1. Identify the half-cells

The cell is: $\mathrm{Pt(s)|H_2(g,1\,atm)|H^+(aq, [H^+]=1)||Fe^{3+}(aq),Fe^{2+}(aq)|Pt(s)}$

Left electrode is the hydrogen electrode: $\mathrm{2H^+ + 2e^- \rightleftharpoons H_2}$ Since $[\mathrm{H^+}] = 1$ and $P_{H_2}=1\,\text{atm}$, it is under standard conditions, so $E_{\text{left}} = 0\,\text{V}$

Right electrode is: $\mathrm{Fe^{3+} + e^- \rightleftharpoons Fe^{2+}}$ with $E^\circ_{\mathrm{Fe^{3+}/Fe^{2+}}} = 0.771\,\text{V}$

2. Write the Nernst equation for the iron electrode

For the reduction $\mathrm{Fe^{3+} + e^- \rightarrow Fe^{2+}}$ we have $n=1$, so $E_{\mathrm{Fe}} = E^\circ_{\mathrm{Fe}} - \frac{0.0591}{1}\log\frac{[\mathrm{Fe^{2+}}]}{[\mathrm{Fe^{3+}}]}$ Thus, $E_{\mathrm{Fe}} = 0.771 - 0.0591\log\frac{[\mathrm{Fe^{2+}}]}{[\mathrm{Fe^{3+}}]}$

3. Use the cell potential expression

Since the right electrode has higher reduction potential, it acts as cathode. The hydrogen electrode is the anode.

Therefore, $E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} = E_{\mathrm{Fe}} - E_{\mathrm{H}}$ But $E_{\mathrm{H}}=0$, so $E_{\text{cell}} = E_{\mathrm{Fe}}$ Given: $E_{\text{cell}} = 0.712\,\text{V}$ Hence, $0.712 = 0.771 - 0.0591\log\frac{[\mathrm{Fe^{2+}}]}{[\mathrm{Fe^{3+}}]}$

4. Solve for the ratio

Rearranging, $0.0591\log\frac{[\mathrm{Fe^{2+}}]}{[\mathrm{Fe^{3+}}]} = 0.771 - 0.712 = 0.059$ So, $\log\frac{[\mathrm{Fe^{2+}}]}{[\mathrm{Fe^{3+}}]} = \frac{0.059}{0.0591} \approx 0.998 \approx 1$ Therefore, $\frac{[\mathrm{Fe^{2+}}]}{[\mathrm{Fe^{3+}}]} \approx 10^1 = 10$

5. Nearest integer

$\boxed{10}$

6. Comparison with stored answer

Stored correct answer = $10$.

Our derived answer matches the stored answer.