Chemistry - Electrochemistry Question with Solution | TestHub

To calculate the electrode potential, we use the Nernst equation:

$$E=E^{\circ }-\frac{0.06}{n}\log Q$$

where $E^{\circ }$ is the standard electrode potential, $n$ is the number of electrons transferred, and $Q$ is the reaction quotient.

For the reaction ${\mathrm{Zn}}^{2+}+2e^{-}\rightleftharpoons \mathrm{Zn}$, $n=2$ and $Q=\frac{1}{[{\mathrm{Zn}}^{2+}]}$.

Given $E^{\circ }=-0.76\mathrm{V}$ and $[{\mathrm{Zn}}^{2+}]=0.1\mathrm{M}$,

$$E=-0.76-\frac{0.06}{2}\log \frac{1}{0.1}=-0.76-0.03\log 10=-0.76-0.03=-0.79\mathrm{V}$$

Therefore, the electrode potential $E_{({\mathrm{Zn}}^{2+}/\mathrm{Zn})}$ is $-0.79\mathrm{V}$.