$\mathrm{Pt}\left(\mathrm{s}\right)\left|{\mathrm{H}}_2\left(\mathrm{g}\right)\left(\mathrm{lbar}\right)\right|{\mathrm{H}}^{+}\left(\mathrm{aq}\right)\left(1\mathrm{M}\right)\parallel {\mathrm{M}}^{3+}\text{ (aq), }{\mathrm{M}}^{2+}\left(\mathrm{aq}\right)\mid \mathrm{Pt}\left(\mathrm{s}\right)$ The E_Cell for the given cell is 0.1115V at 298 K when $\frac{\left[{\mathrm{M}}^{+}\left(\mathrm{aq}\right)\right]}{\left[{\mathrm{M}}^{3+}\left(\mathrm{aq}\right)\right]}={10}^{\mathrm{a}}$ The value of a is …..
Given ${\mathrm{E}}^{\circ }{\mathrm{M}}^{3+}/{\mathrm{M}}^{+}=0.2\mathrm{V}\frac{2.303\mathrm{RT}}{\mathrm{F}}=0.059\mathrm{V}$

$\begin{array}{l}{\mathrm{H}}_2\to 2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\\ {\mathrm{M}}^{3+}2{\mathrm{e}}^{-}\to {\mathrm{M}}^{+}\\ ----------\\ {\mathrm{H}}_2+{\mathrm{M}}^{3+}\to 2{\mathrm{H}}^{+}+{\mathrm{M}}^{+}\end{array}$     ${\mathrm{E}}_{\text{cell }}^0={\mathrm{E}}_{\mathrm{C}}^0-{\mathrm{E}}_{\mathrm{A}}^0=0.2-0=0.2$
$\begin{array}{l}{\mathrm{E}}_{\text{cell }}={\mathrm{E}}_{\text{cell }}^0-\frac{0.0591}{2}\log \frac{\left[{\mathrm{M}}^{+}\right]\left[{\mathrm{H}}^{+}\right]}{\left[{\mathrm{M}}^{3+}\right]}=0.2\\ 0.1115=0.2-\frac{0.059}{2}\log \frac{\left[{\mathrm{M}}^{+}\right]}{\left[{\mathrm{M}}^{3+}\right]}\\ -0.0885=-0.0295\log \frac{\left[{\mathrm{M}}^{+}\right]}{\left[{\mathrm{M}}^{3+}\right]}\\ \log \frac{\left[{\mathrm{M}}^{+}\right]}{\left[{\mathrm{M}}^{3+}\right]}=3\Rightarrow \frac{\left[{\mathrm{M}}^{+}\right]}{\left[{\mathrm{M}}^{3+}\right]}={10}^3=\mathrm{a}=3\end{array}$