Question

What is reduction potential of hydrogen gas electrode when pure hydrogen gas is at 1 atmospheric pressure and platinum electrode is in contact with HCl solution of pH 1 at 298 K?

• A. -0.1184 V
• B. -0.0592 V
• C. -0.0296 V
• D. -0.592 V

Hint:

For Hydrogen electrode: $E = -0.059 \times pH$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The reduction potential of a hydrogen electrode depends on the concentration of \( \text{H}^+ \) ions (and thus pH). We use the Nernst equation to find it.

Step 2: Key Formula or Approach:
For the half-reaction \( 2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2 \):
\[ \text{E}_{\text{red}} = \text{E}^{\circ}_{\text{red}} - \frac{0.0592}{\text{n}} \log \left( \frac{\text{P}_{\text{H}_2}}{[\text{H}^+]^2} \right) \]
Alternatively, simplified at \( 298 \text{ K} \): \( \text{E}_{\text{red}} = -0.0592 \times \text{pH} \).

Step 3: Detailed Explanation:
Given: pH \( = 1 \).
For a hydrogen electrode, \( \text{E}^{\circ}_{\text{red}} = 0.00 \text{ V} \).
Using the simplified relation:
\[ \text{E}_{\text{red}} = -0.0592 \times (1) = -0.0592 \text{ V} \]

Step 4: Final Answer:
The reduction potential is -0.0592 V.