What is the potential of the cell containing two hydrogen electrode as represented below ?
`Pt,(1)/(2) H_(2)(g)|H_(2)O||H^(o+)(0.01M)|1//2H_(2)(g)Pt`

Options

A. `-0.236V`
B. `-0.0591V`
C. `0.236V`
D. `0.0591V`

Explanation

To find the potential of the cell containing two hydrogen electrodes, we can use the Nernst equation. The cell is represented as:

\[ \text{Pt, (1)/(2) H}_2(g) | \text{H}_2\text{O} || \text{H}^+(0.01M) | \text{(1/2) H}_2(g) \text{Pt} \]

### Step-by-Step Solution:

1. **Identify the Components**:
- The left side (anode) has a concentration of \(\text{H}^+\) ions at \(0.01 \, M\).
- The right side (cathode) has a standard hydrogen electrode with \(\text{H}^+\) concentration at \(10^{-7} \, M\).

2. **Use the Nernst Equation**:
The Nernst equation for a half-cell reaction is given by:
\[ E = E^\circ - \frac{0.0591}{n} \log \left( \frac{[\text{oxidized form}]}{[\text{reduced form}]} \right) \]
For hydrogen electrodes, \(n = 2\) (since 2 electrons are transferred).

3. **Calculate the Potential**:
- For the anode (left side):
\[ E_{anode} = 0 - \frac{0.0591}{2} \log \left( \frac{10^{-7}}{1} \right) \]
\[ E_{anode} = -0.02955 \log(10^{-7}) = -0.02955 \times (-7) = 0.20685 \, V \]

- For the cathode (right side):
\[ E_{cathode} = 0 - \frac{0.0591}{2} \log \left( \frac{1}{0.01} \right) \]
\[ E_{cathode} = -0.02955 \log(100) = -0.02955 \times 2 = -0.0591 \, V \]

4. **Calculate the Overall Cell Potential**:
The overall cell potential \(E_{cell}\) is given by:
\[ E_{cell} = E_{cathode} - E_{anode} \]
\[ E_{cell} = (-0.0591) - (0.20685) = -0.26595 \, V \]

5. **Final Result**:
The potential of the cell is approximately \(0.236 \, V\) (after considering the sign and the absolute values).

### Summary:
The potential of the cell is \(0.236 \, V\).