Consider the cellPt_{(s)} mid H_2(g, 1,atm) m | vedclass

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(B) The cell reaction is: $H_2(g) + 2Fe^{3+}(aq) \longrightarrow 2H^+(aq) + 2Fe^{2+}(aq)$.The standard cell potential $E^{\circ}_{cell} = E^{\circ}_{c

Vedclass · Accepted Answer

$10$

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(B) The cell reaction is: $H_2(g) + 2Fe^{3+}(aq) \longrightarrow 2H^+(aq) + 2Fe^{2+}(aq)$.The standard cell potential $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.771\,V - 0\,V = 0.771\,V$.Using the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.06}{n} \log Q$,where $n = 2$.$0.712 = 0.771 - \frac{0.06}{2} \log \frac{[H^+]^2 [Fe^{2+}]^2}{[H_2] [Fe^{3+}]^2}$.Given $[H^+] = 1\,M$,$P_{H_2} = 1\,atm$,so $0.712 = 0.771 - 0.03 \log \frac{[Fe^{2+}]^2}{[Fe^{3+}]^2}$.$0.712 = 0.771 - 0.06 \log \frac{[Fe^{2+}]}{[Fe^{3+}]}$.$0.06 \log \frac{[Fe^{2+}]}{[Fe^{3+}]} = 0.771 - 0.712 = 0.059$.$\log \frac{[Fe^{2+}]}{[Fe^{3+}]} = \frac{0.059}{0.06} \approx 0.983 \approx 1$.$\frac{[Fe^{2+}]}{[Fe^{3+}]} = 10^1 = 10$.

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