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Question

(D) The reactions occurring in the cell are:At anode: $H_{2(g)} \rightarrow 2H^{+}_{(aq)} + 2e^{-}$At cathode: $M^{4+}_{(aq)} + 2e^{-} \rightarrow M^{

Vedclass · Accepted Answer

$2$

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(D) The reactions occurring in the cell are:At anode: $H_{2(g)} ightarrow 2H^{+}_{(aq)} + 2e^{-}$At cathode: $M^{4+}_{(aq)} + 2e^{-} ightarrow M^{2+}_{(aq)}$Overall reaction: $H_{2(g)} + M^{4+}_{(aq)} ightarrow M^{2+}_{(aq)} + 2H^{+}_{(aq)}$The Nernst equation is:$E_{	ext{cell}} = E^0_{	ext{cell}} - \frac{0.059}{n} \log \frac{[M^{2+}][H^{+}]^2}{[M^{4+}][P_{H_2}]}$Given $E^0_{	ext{cell}} = E^0_{M^{4+}/M^{2+}} - E^0_{H^{+}/H_2} = 0.151 - 0 = 0.151 \ V$,$[H^{+}] = 1 \ M$,$P_{H_2} = 1 \ bar$,and $n = 2$:$0.092 = 0.151 - \frac{0.059}{2} \log \frac{[M^{2+}] 	imes (1)^2}{[M^{4+}] 	imes 1}$$0.092 = 0.151 - 0.0295 \log (10^x)$$0.0295 \log (10^x) = 0.151 - 0.092 = 0.059$$\log (10^x) = \frac{0.059}{0.0295} = 2$$x = 2$

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