Q29 / 160 · AP EAMCET 2021 (25 Aug Shift 1) · Chemistry
A hydrogen electrode is made by dipping platinum wire in a solution of nitric acid of $\mathrm{pH}=9$ and passing hydrogen gas around the platinum wire at $1.2 \mathrm{~atm}$ pressure. The oxidation potential of such an electrode equals .V.
- A+0.059
- B-0.0531
- C-0.059
- Correct answerD+0.0531
Solution
For hydrogen electrode, oxidation half reaction is,
$$
\underset{1.2 \mathrm{~atm}}{\mathrm{H}_2} \longrightarrow \underset{\text { (at pH 9) }}{2 \mathrm{H}^{+}}+2 e^{-}
$$
If $\mathrm{pH}=9$ then $\mathrm{H}^{+}$ion $=1 \times 10^{-\mathrm{pH}}=1 \times 10^{-9}$
Using Nernst equation,
$$
E_{\text {cell }}=E_{\text {cell }}^0-\frac{0.0591}{n} \log \frac{\left[\mathrm{H}^{+}\right]^2}{\mathrm{p}_{\mathrm{H}_2}}
$$
For hydrogen electrode, $E_{\text {cell }}^0=0, n=2$
$$
\begin{aligned}
& E_{\text {Cell }}=0-\frac{0.0591}{2} \log \frac{\left(10^{-9}\right)^2}{1.2} \\
& E_{\text {Cell }}=-\frac{0.0591}{2}\left[\log 10^{-18}-\log 1.2\right] \Rightarrow E_{\text {Cell }}=+0.0531
\end{aligned}
$$
Hence, the correct answer is option: 'D'.