If $\mathrm{E}^{\circ}\left(\mathrm{Mg}^{+2}{ }_{(\mathrm{aq})} \mid \mathrm{Mg}_{(\mathrm{s})}\right)=-2.37 \mathrm{~V}$. What is potential for $\mathrm{Mg}_{(\mathrm{s})} \longrightarrow \mathrm{Mg}^{+2}(0.1 \mathrm{M})+2 \overline{\mathrm{e}}$ at 298 K ?

The standard oxidation potential $E^\circ_\text{ox}$ is determined from the given reduction potential $E^\circ(\mathrm{Mg}^{+2}_{(\mathrm{aq})} \mid \mathrm{Mg}_{(\mathrm{s})}) = -2.37\,\mathrm{V}$:

$E^\circ_\text{ox} = -E^\circ_\text{red} = +2.37\,\mathrm{V}$

For the reaction $\mathrm{Mg}_{(\mathrm{s})} \to \mathrm{Mg}^{+2}(0.1\,\mathrm{M}) + 2e^-$, the number of electrons transferred is $n = 2$ and the reaction quotient is $Q = [\mathrm{Mg}^{+2}] = 0.1$.

The Nernst equation gives the oxidation potential as:

$E = E^\circ_\text{ox} - \frac{0.0592}{n} \log Q = 2.37 - \frac{0.0592}{2} \log(0.1)$

Since $\log(0.1) = -1$, the calculation becomes:

$E = 2.37 - 0.0296 \times (-1) = 2.37 + 0.0296 = 2.3996\,\mathrm{V}$

Final answer: $\boxed{\text{A}}$