Given E_{Cu^{2+}/Cu^{+}}^o = 0.15 , V,E_{Cu^{ | vedclass

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(B) For the reaction $Cu^{2+} + e^- 	o Cu^{+}$,$\Delta G_1^o = -n_1 E_1^o F = -1 	imes 0.15 	imes F = -0.15F$. For the reaction $Cu^{2+} + 2e^- 	o

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$0.53$

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(B) For the reaction $Cu^{2+} + e^- 	o Cu^{+}$,$\Delta G_1^o = -n_1 E_1^o F = -1 	imes 0.15 	imes F = -0.15F$. For the reaction $Cu^{2+} + 2e^- 	o Cu$,$\Delta G_2^o = -n_2 E_2^o F = -2 	imes 0.34 	imes F = -0.68F$. To find the potential for $Cu^{+} + e^- 	o Cu$,we subtract the first reaction from the second: $(Cu^{2+} + 2e^- 	o Cu) - (Cu^{2+} + e^- 	o Cu^{+}) \implies Cu^{+} + e^- 	o Cu$. $\Delta G^o = \Delta G_2^o - \Delta G_1^o = -0.68F - (-0.15F) = -0.53F$. Since $\Delta G^o = -n E^o F$,where $n=1$: $-0.53F = -1 	imes E^o 	imes F \implies E^o = 0.53 \, V$.

Given $E_{Cu^{2+}/Cu^{+}}^o = 0.15 \, V$,$E_{Cu^{2+}/Cu}^o = 0.34 \, V$. The standard electrode potential for the half-cell $Cu^{+}/Cu$ is ............. $V$.

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