Given below are the half - cell reactions :

$\begin{array}{cc}{\text{Mn}}^{2+}+2{\text{e}}^{-}\to \text{Mn} ; & {\text{E}}^{\circ }=-\text{1.18 V}\\ 2\left({\text{Mn}}^{3+}+{\text{e}}^{-}\to {\text{Mn}}^{2+}\right) ; & {\text{E}}^{\circ }=+\text{1.51 V}\end{array}$

The ${\text{E}}^{\circ }$ for $3{\text{Mn}}^{2+}\to \text{Mn}+2{\text{Mn}}^{3+}$ will be :

For the given cell; $\mathrm{Cu}\left(\mathrm{s}\right)\left|{\mathrm{Cu}}^{2+}\left({\mathrm{C}}_1\mathrm{M}\right)\right|\left|{\mathrm{Cu}}^{2+}\left({\mathrm{C}}_2\mathrm{M}\right)\right|\mathrm{Cu}\left(\mathrm{s}\right)$

change in Gibbs energy $\left(∆G\right)$ is negative, it :

The variation of molar conductively with concentration of an electrolyte (X) in aqueous solution is shown in the given figure.

The electrolyte X is :

An oxidation-reduction reaction in which 3 electrons are transferred has a ${\mathrm{\Delta G}}^0$ of $17.37 \mathrm{kJ} {\mathrm{mol}}^{-1}$ at $25°\mathrm{C}.$ The value of ${\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}$ (in V) is $\_\_\_\_\_\_\_\times {10}^{-2}$.

$\left(1\mathrm{F}=96500 \mathrm{C} {\mathrm{mol}}^{-1}\right)$