Chemistry - Electrochemistry Question with Solution | TestHub

The given half-cell is represented as an oxidation electrode: $\text{Pt (s)}|{\text{H}}_2\text{(g)}|{\text{H}}^{+}\text{(aq)}$.

The corresponding oxidation half-cell reaction is:
${\text{H}}_2\text{(g)}\to 2{\text{H}}^{+}\text{(aq)}+2{\text{e}}^{-}$

Using the Nernst equation for the oxidation potential:
$E=E_{{\text{H}}_2/{\text{H}}^{+}}^{\circ }-\frac{0.059}{n}\log \frac{[{\text{H}}^{+}{]}^2}{P_{{\text{H}}_2}}$

Given values:
$E_{{\text{H}}_2/{\text{H}}^{+}}^{\circ }=0$ V
$n=2$
$[{\text{H}}^{+}]=0.02$ M (since HCl is a strong monoprotic acid)
$P_{{\text{H}}_2}=2$ atm

Substituting the values into the Nernst equation:
$E=0-\frac{0.059}{2}\log \frac{(0.02{)}^2}{2}$

$E=-0.0295\log \frac{4\times 10^{-4}}{2}$

$E=-0.0295\log (2\times 10^{-4})$

$E=-0.0295(\log 2+\log 10^{-4})$

$E=-0.0295(0.3010-4)$

$E=-0.0295\times (-3.699)$

$E=0.10912$ V $\approx 0.109$ V

Answer: $0.109$ V