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

Options

A. `-0.59V`
B. `0.00V`
C. `+0.59V`
D. `0.059V`

Explanation

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

### Step-by-Step Solution:

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

2. **Use the Nernst Equation**:
The Nernst equation for the oxidation potential can be expressed as:
\[ E_{\text{ox}} = E^\circ_{\text{ox}} - \frac{0.059}{n} \log \left( \frac{[H^+]}{p_{H_2}} \right) \]
where:
- \( E^\circ_{\text{ox}} \) is the standard oxidation potential (which is 0 V for the hydrogen electrode),
- \( n \) is the number of electrons transferred (which is 1 for this reaction),
- \([H^+]\) is the concentration of hydrogen ions,
- \( p_{H_2} \) is the partial pressure of hydrogen.

3. **Substituting Known Values**:
Given that \( pH = 10 \), we can find \([H^+]\):
\[ [H^+] = 10^{-pH} = 10^{-10} \, \text{mol/L} \]
Since \( p_{H_2} = 1 \, \text{atm} \), we can substitute these values into the Nernst equation:
\[ E_{\text{ox}} = 0 - \frac{0.059}{1} \log \left( \frac{10^{-10}}{1} \right) \]

4. **Calculating the Logarithm**:
The logarithm simplifies as follows:
\[ \log(10^{-10}) = -10 \]
Therefore, substituting this back into the equation gives:
\[ E_{\text{ox}} = -0.059 \times (-10) = 0.59 \, \text{V} \]

5. **Final Result**:
Thus, the oxidation potential of the hydrogen electrode at \( pH = 10 \) and \( p_{H_2} = 1 \, \text{atm} \) is:
\[ E_{\text{ox}} = 0.59 \, \text{V} \]