Question

The oxidation potential of hydrogen electrode at $\mathrm{pH}=10\mathrm{and}{\mathrm{pH}}_2=1\mathrm{atm}$ is ________ .

• A. $0.51\mathrm{V}$
• B. $0.00\mathrm{V}$
• C. $+0.59\mathrm{V}$
• D. $-0.059\mathrm{V}$

Answer 1

The correct option is C

$+0.59\mathrm{V}$

Explanation for correct option:

(C) $+0.59\mathrm{V}$

Step 1: Concentration of ${\mathrm{H}}^{+}$

• Given that, $\mathrm{pH}=10$
• The concentration of $[\mathrm{H}+]={10}^{-10}\mathrm{M}$

Step 2: Oxidation reaction of hydrogen:

• Oxidation reaction of hydrogen:

$\frac{1}{2}{\mathrm{H}}_2\left(\mathrm{g}\right)\to {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{e}}^{-}$

Step 3: Nernst equation:

• Using the Nernst equation,

$$\begin{gathered} {\mathrm{E}}_{\mathrm{Ox}}={{\mathrm{E}}_{\mathrm{Ox}}}^{\circ }-\frac{0.0591}{\mathrm{n}}\log \frac{\left[{\mathrm{H}}^{+}\right]}{{\mathrm{pH}}_2} \\ \mathrm{n}=1(\mathrm{Number}\mathrm{of}\mathrm{electrons}\mathrm{involved}) \\ {\mathrm{E}}_{\mathrm{Ox}}=0-\frac{0.0591}{1}\log \frac{{10}^{-10}}{1} \\ {\mathrm{E}}_{\mathrm{Ox}}=+0.59\mathrm{V}. \end{gathered}$$

Explanation for incorrect options:

The oxidation potential of the hydrogen electrode is $+0.59\mathrm{V}$. Thus, options A, B, and D are incorrect.

Hence, the correct option is (C) $+0.59\mathrm{V}$.