {Anode }}^{\circ} \\ & =-0.403-(-0.763) \\ & =0.763-0.403 \\ & =0.36 \mathrm{~V}\end{aligned}$
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