Chemistry - Electrochemistry Question with Solution | TestHub

The given standard reduction potential is $E_{\text{red}}^{\circ }=-2.37\text{ V}$.

The given half-cell reaction is an oxidation reaction:
${\mathrm{Mg}}_{(\mathrm{s})}⟶{\mathrm{Mg}}^{+2}(0.1\mathrm{M})+2{\mathrm{e}}^{-}$

The standard oxidation potential is:
$E_{\text{ox}}^{\circ }=-E_{\text{red}}^{\circ }=-(-2.37\text{ V})=+2.37\text{ V}$

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

Here, the number of electrons transferred is $n=2$ and the concentration is $[{\mathrm{Mg}}^{+2}]=0.1\text{ M}$.

Substituting the values:
$E_{\text{ox}}=2.37-\frac{0.0592}{2}{\log }_{10}(0.1)$

$E_{\text{ox}}=2.37-0.0296\times (-1)$

$E_{\text{ox}}=2.37+0.0296=+2.3996\text{ V}$

Answer: $+2.3996\mathrm{V}$