Chemistry - Electrochemistry Question with Solution | TestHub

The given reaction is an oxidation reaction: ${\mathrm{Ag}}_{(\mathrm{s})}\to {\mathrm{Ag}}_{(\mathrm{aq})}^{+}+{\mathrm{e}}^{-}$

The standard oxidation potential is the negative of the standard reduction potential:
${\mathrm{E}}_{\text{ox}}^{\circ }=-{\mathrm{E}}_{\text{red}}^{\circ }=-0.80\mathrm{V}$

According to the Nernst equation for the oxidation half-cell at $298\mathrm{K}$:
${\mathrm{E}}_{\text{ox}}={\mathrm{E}}_{\text{ox}}^{\circ }-\frac{0.0592}{n}{\log }_{10}[{\mathrm{Ag}}^{+}]$

Here, the number of electrons transferred is $n=1$ and the concentration is $[{\mathrm{Ag}}^{+}]=0.01\mathrm{M}=10^{-2}\mathrm{M}$. Substituting these values:
${\mathrm{E}}_{\text{ox}}=-0.80-\frac{0.0592}{1}{\log }_{10}(10^{-2})$

${\mathrm{E}}_{\text{ox}}=-0.80-0.0592\times (-2)$

${\mathrm{E}}_{\text{ox}}=-0.80+0.1184$

${\mathrm{E}}_{\text{ox}}=-0.6816\mathrm{V}$

Answer: $-0.6816\mathrm{V}$