The electrode potential of the following half cell at 298 K $\mathrm{X}\left|{\mathrm{X}}^{2+}(0.001\mathrm{M})\parallel {\mathrm{Y}}^{2+}(0.01\mathrm{M})\right|\mathrm{Y}$ is________ x 10^-2 V (Nearest integer).
Given ${\mathrm{E}}_{{\mathrm{x}}^{2+}/\mathrm{x}}^0=-2.36\mathrm{V}$
${\mathrm{E}}_{{\mathrm{y}}^{2+}/\mathrm{y}}^0=+0.36\mathrm{V}\frac{2.303\mathrm{RT}}{\mathrm{F}}=0.06\mathrm{V}$

$$\begin{gathered} \begin{array}{lc}{\mathrm{X}}_{\left(\mathrm{s}\right)}\to {\mathrm{X}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-} {\mathrm{Y}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Y}\left(\mathrm{s}\right) \\ {\mathrm{Y}}^{2+}+{\mathrm{X}}_{\left(\mathrm{s}\right)}\to \mathrm{Y}\left(\mathrm{s}\right)+{\mathrm{X}}^{2+}\left(\mathrm{aq}\right) {\mathrm{E}}_{\text{cell }}={\mathrm{\epsilon }}^0-\frac{0.06}{\mathrm{n}}\log \left(\frac{\left({\mathrm{X}}^{2+}\right)}{\left({\mathrm{Y}}^{2+}\right)}\right) \\ {\mathrm{\epsilon }}_{\text{cell }}^0=36-(-2.36)=2.72{\mathrm{\epsilon }}_{\text{cell }}=2.72-\frac{0.06}{2}\log \left(\frac{{10}^{-3}}{{10}^{-2}}\right)& \\ =2.72-\frac{0.06}{2}\log \left({10}^{-1}\right)=2.72+0.03=2.75\Rightarrow 275\times {10}^{-2}& \end{array} \end{gathered}$$