`Pt(H_(2))(p_(1))|H^(o+)(1M)|(H_(2))(p_(2)),Pt` cell reaction will be exergonic if

Options

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

Explanation

To determine the conditions under which the cell reaction \( Pt(H_2)(p_1)|H^+(1M)|(H_2)(p_2),Pt \) will be exergonic, we can follow these steps:

### Step 1: Identify the half-cell reactions
In the given cell, we have two half-cell reactions:
1. **Anode (oxidation)**:
\[ H_2 (p_1) \rightarrow 2H^+ + 2e^- \]
2. **Cathode (reduction)**:
\[ 2H^+ + 2e^- \rightarrow H_2 (p_2) \]

### Step 2: Write the overall cell reaction
Combining the two half-reactions, we get the overall cell reaction:
\[ H_2 (p_1) \rightarrow H_2 (p_2) + 2H^+ \]

### Step 3: Use the Nernst equation
The Nernst equation relates the cell potential to the concentrations and pressures of the reactants and products:
\[ E_{cell} = E^0_{cell} - \frac{0.0591}{n} \log Q \]
where \( n \) is the number of electrons transferred (which is 2 in this case), and \( Q \) is the reaction quotient.

### Step 4: Calculate the reaction quotient \( Q \)
For our reaction, the reaction quotient \( Q \) is given by:
\[ Q = \frac{[H^+]^2}{p_1/p_2} \]
Since the concentration of \( H^+ \) is 1 M, we can substitute:
\[ Q = \frac{(1)^2}{\frac{p_2}{p_1}} = \frac{p_1}{p_2} \]

### Step 5: Substitute \( Q \) into the Nernst equation
Substituting \( Q \) into the Nernst equation gives:
\[ E_{cell} = E^0_{cell} - \frac{0.0591}{2} \log \left( \frac{p_1}{p_2} \right) \]

### Step 6: Determine the conditions for exergonic reaction
For the reaction to be exergonic, we need:
1. \( \Delta G < 0 \) which implies \( E_{cell} > 0 \)
2. From the Nernst equation, we need:
\[ E^0_{cell} - \frac{0.0591}{2} \log \left( \frac{p_1}{p_2} \right) > 0 \]

### Step 7: Analyze the logarithmic term
Rearranging gives:
\[ \frac{0.0591}{2} \log \left( \frac{p_1}{p_2} \right) < E^0_{cell} \]
This means that for \( E_{cell} \) to be positive, \( \log \left( \frac{p_1}{p_2} \right) \) must be positive, which occurs when:
\[ p_1 > p_2 \]

### Conclusion
Thus, the cell reaction will be exergonic if:
\[ p_1 > p_2 \]