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Chemistry Question on Nernst Equation

The electrode potentials for Cu2+(aq)+e−−>Cu+(aq){Cu^{2+}(aq)+e^-->Cu^+(aq)} and Cu+(aq)+e−−>Cu(s){Cu^{+}(aq)+e^-->Cu(s)} are +0.15 V+ 0.15\, V and +0.50 V+ 0.50 \,V, respectively. The value of ECu2+/Cu?E^?_{Cu^{2+}/Cu} will be :

A

0.500 V0.500 \,V

B

0352 V0352 \,V

C

0.650 V0.650 \,V

D

0.150 V0.150 \,V

Answer

0352 V0352 \,V

Explanation

Solution

cu2++le−→Cu+ ; ΔG1∘=−n1E1∘F]cu^{2+}+le^{-} \rightarrow Cu^{+}\,; \,\Delta G^{\circ}_{1}=-n_{1}E_{1}^{\circ}F]
cu++le−→Cu ; ΔG2∘=−n2E2∘Fcu2++2e−→Cu ; ΔG∘=ΔG1∘=ΔG2∘\frac{cu^{+}+le^{-} \rightarrow Cu\,; \,\Delta G^{\circ }_{2}=-n_{2}E_{2}^{\circ }F}{cu^{2+}+2e^{-} \rightarrow Cu\,; \,\Delta G^{\circ}=\Delta G^{\circ }_{1}=\Delta G^{\circ}_{2}}
−nE?F=1 n1 E?F+(−1)n2 E2?F-nE^{?}F=1\,n_{1}\,E^{?}F+\left(-1\right)n_{2}\,E^{?}_{2}F
−nE? F=−F(n1 E1?+n2 E2?F)-nE^{?}\,F=-F\left(n_{1}\,E^{?}_{1}+n_{2}\,E^{?}_{2}F\right)
EV=n1E1?+n2E2?n=0.15×1+0.50×12=0.325E^{V}=\frac{n_{1}E^{?}_{1}+n_{2}E^{?}_{2}}{n}=\frac{0.15\times1+0.50\times1}{2}=0.325