A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of $\mathrm{pH}=10$ and by…

A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of $\mathrm{pH}=10$ and by passing hydrogen gas around the platinum wire at $1 \mathrm{~atm}$ pressure. The oxidation potential of electrode would be
  1. $0.059 \mathrm{~V}$
  2. $0.59 \mathrm{~V}$
  3. $0.118 \mathrm{~V}$
  4. $1.18 \mathrm{~V}$

Solution

For hydrogen electrode, oxidation half reaction is
$\underset{(1 \mathrm{~atm})}{\mathrm{H}_2} \longrightarrow \underset{(\text{At } \mathrm{pH~} 10)}{2 \mathrm{H}^{+}}+2 \mathrm{e}^{-}$
If $\mathrm{pH}=10$
$\mathrm{H}^{+}=1 \times 10^{-\mathrm{pH}}=1 \times 10^{-10}$
From Nernst equation,
$E_{\text {cell }}=E_{\text {cell }}^0-\frac{0.0591}{2} \log \frac{\left[\mathrm{H}^{+}\right]^2}{P_{\mathrm{H}_2}}$
For hydrogen electrode, $E_{\text {cell }}^{\circ}=0$
$\begin{aligned}
E_{\text {cell }} & =-\frac{0.0591}{2} \log \frac{\left(10^{-10}\right)^2}{1} \\
& =+\frac{0.0591 \times 2}{2} \log \frac{1}{10^{-10}} \\
& =0.0591 \log 10^{10} \\
& =0.0591 \times 10=0.591 \mathrm{~V}
\end{aligned}$

Asked in: NEET 2013 (All India)

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