A reaction, $\mathrm{Ni}_{(\mathrm{s})}+\mathrm{Cu}_{(\mathrm{IM})}^{++} \rightarrow…
A reaction, $\mathrm{Ni}_{(\mathrm{s})}+\mathrm{Cu}_{(\mathrm{IM})}^{++} \rightarrow \mathrm{Ni}_{(\mathrm{IM} 2}^{++}+\mathrm{Cu}_{(\mathrm{s})}$ occurs in a cell. Calculate $\mathrm{E}_{\mathrm{cell}}^6$ if $\mathrm{E}_{\mathrm{Cu}}^0=0.337 \mathrm{~V}$ and $\mathrm{E}_{\mathrm{Ni}}^{\mathrm{o}}=-0.257 \mathrm{~V}$
$0.594 \mathrm{~V}$
$-0.594 \mathrm{~V}$
$-0.08 \mathrm{~V}$
$0.08 \mathrm{~V}$
Solution
The standard cell potential is given by
$\begin{aligned}
\mathrm{E}_{\text {cell }}^o & =\mathrm{E}_{\text {cathode }}^{\circ}-\mathrm{E}_{\text {anode }}^o \\
\mathrm{E}_{\text {cell }}^{\circ} & =\mathrm{E}_{\mathrm{Cu}}^{\circ}-\mathrm{E}_{\mathrm{Ni}}^{\circ} \\
& =(0.337 \mathrm{~V})-(-0.257 \mathrm{~V}) \\
& =0.337 \mathrm{~V}+0.257 \mathrm{~V} \\
& =0.594 \mathrm{~V}
\end{aligned}$