Q5 / 160 · WBJEE 2010 · Chemistry
The potential of a hydrogen electrode at $\mathrm{pH}=10$ is
- A$0.59 \mathrm{~V}$
- B$0.00 \mathrm{~V}$
- Correct answerC$-0.59 \mathrm{~V}$
- D-0.059
Solution
Hints : $\mathrm{H}^{+}(\mathrm{pH}=10)\left|\mathrm{H}_2(1 \mathrm{~atm})\right| \operatorname{Pt}(\mathrm{s})$
Reaction: $2 \mathrm{H}^{+}\left(\mathrm{p}^{\mathrm{H}}=10\right)+2 \mathrm{e} \rightarrow \mathrm{H}_2(1 \mathrm{~atm})$
$$
\begin{aligned}
& \mathrm{E}=\mathrm{E}^0-\frac{0.0591}{2} \log \left(\frac{\mathrm{P}_{\mathrm{H}_2}}{\left[\mathrm{H}^{+}\right]^2}\right) \\
& =0-\frac{0.0591}{2} \log \frac{1}{\left(10^{-10}\right)^2}=-\frac{0.0591}{2} \times 2 \log \frac{1}{10^{-10}}=-0.0591 \times 10=-0.591 \\
& \text { i.e. } E=-0.591 \mathrm{~V} \\
&
\end{aligned}
$$
Hence, the correct answer is option: 'C'.