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(D) The given reaction is the reduction of $Cu^{2+}$ to $Cu_{(s)}$: $Cu^{2+}_{(aq)} + 2e^{-} \rightarrow Cu_{(s)}$ with $E^{\circ} = +0.34 \ V$. The r

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$-0.34 \ V$

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(D) The given reaction is the reduction of $Cu^{2+}$ to $Cu_{(s)}$: $Cu^{2+}_{(aq)} + 2e^{-} \rightarrow Cu_{(s)}$ with $E^{\circ} = +0.34 \ V$. The requested reaction is the oxidation of $Cu_{(s)}$ to $Cu^{2+}_{(aq)}$: $2Cu_{(s)} \rightarrow 2Cu^{2+}_{(aq)} + 4e^{-}$. This is the reverse of the reduction reaction multiplied by a factor of $2$. Since the standard electrode potential $(E^{\circ})$ is an intensive property,it does not depend on the stoichiometric coefficients of the reaction. Therefore,reversing the reaction changes the sign of the potential,but multiplying the coefficients does not change the value of $E^{\circ}$. Thus,the potential for the oxidation reaction $Cu_{(s)} \rightarrow Cu^{2+}_{(aq)} + 2e^{-}$ is $-0.34 \ V$. The potential for $2Cu_{(s)} \rightarrow 2Cu^{2+}_{(aq)} + 4e^{-}$ remains $-0.34 \ V$.

Standard potential of electrode reaction $Cu^{2+}_{(aq)} + 2e^{-} \rightarrow Cu_{(s)}$ is $+0.34 \ V$. What is the standard potential of the reaction $2Cu_{(s)} \rightarrow 2Cu^{2+}_{(aq)} + 4e^{-}$?

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