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 use the Nernst equation:

### Step-by-Step Solution:

1. **Understanding the Reaction**: The reduction reaction for the hydrogen half-cell can be represented as:
\[ \text{H}^+ + e^- \rightarrow \frac{1}{2} \text{H}_2 \]
This indicates that hydrogen ions (H⁺) gain electrons to form hydrogen gas (H₂).

2. **Nernst Equation**: The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ - \frac{0.059}{n} \log Q \]
Where:
- \(E_{\text{cell}}\) is the cell potential.
- \(E^\circ\) is the standard reduction potential (0 V for the hydrogen half-cell).
- \(n\) is the number of moles of electrons transferred (1 for hydrogen).
- \(Q\) is the reaction quotient.

3. **Setting Up the 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 partial pressure of hydrogen gas and \([\text{H}^+]\) is the concentration of hydrogen ions.

4. **Substituting into the Nernst Equation**:
Given that \(E^\circ = 0\) for the hydrogen half-cell, we can rewrite the equation as:
\[ E_{\text{cell}} = -\frac{0.059}{1} \log Q \]
This simplifies to:
\[ E_{\text{cell}} = -0.059 \log Q \]

5. **Analyzing Conditions for Negative Potential**: For \(E_{\text{cell}}\) to be negative:
\[ -0.059 \log Q < 0 \implies \log Q > 0 \implies Q < 1 \]
This means that the partial pressure of hydrogen gas must be less than the square of the concentration of hydrogen ions:
\[ P_{\text{H}_2} < [\text{H}^+]^2 \]

6. **Evaluating Given Options**:
- If we consider different scenarios with varying pressures and concentrations, we can calculate \(Q\) for each case to determine if \(E_{\text{cell}}\) is negative.
- For example, if \(P_{\text{H}_2} = 2 \, \text{atm}\) and \([\text{H}^+] = 1 \, \text{M}\), then:
\[ Q = \frac{2}{1^2} = 2 \quad (\text{which gives a negative } E_{\text{cell}}) \]
- If \(P_{\text{H}_2} = 1 \, \text{atm}\) and \([\text{H}^+] = 2 \, \text{M}\), then:
\[ Q = \frac{1}{2^2} = 0.25 \quad (\text{which gives a positive } E_{\text{cell}}) \]

7. **Conclusion**: The reduction potential of the hydrogen half-cell will be negative when the partial pressure of hydrogen gas is greater than the square of the concentration of hydrogen ions.

### Final Answer:
The reduction potential of the hydrogen half-cell will be negative if the pressure of hydrogen gas is sufficiently high relative to the concentration of hydrogen ions.