The reduction potentail of hydrogen half -cell will be negative if :

Options

A. `P(H_2) = 1 atm and [H^(+)] = 2.0 M`
B. `P (H_2) = 1atm and (H^(+) = 1.0| M`
C. `(PH_2) = 2atm and [H(+)] = 1.0M`
D. `P(H_2) = 2 atm and [H^(+)] = 2.0 M`

Explanation

To determine when the reduction potential of the hydrogen half-cell will be negative, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Reaction**:
The half-reaction for the hydrogen electrode can be represented as:
\[ \text{H}^+ + e^- \leftrightarrow \frac{1}{2} \text{H}_2 \]

2. **Standard Conditions**:
Under standard conditions, the standard reduction potential \(E^\circ\) for the hydrogen half-cell is defined as 0 V. This is the reference point.

3. **Nernst Equation**:
The Nernst equation relates the cell potential under non-standard conditions to the standard potential:
\[ E = E^\circ - \frac{0.0591}{n} \log Q \]
Where:
- \(E\) = cell potential
- \(E^\circ\) = standard cell potential (0 V for hydrogen)
- \(n\) = number of moles of electrons transferred (1 for hydrogen)
- \(Q\) = reaction quotient

4. **Reaction Quotient \(Q\)**:
For the hydrogen half-cell, the reaction quotient \(Q\) can be expressed as:
\[ Q = \frac{P_{\text{H}_2}}{[\text{H}^+]^2} \]
Where \(P_{\text{H}_2}\) is the pressure of hydrogen gas and \([\text{H}^+]\) is the concentration of hydrogen ions.

5. **Condition for Negative Potential**:
To find when \(E\) becomes negative, we need:
\[ E < 0 \implies 0 - \frac{0.0591}{1} \log Q < 0 \]
This simplifies to:
\[ \log Q > 0 \implies Q > 1 \]
Therefore, for \(Q > 1\), we need:
\[ \frac{P_{\text{H}_2}}{[\text{H}^+]^2} > 1 \implies P_{\text{H}_2} < [\text{H}^+]^2 \]

6. **Conclusion**:
The reduction potential of the hydrogen half-cell will be negative if the pressure of hydrogen gas is less than the square of the concentration of hydrogen ions.