Chemistry - Electrochemistry Question with Solution | TestHub

The given standard reduction potential is $E^{\circ }({\mathrm{Zn}}^{+2}\mid \mathrm{Zn})=-0.76\mathrm{V}$.

The standard oxidation potential is $E_{\mathrm{ox}}^{\circ }=-E_{\mathrm{red}}^{\circ }=+0.76\mathrm{V}$.

The oxidation half-reaction is:
${\mathrm{Zn}}_{(\mathrm{s})}\to {\mathrm{Zn}}_{(\mathrm{aq})}^{+2}+2{\mathrm{e}}^{-}$

Here, the number of electrons transferred is $n=2$ and the concentration of ${\mathrm{Zn}}^{+2}$ is $0.01\mathrm{M}=10^{-2}\mathrm{M}$.

Using the Nernst equation for the oxidation potential at $298\mathrm{K}$:
$E_{\mathrm{ox}}=E_{\mathrm{ox}}^{\circ }-\frac{0.0592}{n}{\log }_{10}[{\mathrm{Zn}}^{+2}]$

Substituting the values:
$E_{\mathrm{ox}}=0.76-\frac{0.0592}{2}{\log }_{10}(10^{-2})$

$E_{\mathrm{ox}}=0.76-0.0296\times (-2)$

$E_{\mathrm{ox}}=0.76+0.0592=+0.8192\mathrm{V}$

Answer: $+0.8192\mathrm{V}$