Calculate $\mathrm{E}_{\text {cell }}^{\circ}$ in which following reaction occurs.…
Calculate $\mathrm{E}_{\text {cell }}^{\circ}$ in which following reaction occurs. $\mathrm{Mg}_{(\mathrm{s})}+2 \mathrm{Ag}_{(1 \mathrm{M})}^{+} \rightarrow \mathrm{Mg}_{(1 \mathrm{M})}^{++}+2 \mathrm{Ag}_{(\mathrm{s})}$ if $\mathrm{E}_{\mathrm{Ag}}^{\circ}=0.8 \mathrm{~V}$ and $\mathrm{E}_{\mathrm{Mg}}^{\circ}=-2.37 \mathrm{~V}$
$-3.17 \mathrm{~V}$
$3.17 \mathrm{~V}$
$-1.57 \mathrm{~V}$
$1.57 \mathrm{~V}$
Solution
For the given cell reaction, anode is $\mathrm{Mg}$ and cathode is Ag.
$\begin{aligned}
\mathrm{E}_{\text {cell }}^0 & =\mathrm{E}_{\text {cathode }}^0-\mathrm{E}_{\text {anode }}^0 \\
& =0.8-(-2.37) \\
& =3.17 \mathrm{~V}
\end{aligned}$