Question

Calculate the reduction potential for the following half cell at ${25}^{o}C$.

$Mg\left(s\right)|M{g}^{2+}(aq,1\times {10}^{-4}M)E_{Mg/M{g}^{{}^{2+}}}^0=+2.36V$

• A. $E_{M{g}^{2+}/Mg}=1.34V$
• B. $E_{M{g}^{2+}/Mg}=-1.34V$
• C. $E_{M{g}^{2+}/Mg}=2.48V$
• D. $E_{M{g}^{2+}/Mg}=-2.48V$

Answer 1

The correct option is D $E_{M{g}^{2+}/Mg}=-2.48V$

Given,
The standard oxidation potential of the electrode $Mg/M{g}^{2+}$ is $2.36V$.
The reduction potential of the electrode $M{g}^{2+}/Mg$ is $-2.36V$.
The reduction reaction of the electrode is
$M{g}^{2+}\left(aq\right)+2e\rightleftharpoons Mg\left(s\right)\left(reduction\right)$

Nernst equation for the half cell is

$E_{M^{n+}/M}=E_{M^{n+}/M}^0-\frac{2.303RT}{nF}log\frac{\left[M\right(s\left)\right]}{\left[M^{n+}\right(aq\left)\right]}$

n is the number of electrons involved in the half cell reaction.
$E_{M{g}^{2+}/Mg}=E_{M{g}^{2+}/Mg}^o-\frac{0.0591}{2}log\frac{1}{{10}^{-4}}$

$E_{M{g}^{2+}/Mg}=E_{M{g}^{2+}/Mg}^o+\frac{0.0591}{2}log{10}^{-4}$

$E_{M{g}^{2+}/Mg}=-2.36-\frac{4\times 0.0591}{2}$
$E_{M{g}^{2+}/Mg}=-2.48V$