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

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(B) Given:$(1) \ Cu^{2+}_{(aq)} + e^- 	o Cu^{+}_{(aq)} \quad \Delta G_1^{\circ} = -1 	imes 0.15 	imes F$$(2) \ Cu^{+}_{(aq)} + e^- 	o Cu_{(s)} \qu

Vedclass · Accepted Answer

$0.325$

Vedclass · Answer

(B) Given:$(1) \ Cu^{2+}_{(aq)} + e^- 	o Cu^{+}_{(aq)} \quad \Delta G_1^{\circ} = -1 	imes 0.15 	imes F$$(2) \ Cu^{+}_{(aq)} + e^- 	o Cu_{(s)} \quad \Delta G_2^{\circ} = -1 	imes 0.50 	imes F$Adding $(1)$ and $(2)$ gives the overall reaction:$Cu^{2+}_{(aq)} + 2e^- 	o Cu_{(s)} \quad \Delta G_3^{\circ} = -2 	imes F 	imes E^{\circ}_{Cu^{2+}/Cu}$Since $\Delta G_3^{\circ} = \Delta G_1^{\circ} + \Delta G_2^{\circ}$,we have:$-2 	imes F 	imes E^{\circ}_{Cu^{2+}/Cu} = (-1 	imes 0.15 	imes F) + (-1 	imes 0.50 	imes F)$$-2 	imes E^{\circ}_{Cu^{2+}/Cu} = -0.65$$E^{\circ}_{Cu^{2+}/Cu} = \frac{0.65}{2} = 0.325 \, V$.

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

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