Consider the cell Pt(s)|H$$_2$$(s)(1atm)|H$$^+$$(aq, [H$$^+$$] = 1)||Fe$$^{3+}$$(aq), Fe$$^{2+}$$(aq)|Pt(s)
Given: E$$_{Fe^{3+}/Fe^{2+}}^\circ$$ = 0.771 V and E$$_{H^+/\frac{1}{2}H_2}^\circ$$ = 0 V, T = 298 K
If the potential of the cell is 0.712 V the ratio of concentration of Fe$$^{2+}$$ to Fe$$^{3+}$$ is _____ (Nearest integer)

Given the electrochemical cell: Pt(s) | H$$_2$$(g)(1 atm) | H$$^+$$(aq, [H$$^+$$] = 1) || Fe$$^{3+}$$(aq), Fe$$^{2+}$$(aq) | Pt(s)

$$E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = 0.771$$ V, $$E_{\text{cell}} = 0.712$$ V.

Anode (oxidation): $$\frac{1}{2}\text{H}_2 \to \text{H}^+ + e^-$$

Cathode (reduction): $$\text{Fe}^{3+} + e^- \to \text{Fe}^{2+}$$

Here $$n = 1$$ (one electron transferred).

$$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q$$

Since the anode is SHE (standard), $$E^\circ_{\text{cell}} = 0.771 - 0 = 0.771$$ V.

The reaction quotient for the cathode half-cell: $$Q = \dfrac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}$$.

$$0.712 = 0.771 - 0.0591 \times \log\left(\frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}\right)$$

$$0.0591 \times \log\left(\frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}\right) = 0.771 - 0.712 = 0.059$$

$$\log\left(\frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}\right) = \frac{0.059}{0.0591} \approx 0.9983$$

$$\frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]} = 10^{0.9983} \approx 10$$

The ratio of $$[\text{Fe}^{2+}]$$ to $$[\text{Fe}^{3+}]$$ is $$\mathbf{10}$$.