Chemistry - Electrochemistry Question with Solution | TestHub

The given standard reduction potential is ${\mathrm{E}}_{\mathrm{red}}^{\circ }=-2.37\mathrm{V}$.

The standard oxidation potential is ${\mathrm{E}}_{\mathrm{ox}}^{\circ }=-{\mathrm{E}}_{\mathrm{red}}^{\circ }=+2.37\mathrm{V}$.

The given reaction is an oxidation half-reaction:
${\mathrm{Mg}}_{(\mathrm{s})}\to {\mathrm{Mg}}^{+2}(0.01\mathrm{M})+2{\mathrm{e}}^{-}$

Here, the number of electrons transferred is $n=2$ and the concentration of ${\mathrm{Mg}}^{+2}$ is $0.01\mathrm{M}=10^{-2}\mathrm{M}$.

Using the Nernst equation for the oxidation potential at $298\mathrm{K}$:
${\mathrm{E}}_{\mathrm{ox}}={\mathrm{E}}_{\mathrm{ox}}^{\circ }-\frac{0.0592}{n}{\log }_{10}[{\mathrm{Mg}}^{+2}]$

Substituting the values:
${\mathrm{E}}_{\mathrm{ox}}=2.37-\frac{0.0592}{2}{\log }_{10}(10^{-2})$

${\mathrm{E}}_{\mathrm{ox}}=2.37-0.0296\times (-2)$

${\mathrm{E}}_{\mathrm{ox}}=2.37+0.0592=+2.4292\mathrm{V}$

Answer: $+2.4292\mathrm{V}$