1. Identify the cell reaction and standard cell potential

The cell is $\mathrm{Zn}|\mathrm{Zn}^{2+}(aq)||\mathrm{Sn}^{x+}|\mathrm{Sn}$

So, zinc acts as the anode and tin ion acts as the cathode.

Half-reactions:

• Anode: $\mathrm{Zn} \to \mathrm{Zn}^{2+} + 2e^-$
• Cathode: $\mathrm{Sn}^{x+} + xe^- \to \mathrm{Sn}$

Given standard reduction potentials: $E^\circ_{\mathrm{Zn}^{2+}/\mathrm{Zn}}=-0.763\,\text{V}$ $E^\circ_{\mathrm{Sn}^{x+}/\mathrm{Sn}}=+0.008\,\text{V}$

Therefore, $E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}$ $E^\circ_{\text{cell}}=0.008-(-0.763)=0.771\,\text{V}$

2. Apply the Nernst equation

At $298\,\text{K}$, $E_{\text{cell}}=E^\circ_{\text{cell}}-\frac{0.06}{n}\log Q$

Given: $E_{\text{cell}}=0.801\,\text{V}, \quad Q=10^{-2}$

Since $\log(10^{-2})=-2$

So, $0.801=0.771-\frac{0.06}{n}(-2)$ $0.801=0.771+\frac{0.12}{n}$

3. Solve for $n$

$0.801-0.771=\frac{0.12}{n}$ $0.030=\frac{0.12}{n}$

Thus, $n=\frac{0.12}{0.03}=4$

4. Final answer

The number of electrons involved in the cell reaction is $\boxed{4}$