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Q.

Consider the cell
$Pt (s) | H_{2} (g) (1 atm) | H^{+} (aq . [H^{+}] = 1) | | {Fe}^{3 +} (aq), {Fe}^{2 +} (aq) | | Pt (s)$
Given $E_{F e^{+ 3} / F e^{+ 2}}^{0} = 0.771 and E_{H^{+} / 1 / 2 H_{2}}^{0} = 0 V, T = 298 K$
If the potential of the cell is 0.712 V, the ratio of concentration of ${Fe}^{2 +}$ to ${Fe}^{3 +}$ is____(Nearest integer)

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answer is 10.

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Detailed Solution

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$\begin{array}{l} AR : \frac{1}{2} H_{2} \to H^{+} + e^{-} \\ CR : {Fe}^{+ 3} + e^{-} \to {Fe}^{+ 2} \\ NR : \frac{1}{2} H_{2} + {Fe}^{+ 3} \to H^{+} + {Fe}^{+ 2} \\ E = E_{(Cℓ)}^{0} - \frac{0.059}{1} \log \frac{[{Fe}^{+ 2}]}{[{Fe}^{+ 3}]} \times \frac{[He]}{{[P_{H_{2}}]}^{1 / 2}} \\ 0.712 = 0.771 - 0.059 \log \frac{[{Fe}^{+ 2}]}{[{Fe}^{+ 3}]} \\ \log \frac{[{Fe}^{+ 2}]}{[{Fe}^{+ 3}]} = 1 \Rightarrow \frac{[{Fe}^{+ 2}]}{[{Fe}^{+ 3}]} = 10 \end{array}$

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Consider the cell Pt(s)|H2(g)(1 atm)|H+(aq.[H+]=1)||Fe3+(aq),Fe2+(aq)||Pt(s)Given EFe+3/Fe+20=0.771 and EH+/1/2H20=0 V,T=298 KIf the potential of the cell is 0.712 V, the ratio of concentration of Fe2+ to Fe3+ is____(Nearest integer)