Chemistry - Electrochemistry Question with Solution | TestHub

The reduction half-reaction for the magnesium electrode is:

${\text{Mg}}^{2+}+2e^{-}\to \text{Mg}$

Here, the number of electrons involved, $n=2$.

Using the Nernst equation for the electrode potential:

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

Substituting the given values ($E^{\circ }=-2.36$ V, $[{\text{Mg}}^{2+}]=0.01$ M):

$E=-2.36-\frac{0.0591}{2}\log \frac{1}{0.01}$

$E=-2.36-0.02955\log 100$

$E=-2.36-0.02955\times 2$

$E=-2.36-0.0591$

$E=-2.4191$ V $\approx -2.42$ V

Answer: $-2.42$ V