For the following concentration cell, to be spontancous `Pt(H_2)P_1` atm. `|HCl |Pt(H_2) P_2 ` atm which of the following is correct ?

Options

A. `P_1 = P_2 `
B. `P_1 lt P_2 `
C. `P_1 gt P_2 `
D. can't be predicted.

Explanation

To solve the problem regarding the spontaneous reaction in the given concentration cell, we will follow these steps:

### Step 1: Understand the Cell Configuration
The cell is represented as:
\[ \text{Pt}(H_2) | P_1 \text{ atm} \, \text{HCl} | \text{Pt}(H_2) | P_2 \text{ atm} \]

This indicates that we have two hydrogen electrodes at different pressures (P1 and P2) with HCl solution in between.

### Step 2: Identify the Reactions
- **Left Electrode (Reduction)**: At the left electrode, hydrogen ions (H⁺) from HCl are reduced to hydrogen gas (H₂):
\[ 2H^+ + 2e^- \rightarrow H_2 \quad \text{(reduction)} \]

- **Right Electrode (Oxidation)**: At the right electrode, hydrogen gas (H₂) is oxidized to hydrogen ions (H⁺) and releases electrons:
\[ H_2 \rightarrow 2H^+ + 2e^- \quad \text{(oxidation)} \]

### Step 3: Write the Nernst Equation
The Nernst equation for the cell can be expressed as:
\[ E = E^\circ - \frac{RT}{nF} \ln Q \]
Where:
- \(E^\circ\) is the standard cell potential (0 for hydrogen electrodes),
- \(R\) is the gas constant (8.314 J/(K·mol)),
- \(T\) is the temperature in Kelvin (assumed to be 298 K),
- \(n\) is the number of moles of electrons transferred (2 in this case),
- \(F\) is Faraday's constant (96500 C/mol),
- \(Q\) is the reaction quotient.

### Step 4: Substitute Values into the Nernst Equation
Substituting the values into the Nernst equation:
\[ E = 0 - \frac{0.0591}{n} \log \left( \frac{[H^+]_{P_1}}{[H^+]_{P_2}} \right) \]
Since \(Q = \frac{P_1}{P_2}\) (where \(P_1\) and \(P_2\) are the pressures of hydrogen gas at the two electrodes):
\[ E = -\frac{0.0591}{2} \log \left( \frac{P_1}{P_2} \right) \]

### Step 5: Determine Conditions for Spontaneity
For the reaction to be spontaneous, the cell potential \(E\) must be positive:
\[ -\frac{0.0591}{2} \log \left( \frac{P_1}{P_2} \right) > 0 \]
This implies:
\[ \log \left( \frac{P_1}{P_2} \right) < 0 \quad \Rightarrow \quad \frac{P_1}{P_2} < 1 \quad \Rightarrow \quad P_1 < P_2 \]

### Conclusion
Thus, for the concentration cell to be spontaneous, the correct relationship is:
\[ P_1 < P_2 \]

### Final Answer
The correct option is that \( P_1 < P_2 \).

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