1. Write the given reaction and identify electron transfer

The reaction is $\mathrm{Zn(s) + Sn^{2+}(aq) \rightleftharpoons Zn^{2+}(aq) + Sn(s)}$

Half-reactions:

• Oxidation: $\mathrm{Zn \to Zn^{2+} + 2e^-}$
• Reduction: $\mathrm{Sn^{2+} + 2e^- \to Sn}$

So, the number of electrons transferred is $n=2$

2. Use the relation between standard cell potential and equilibrium constant

At equilibrium, $E^\circ_{\text{cell}} = \frac{0.059}{n}\log K$

Given:

• $K = 1 \times 10^{20}$
• $n=2$

Thus, $E^\circ_{\text{cell}} = \frac{0.059}{2}\log(10^{20})$ $E^\circ_{\text{cell}} = \frac{0.059}{2}\times 20$ $E^\circ_{\text{cell}} = 0.59\ \text{V}$

3. Relate cell potential to electrode potentials

For the reaction,

• Cathode: $\mathrm{Sn^{2+}/Sn}$
• Anode: $\mathrm{Zn^{2+}/Zn}$

Therefore, $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ $0.59 = E^\circ_{\mathrm{Sn^{2+}/Sn}} - (-0.76)$ $0.59 = E^\circ_{\mathrm{Sn^{2+}/Sn}} + 0.76$

So, $E^\circ_{\mathrm{Sn^{2+}/Sn}} = 0.59 - 0.76 = -0.17\ \text{V}$

Hence, the magnitude is $|E^\circ_{\mathrm{Sn^{2+}/Sn}}| = 0.17\ \text{V} = 17 \times 10^{-2}\ \text{V}$

4. Nearest integer

The required value is $\boxed{17}$