The electrode potentials for Cu^{2+}_{(aq)} + | vedclass

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(B) For the reaction $Cu^{2+} + e^- ightarrow Cu^{+}$,$E^0_1 = 0.15 \ V$. The Gibbs free energy change is $\Delta G^0_1 = -n_1 F E^0_1 = -1 	imes F

Vedclass · Accepted Answer

$0.325$

Vedclass · Answer

(B) For the reaction $Cu^{2+} + e^- ightarrow Cu^{+}$,$E^0_1 = 0.15 \ V$. The Gibbs free energy change is $\Delta G^0_1 = -n_1 F E^0_1 = -1 	imes F 	imes 0.15 = -0.15 F$.For the reaction $Cu^{+} + e^- ightarrow Cu$,$E^0_2 = 0.50 \ V$. The Gibbs free energy change is $\Delta G^0_2 = -n_2 F E^0_2 = -1 	imes F 	imes 0.50 = -0.50 F$.For the overall reaction $Cu^{2+} + 2e^- ightarrow Cu$,the total Gibbs free energy change is $\Delta G^0 = \Delta G^0_1 + \Delta G^0_2 = -0.15 F - 0.50 F = -0.65 F$.Using the relation $\Delta G^0 = -n F E^0_{Cu^{2+}/Cu}$,where $n = 2$:$-2 F E^0_{Cu^{2+}/Cu} = -0.65 F$.Therefore,$E^0_{Cu^{2+}/Cu} = \frac{0.65}{2} = 0.325 \ V$.

The electrode potentials for $Cu^{2+}_{(aq)} + e^- \rightarrow Cu^{+}_{(aq)}$ and $Cu^{+}_{(aq)} + e^- \rightarrow Cu_{(s)}$ are $+0.15 \ V$ and $+0.50 \ V$ respectively. The value of $E^o_{Cu^{2+}/Cu}$ will be $........ \ V$.

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