Chemistry - Electrochemistry Question with Solution | TestHub

The given standard reduction potential is $E_{\mathrm{red}}^{\circ }=-1.66\mathrm{V}$.

The standard oxidation potential is $E_{\mathrm{ox}}^{\circ }=-E_{\mathrm{red}}^{\circ }=+1.66\mathrm{V}$.

The given half-cell reaction is an oxidation reaction:

${\mathrm{Al}}_{(\mathrm{s})}\to {\mathrm{Al}}^{+3}(0.1\mathrm{M})+3{\mathrm{e}}^{-}$

Here, the number of electrons involved is $n=3$ and the concentration of ${\mathrm{Al}}^{+3}$ is $0.1\mathrm{M}$.

Using the Nernst equation for the oxidation potential:

$E_{\mathrm{ox}}=E_{\mathrm{ox}}^{\circ }-\frac{0.0592}{n}{\log }_{10}[{\mathrm{Al}}^{+3}]$

Substituting the values:

$E_{\mathrm{ox}}=1.66-\frac{0.0592}{3}{\log }_{10}(0.1)$

$E_{\mathrm{ox}}=1.66-0.0197\times (-1)$

$E_{\mathrm{ox}}=1.66+0.0197=+1.6797\mathrm{V}$

Rounding to three decimal places, we get $+1.679\mathrm{V}$.

Answer: $+1.679\mathrm{V}$