If the pressure of hydrogen gas is increased from 1 atm. To 100 atm, keeping the hydrogen ion concentration constant at 1 M, the reduction potential of the hydrogen half cell is at `25^(@)C` will be

Options

A. 0.059 V
B. `-0.059V`
C. `0.295V`
D. `0.118V`

Explanation

To solve the problem, we need to determine the reduction potential of the hydrogen half-cell when the pressure of hydrogen gas is increased from 1 atm to 100 atm, while keeping the hydrogen ion concentration constant at 1 M.

### Step-by-Step Solution:

1. **Identify the Standard Reduction Potential**:
The standard reduction potential (E°) for the hydrogen half-cell reaction is defined as:
\[ \text{H}^+ + e^- \rightarrow \frac{1}{2} \text{H}_2(g) \]
At standard conditions (1 atm, 1 M), E° = 0 V.

2. **Use the Nernst Equation**:
The Nernst equation for the hydrogen half-cell can be written as:
\[ E = E° - \frac{0.0591}{n} \log \left( \frac{P_{\text{H}_2}^{1/2}}{[\text{H}^+]}\right) \]
Here, \( n \) is the number of electrons transferred (which is 1 for the hydrogen half-cell), \( P_{\text{H}_2} \) is the pressure of hydrogen gas, and \([\text{H}^+]\) is the concentration of hydrogen ions.

3. **Substituting Values for Initial Conditions**:
For the initial condition where \( P_{\text{H}_2} = 1 \text{ atm} \) and \([\text{H}^+] = 1 \text{ M}\):
\[ E_1 = 0 - \frac{0.0591}{1} \log \left( \frac{1^{1/2}}{1} \right) \]
Since \(\log(1) = 0\):
\[ E_1 = 0 \text{ V} \]

4. **Substituting Values for Final Conditions**:
Now, for the condition where \( P_{\text{H}_2} = 100 \text{ atm} \):
\[ E_2 = 0 - \frac{0.0591}{1} \log \left( \frac{100^{1/2}}{1} \right) \]
Simplifying further:
\[ E_2 = -0.0591 \log(10) \]
Since \(\log(10) = 1\):
\[ E_2 = -0.0591 \text{ V} \]

5. **Final Result**:
Therefore, the reduction potential of the hydrogen half-cell at 100 atm is:
\[ E_2 = -0.0591 \text{ V} \]

### Conclusion:
The reduction potential of the hydrogen half-cell when the pressure of hydrogen gas is increased to 100 atm, while keeping the hydrogen ion concentration constant at 1 M, is \(-0.0591 \text{ V}\).