The oxidation potential of a hydrogen electrode at `pH = 10` and `P_(H_(2)) =1` is

Explanation

To calculate the oxidation potential of a hydrogen electrode at pH 10 and \( P_{H_2} = 1 \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The oxidation reaction for the hydrogen electrode can be written as:
\[ \frac{1}{2} H_2 \rightarrow H^+ + e^- \]

2. **Use the Nernst Equation**:
The Nernst equation for this reaction is:
\[ E = E^\circ - \frac{0.0591}{n} \log \left( \frac{[H^+]}{[H_2]^{1/2}} \right) \]
where \( E \) is the oxidation potential, \( E^\circ \) is the standard electrode potential, \( n \) is the number of electrons transferred, and \([H^+]\) and \([H_2]\) are the concentrations of hydrogen ions and hydrogen gas, respectively.

3. **Determine Standard Potential**:
For the hydrogen electrode, the standard potential \( E^\circ \) is 0 V.

4. **Calculate Concentration of \( H^+ \)**:
Given that \( pH = 10 \):
\[ [H^+] = 10^{-pH} = 10^{-10} \, \text{mol/L} \]

5. **Concentration of \( H_2 \)**:
The pressure of hydrogen gas \( P_{H_2} = 1 \) atm can be considered as its concentration:
\[ [H_2] = 1 \, \text{atm} \]

6. **Substitute Values into Nernst Equation**:
Now, substituting the values into the Nernst equation:
\[ E = 0 - \frac{0.0591}{1} \log \left( \frac{10^{-10}}{1^{1/2}} \right) \]
Simplifying this gives:
\[ E = -0.0591 \log(10^{-10}) = -0.0591 \times (-10) = 0.591 \, \text{V} \]

7. **Final Result**:
Therefore, the oxidation potential of the hydrogen electrode at pH 10 is:
\[ E = 0.591 \, \text{V} \]