`Pt(Cl_(2))(p_(1))|HCl(0.1M)|(Cl_(2))(p_(2)),Pt` cell reaction will be endergonic if

Options

A. `p_(1)=p_(2)`
B. `p_(1)gtp_(2)`
C. `p_(2)gtp_(1)`
D. `p_(1)=p_(2)=1atm`

Explanation

To determine the conditions under which the cell reaction \( \text{Pt(Cl}_2)(p_1)|\text{HCl}(0.1M)|(\text{Cl}_2)(p_2),\text{Pt} \) will be endergonic, we need to analyze the relationship between the Gibbs free energy change (\( \Delta G \)) and the cell potential (\( E_{cell} \)).

### Step-by-Step Solution:

1. **Understanding Endergonic Reaction**:
- An endergonic reaction is characterized by a positive Gibbs free energy change (\( \Delta G > 0 \)). This indicates that the reaction is non-spontaneous under the given conditions.

2. **Relation Between Gibbs Free Energy and Cell Potential**:
- The relationship between Gibbs free energy change and cell potential is given by the equation:
\[ \Delta G = -nFE_{cell} \]
- Where:
- \( n \) = number of moles of electrons transferred in the reaction,
- \( F \) = Faraday's constant (approximately 96485 C/mol),
- \( E_{cell} \) = cell potential.

3. **Condition for Endergonic Reaction**:
- For the reaction to be endergonic, we need:
\[ \Delta G > 0 \implies -nFE_{cell} > 0 \implies E_{cell} < 0 \]
- Therefore, the cell potential must be negative.

4. **Determining the Cell Potential**:
- The cell potential for the given cell can be calculated using the Nernst equation:
\[ E_{cell} = E^{\circ} - \frac{0.0591}{n} \log \left( \frac{P_{2}}{P_{1}} \right) \]
- Here, \( P_1 \) is the pressure of \( Cl_2 \) in the first half-cell, and \( P_2 \) is the pressure of \( Cl_2 \) in the second half-cell.

5. **Analyzing the Nernst Equation**:
- For \( E_{cell} \) to be negative:
\[ E^{\circ} - \frac{0.0591}{n} \log \left( \frac{P_{2}}{P_{1}} \right) < 0 \]
- Rearranging gives:
\[ \frac{0.0591}{n} \log \left( \frac{P_{2}}{P_{1}} \right) > E^{\circ} \]

6. **Condition on Pressures**:
- This implies that if \( P_2 \) is greater than \( P_1 \), the logarithmic term becomes positive, which can lead to \( E_{cell} < 0 \) if \( E^{\circ} \) is not excessively positive.
- Therefore, for the reaction to be endergonic, we conclude:
\[ P_{2} > P_{1} \]

### Final Conclusion:
The cell reaction will be endergonic if \( P_{2} > P_{1} \).