For the given reactions
$$Sn^{2+} + 2e^- \rightarrow Sn$$
$$Sn^{4+} + 4e^- \rightarrow Sn$$
the electrode potentials are; $$E^\circ_{Sn^{2+}/Sn} = -0.140$$ V and $$E^\circ_{Sn^{4+}/Sn} = 0.010$$ V. The magnitude of standard electrode potential for $$Sn^{4+}/Sn^{2+}$$ i.e. $$E^\circ_{Sn^{4+}/Sn^{2+}}$$ is ______ $$\times 10^{-2}$$ V (Nearest integer)

We need to find the standard electrode potential for $$\text{Sn}^{4+}/\text{Sn}^{2+}$$. The known electrode potentials are $$E^\circ_{\text{Sn}^{2+}/\text{Sn}} = -0.140 \text{ V}$$ for the reaction $$\text{Sn}^{2+} + 2e^- \rightarrow \text{Sn}$$ and $$E^\circ_{\text{Sn}^{4+}/\text{Sn}} = 0.010 \text{ V}$$ for $$\text{Sn}^{4+} + 4e^- \rightarrow \text{Sn}$$.

Since electrode potentials are not additive, we use the relation $$\Delta G^\circ = -nFE^\circ$$. For reaction (i), $$\Delta G_1^\circ = -2F(-0.140) = 0.280F$$, and for reaction (ii), $$\Delta G_2^\circ = -4F(0.010) = -0.040F$$.

The target reaction is $$\text{Sn}^{4+} + 2e^- \rightarrow \text{Sn}^{2+}$$, which can be obtained by subtracting reaction (i) from reaction (ii). Thus, $$\Delta G_3^\circ = \Delta G_2^\circ - \Delta G_1^\circ = -0.040F - 0.280F = -0.320F$$.

With n = 2 for this reaction, $$\Delta G_3^\circ = -nFE^\circ_{\text{Sn}^{4+}/\text{Sn}^{2+}}$$ gives $$-0.320F = -2F \times E^\circ_{\text{Sn}^{4+}/\text{Sn}^{2+}}$$ and hence $$E^\circ_{\text{Sn}^{4+}/\text{Sn}^{2+}} = \frac{0.320}{2} = 0.160 \text{ V}$$.

Finally, $$|E^\circ_{\text{Sn}^{4+}/\text{Sn}^{2+}}| = 0.160 \text{ V} = 16 \times 10^{-2} \text{ V}$$. Hence, the answer is 16.