Question

A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of pH = 10 and by passing hydrogen gas around the platinum wire at one atm pressure. The oxidation potential of electrode would be

• A. 0.059 V
• B. 0.59 V
• C. 0.118 V
• D. 1.18 V

Answer 1

The Correct Option is B

To determine the oxidation potential of the hydrogen gas electrode under the given conditions, we need to use the Nernst Equation. The hydrogen electrode is set up in a solution of HCl with a pH of 10, which means the concentration of H⁺ ions is very low.

The pH is related to the concentration of hydrogen ions \([H^+]\) by the formula:

[\text{H}^+] = 10^{-\text{pH}}

Given that the pH is 10:

[\text{H}^+] = 10^{-10} \, \text{mol/L}

The Nernst equation for the hydrogen electrode reaction, \text{H}_2(g) \rightleftharpoons 2\text{H}^+(aq) + 2e^-\hspace{-0.25em}, is as follows:

\text{E} = \text{E}^\circ - \frac{0.0591}{2} \log \frac{1}{[\text{H}^+]^2}

Since it is a standard hydrogen electrode:

• The standard electrode potential, \text{E}^\circ = 0 \, \text{V}
• The pressure of \text{H}_2 is at 1 atm.

Substituting these values into the Nernst Equation, we get:

\text{E} = 0 - \frac{0.0591}{2} \log \frac{1}{(10^{-10})^2}

Simplifying the logarithmic term:

\log \frac{1}{10^{-20}} = \log(10^{20}) = 20

Thus, the oxidation potential becomes:

\text{E} = - \frac{0.0591}{2} \times 20 = - 0.0591 \times 10 = - 0.591 \, \text{V}

The negative sign indicates the electrode is functioning as an anodic half-cell, hence the oxidation potential is 0.59 \, \text{V}, which matches the given correct answer.