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(A) The standard Gibbs free energy change $(\Delta G^{\circ})$ is related to the standard electrode potential $(E^{\circ})$ by the equation $\Delta G^

Vedclass · Accepted Answer

$0.158$

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(A) The standard Gibbs free energy change $(\Delta G^{\circ})$ is related to the standard electrode potential $(E^{\circ})$ by the equation $\Delta G^{\circ} = -nFE^{\circ}$.For the reaction $Cu^{2+} + 2e^{-} \longrightarrow Cu$,$\Delta G^{\circ}_{1} = -2F(0.34) = -0.68F$.For the reaction $Cu^{+} + e^{-} \longrightarrow Cu$,$\Delta G^{\circ}_{2} = -1F(0.522) = -0.522F$.We want to find $E^{\circ}$ for the reaction $Cu^{2+} + e^{-} \longrightarrow Cu^{+}$.This reaction can be obtained by subtracting the second reaction from the first: $(Cu^{2+} + 2e^{-}$ $\longrightarrow Cu) - (Cu^{+} + e^{-}$ $\longrightarrow Cu) \implies Cu^{2+} + e^{-}$ $\longrightarrow Cu^{+}$.Therefore,$\Delta G^{\circ}_{3} = \Delta G^{\circ}_{1} - \Delta G^{\circ}_{2} = -0.68F - (-0.522F) = -0.158F$.Since $\Delta G^{\circ}_{3} = -nFE^{\circ}_{3}$ where $n=1$,we have $-0.158F = -1F(E^{\circ}_{3})$.Thus,$E^{\circ}_{3} = 0.158 \ V$.

Given that the standard potentials $(E^{\circ})$ of $Cu^{2+}/Cu$ and $Cu^{+}/Cu$ are $0.34 \ V$ and $0.522 \ V$ respectively,the $E^{\circ}$ of $Cu^{2+}/Cu^{+}$ is (in $V$)

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