Question

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

• A. $0.500 \,V$
• B. $0352 \,V$
• C. $0.650 \,V$
• D. $0.150 \,V$

Answer 1

Answer: (B)

$cu^{2+}+le^{-} \rightarrow Cu^{+}\,; \,\Delta G^{\circ}_{1}=-n_{1}E_{1}^{\circ}F]$
$\frac{cu^{+}+le^{-} \rightarrow Cu\,; \,\Delta G^{\circ }_{2}=-n_{2}E_{2}^{\circ }F}{cu^{2+}+2e^{-} \rightarrow Cu\,; \,\Delta G^{\circ}=\Delta G^{\circ }_{1}=\Delta G^{\circ}_{2}}$
$-nE^{°}F=1\,n_{1}\,E^{°}F+\left(-1\right)n_{2}\,E^{°}_{2}F$
$-nE^{°}\,F=-F\left(n_{1}\,E^{°}_{1}+n_{2}\,E^{°}_{2}F\right)$
$E^{V}=\frac{n_{1}E^{°}_{1}+n_{2}E^{°}_{2}}{n}=\frac{0.15\times1+0.50\times1}{2}=0.325$