Consider the cell Pt _ (s) | H _ 2 (g, 1atm)| H ^ + (aq, 1M) | | Fe...

Given cell reaction:
${Pt}_{\left(s\right)}\left|{H}_{2}\left(g, 1atm\right)\right|{H}^{+}\left(aq, 1M\right) | | {Fe}^{3 +}\left(aq\right), {Fe}^{2 +}\left(aq\right) Pt \left(s\right)$

at anode(oxidation) ${H}_{2} ⟶ 2{H}^{+} + 2{e}^{-}$

At cathode(reduction) ${Fe}_{aq}^{3 +} + {e}^{-} ⟶ {Fe}_{aq}^{2 +}$

$E{°}_{cell} = {E}_{cathode}^{o} - {E}_{anode}^{o} = {E}^{o}{H}_{2} | {H}^{+} + {E}^{o}{Fe}^{3 +} ∣ {Fe}^{2 +} = 0 . 771V$

Thus, using Nernst equation,
$⇒ E = E ° - \dfrac{0 \cdot 06}{1}log\dfrac{{Fe}^{2 +}}{{Fe}^{3 +}}$

$⇒ 0 . 712 = \left(0 + 0 . 771\right) - \dfrac{0 . 06}{1}log\dfrac{{Fe}^{2 +}}{{Fe}^{3 +}}$

$⇒ log\dfrac{{Fe}^{2 +}}{{Fe}^{3 +}} = \dfrac{0 . 059}{0 . 06} \approx 1$

$\dfrac{{Fe}^{2 +}}{{Fe}^{3 +}} = 10$