Given:`Pt(s)"|"underset(P_1atm)Cl_2(g)"|"Cl^(-)(C_1)"||"Cl^(-)(C_2)"|"underset(P_2atm)``(Cl)_(2)`(g)| `Pt(s)`
identify in which of following condition working of cell takes place:

Options

A. `C_1gtC_2"and"P_1=P_2`
B. `P_1gtP_2"and"C_1=C_2`
C. `C_2gtC_1"and"P_1=P_2`
D. `P_1ltP_2"and"C_1=C_2`

Answer

To solve the problem, we need to analyze the given electrochemical cell representation and determine the conditions under which the cell operates effectively. Here's the step-by-step solution:

### Step 1: Write the Cell Representation
The cell is represented as:
\[ \text{Pt(s)} | \text{Cl}_2(g) | \text{Cl}^-(C_1) || \text{Cl}^-(C_2) | \text{Cl}_2(g) | \text{Pt(s)} \]

### Step 2: Identify the Anode and Cathode
In this cell:
- The anode is where oxidation occurs, and the cathode is where reduction occurs.
- Chlorine gas (\( \text{Cl}_2 \)) is involved in both oxidation and reduction reactions.

### Step 3: Write the Half-Reactions
1. **Oxidation at Anode**:
\[ \text{2Cl}^- \rightarrow \text{Cl}_2 + 2e^- \]
(This occurs at the anode)

2. **Reduction at Cathode**:
\[ \text{Cl}_2 + 2e^- \rightarrow \text{2Cl}^- \]
(This occurs at the cathode)

### Step 4: Overall Reaction
Combining the half-reactions gives:
\[ \text{2Cl}^- (C_1) + \text{Cl}_2 (P_2) \rightarrow \text{Cl}_2 (P_1) + \text{2Cl}^- (C_2) \]

### Step 5: Calculate Reaction Quotient (Q)
The reaction quotient \( Q \) is defined as:
\[ Q = \frac{[\text{Cl}^-]^2 \cdot P_1}{[\text{Cl}^-]^2 \cdot P_2} = \frac{C_2^2 \cdot P_1}{C_1^2 \cdot P_2} \]

### Step 6: Determine Standard Cell Potential (E°cell)
Since both half-reactions involve the same species (chlorine), the standard cell potential \( E°_{cell} \) is:
\[ E°_{cell} = E°_{cathode} - E°_{anode} = 0 \]

### Step 7: Apply the Nernst Equation
Using the Nernst equation:
\[ E_{cell} = E°_{cell} - \frac{0.059}{n} \log Q \]
Substituting \( E°_{cell} = 0 \) and \( n = 2 \):
\[ E_{cell} = -\frac{0.059}{2} \log \left( \frac{C_2^2 \cdot P_1}{C_1^2 \cdot P_2} \right) \]

### Step 8: Conditions for Cell to Work
For the cell to work, \( E_{cell} \) must be positive:
\[ -\frac{0.059}{2} \log \left( \frac{C_2^2 \cdot P_1}{C_1^2 \cdot P_2} \right) > 0 \]
This implies:
\[ \log \left( \frac{C_2^2 \cdot P_1}{C_1^2 \cdot P_2} \right) < 0 \]
Thus, \( \frac{C_2^2 \cdot P_1}{C_1^2 \cdot P_2} < 1 \).

### Step 9: Analyze Conditions
1. If \( C_1 = C_2 \), then \( P_2 > P_1 \).
2. If \( P_1 = P_2 \), then \( C_1 > C_2 \).

### Conclusion
The conditions under which the cell operates effectively are:
1. \( C_1 = C_2 \) and \( P_2 > P_1 \)
2. \( P_1 = P_2 \) and \( C_1 > C_2 \)