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(C) Given:$(1) \ Cu^{2+} + e^- 	o Cu^{+} ; E_1^o = 0.15 \ V, \Delta G_1^o = -n_1 F E_1^o, n_1 = 1$$(2) \ Cu^{2+} + 2e^- 	o Cu ; E_2^o = 0.34 \ V, \D

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$+0.53$

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(C) Given:$(1) \ Cu^{2+} + e^- 	o Cu^{+} ; E_1^o = 0.15 \ V, \Delta G_1^o = -n_1 F E_1^o, n_1 = 1$$(2) \ Cu^{2+} + 2e^- 	o Cu ; E_2^o = 0.34 \ V, \Delta G_2^o = -n_2 F E_2^o, n_2 = 2$We need to find $E_3^o$ for:$(3) \ Cu^{+} + e^- 	o Cu ; E_3^o = ?, \Delta G_3^o = -n_3 F E_3^o, n_3 = 1$Reaction $(3)$ can be obtained by subtracting reaction $(1)$ from reaction $(2)$:$(Cu^{2+} + 2e^- 	o Cu) - (Cu^{2+} + e^- 	o Cu^{+}) \implies Cu^{+} + e^- 	o Cu$Since $\Delta G^o$ is an extensive property,$\Delta G_3^o = \Delta G_2^o - \Delta G_1^o$$-n_3 F E_3^o = -n_2 F E_2^o - (-n_1 F E_1^o)$$-1 	imes F 	imes E_3^o = -(2 	imes F 	imes 0.34) + (1 	imes F 	imes 0.15)$$-E_3^o = -0.68 + 0.15$$-E_3^o = -0.53$$E_3^o = +0.53 \ V$

$E^o$ values for the half-cell reactions are given below:
$Cu^{2+} + e^- \to Cu^{+} \quad E^o = 0.15 \ V$
$Cu^{2+} + 2e^- \to Cu \quad E^o = 0.34 \ V$
What will be the $E^o$ for the half-cell reaction:
$Cu^{+} + e^- \to Cu$ ? $\dots \ V$

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$E^o$ values for the half-cell reactions are given below:$Cu^{2+} + e^- \to Cu^{+} \quad E^o = 0.15 \ V$$Cu^{2+} + 2e^- \to Cu \quad E^o = 0.34 \ V$What will be the $E^o$ for the half-cell reaction:$Cu^{+} + e^- \to Cu$ ? $\dots \ V$