Consider the cell:Pt_{(s)} | H_{2(g)}(1 atm) | vedclass

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(B) The cell reaction is: $\frac{1}{2}H_{2(g)} + Fe^{3+}_{(aq)} \longrightarrow H^{+}_{(aq)} + Fe^{2+}_{(aq)}$The Nernst equation for the cell is: $E_

Vedclass · Accepted Answer

$10$

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(B) The cell reaction is: $\frac{1}{2}H_{2(g)} + Fe^{3+}_{(aq)} \longrightarrow H^{+}_{(aq)} + Fe^{2+}_{(aq)}$The Nernst equation for the cell is: $E_{cell} = E^0_{cell} - \frac{0.0591}{n} \log \frac{[H^{+}][Fe^{2+}]}{[Fe^{3+}](P_{H_2})^{1/2}}$Here,$n = 1$,$[H^{+}] = 1 \ M$,$P_{H_2} = 1 \ atm$,and $E^0_{cell} = E^0_{cathode} - E^0_{anode} = 0.771 \ V - 0 \ V = 0.771 \ V$.Substituting the values: $0.712 = 0.771 - 0.0591 \log \frac{[Fe^{2+}]}{[Fe^{3+}]}$$0.0591 \log \frac{[Fe^{2+}]}{[Fe^{3+}]} = 0.771 - 0.712 = 0.059$$\log \frac{[Fe^{2+}]}{[Fe^{3+}]} = \frac{0.059}{0.0591} \approx 1$Therefore,$\frac{[Fe^{2+}]}{[Fe^{3+}]} = 10^1 = 10$.

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