Chemistry - Electrochemistry Question with Solution | TestHub

The reduction half-reaction is ${\text{Cu}}^{+2}+2e^{-}\to \text{Cu}$.

Using the Nernst equation for the electrode potential:

$E=E^{\circ }-\frac{0.0591}{n}\log \frac{1}{[{\text{Cu}}^{+2}]}$

Given $E^{\circ }=0.34$ V, $n=2$, and $[{\text{Cu}}^{+2}]=0.1$ M.

$E=0.34-\frac{0.0591}{2}\log \frac{1}{0.1}$

$E=0.34-0.0295\log 10$

$E=0.34-0.0295\times 1$

$E=0.3105$ V $\approx 0.31$ V.

Answer: $0.31$ V