Consider the cell
Pt, H$$_2$$(g, 1 atm) | H$$^+$$ aq, 1M | Fe$$^{3+}$$ aq, Fe$$^{2+}$$ aq | Pt s
When the potential of the cell is 0.712 V at 298 K, the ratio Fe$$^{2+}$$/Fe$$^{3+}$$ is ______ (Nearest integer)
Given: Fe$$^{3+}$$ + e$$^-$$ = Fe$$^{2+}$$, E° Fe$$^{3+}$$, Fe$$^{2+}$$ Pt = 0.771, $$\frac{2.303RT}{F} = 0.06$$ V

The cell reaction is

$$H_2(g)+2Fe^{3+}(aq)\rightarrow2H^+(aq)+2Fe^{2+}(aq).$$

The standard cell potential is

$$E_{\text{cell}}^\circ=E_{\text{cathode}}^\circ-E_{\text{anode}}^\circ=0.771-0=0.771\ \text{V}.$$

Using the Nernst equation,

$$E_{\text{cell}}=E_{\text{cell}}^\circ-\frac{0.06}{n}\log Q,$$

where (n=2) and

$$Q=\frac{[H^+]^2[Fe^{2+}]^2}{[H_2][Fe^{3+}]^2}.$$

Since ([H^+]=1\ \text{M}) and (P_{H_2}=1\ \text{atm}),

$$Q=\left(\frac{[Fe^{2+}]}{[Fe^{3+}]}\right)^2.$$

Substituting the given values,

$$0.712=0.771-\frac{0.06}{2}\log\left(\frac{[Fe^{2+}]}{[Fe^{3+}]}\right)^2.$$

Simplifying,

$$0.712=0.771-0.06\log\left(\frac{[Fe^{2+}]}{[Fe^{3+}]}\right).$$

Therefore,

$$0.06\log\left(\frac{[Fe^{2+}]}{[Fe^{3+}]}\right)=0.771-0.712=0.059,$$

$$\log\left(\frac{[Fe^{2+}]}{[Fe^{3+}]}\right)=\frac{0.059}{0.06}\approx1.$$

Taking the antilogarithm,

$$\frac{[Fe^{2+}]}{[Fe^{3+}]}=10^1=10.$$

Hence,

$$\boxed{\frac{[Fe^{2+}]}{[Fe^{3+}]}=10}.$$