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 ?

Options

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

Explanation

To find the oxidation potential of the hydrogen gas electrode made by dipping a platinum wire in a solution of HCl with a pH of 10 and passing hydrogen gas around it at one atm pressure, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Reaction**: The hydrogen electrode reaction can be represented as:
\[ 2H^+ + 2e^- \leftrightarrow H_2 \]
Here, \(H^+\) ions are reduced to hydrogen gas.

2. **Identify Standard Electrode Potential**: The standard electrode potential (\(E^\circ\)) for the standard hydrogen electrode (SHE) is defined as 0 V.

3. **Use the Nernst Equation**: The Nernst equation for the hydrogen electrode can be expressed as:
\[ E = E^\circ - \frac{0.059}{n} \log \left( \frac{P_{H_2}}{[H^+]^2} \right) \]
Where:
- \(E\) is the electrode potential.
- \(E^\circ\) is the standard electrode potential (0 V for SHE).
- \(n\) is the number of moles of electrons transferred (which is 2 for this reaction).
- \(P_{H_2}\) is the pressure of hydrogen (1 atm).
- \([H^+]\) is the concentration of hydrogen ions.

4. **Calculate the Concentration of \(H^+\)**: Given that the pH of the solution is 10, we can find the concentration of \(H^+\) ions:
\[ [H^+] = 10^{-pH} = 10^{-10} \, \text{M} \]

5. **Insert Values into Nernst Equation**: Now we can substitute the values into the Nernst equation:
- \(E^\circ = 0\)
- \(n = 2\)
- \(P_{H_2} = 1 \, \text{atm}\)
- \([H^+] = 10^{-10} \, \text{M}\)

Thus, the equation becomes:
\[ E = 0 - \frac{0.059}{2} \log \left( \frac{1}{(10^{-10})^2} \right) \]

6. **Calculate the Logarithm**: Simplifying the logarithm:
\[ \log \left( \frac{1}{(10^{-10})^2} \right) = \log \left( 10^{20} \right) = 20 \]

7. **Final Calculation**: Substitute back into the equation:
\[ E = -\frac{0.059}{2} \times 20 = -0.059 \times 10 = -0.59 \, \text{V} \]

8. **Determine Oxidation Potential**: Since we are looking for the oxidation potential, we take the negative of this value:
\[ \text{Oxidation Potential} = 0.59 \, \text{V} \]

### Final Answer:
The oxidation potential of the electrode is **0.59 V**.

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