Consider the cell Pt ( s )| H_2( g )(1 atm )| H^+( aq,[ H^+]=1) ∥ Fe^3+( aq ), Fe^2+( aq ) ∣ Pt ( s ) Given E_Fe^3+ / Fe^2+^°=0.771 V and E_H^+ / 1/2…

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