For the given reactions Sn^2++2 e^- → Sn; Sn^4++4 e^- → Sn the electrode potentials are; E_Sn^2+ / Sn^o =-0.140 V and E_Sn^4+ / Sn^o =0.010 V. The…

For the given reactions $$\begin{aligned} & \mathrm{Sn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Sn} \\ & \mathrm{Sn}^{4+}+4 \mathrm{e}^{-} \rightarrow \mathrm{Sn} \end{aligned} $$ the electrode potentials are $\displaystyle ; \mathrm{E}_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^{\mathrm{o}}=-0.140 \mathrm{~V}$ and $\displaystyle \mathrm{E}_{\mathrm{Sn}^{4+} / \mathrm{Sn}}^{\mathrm{o}}=0.010 \mathrm{~V}$. The magnitude of standard electrode potential for $\displaystyle \mathrm{Sn}^{4+} / \mathrm{Sn}^{2+}$ i.e. $\displaystyle \mathrm{E}_{\mathrm{Sn}^{4+} / \mathrm{Sn}^{2+}}^{\mathrm{o}}$ is \_\_\_\_ $\displaystyle \times 10^{-2} \mathrm{~V}$. (Nearest integer)