STATEMENT-1: `E_(cell)^(@)=0` for a chloride ion concentration cell.
STATEMENT-2: For this concentration cell where `E_(cell)=(RT)/(nF)In[Cl^-]_(LHS)/([Cl^-]_(RHS))`

Options

A. If both the statements are TRUE and STATEMENTS-2 is the correct explantion of STATEMENTS-10
B. If both the statements are TRUE but STATEMENTS-2 is NOT the correct explanation of STATEMENTS-12
C. If STATEMENTS-1 is TRUE and STATEMENTS-2 is FALSE
D. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE

Answer

To solve the question regarding the statements about a chloride ion concentration cell, we will break down the solution step by step.

### Step 1: Understanding Statement 1
**Statement 1**: \( E_{\text{cell}}^\circ = 0 \) for a chloride ion concentration cell.

- A concentration cell consists of two half-cells that have the same chemical species but different concentrations. In this case, we have chloride ions (\( \text{Cl}^- \)) at different concentrations in each half-cell.
- The standard electrode potential (\( E_{\text{cell}}^\circ \)) for a concentration cell is zero because the two half-reactions involve the same species. The oxidation potential at the anode is equal in magnitude but opposite in sign to the reduction potential at the cathode.

**Conclusion for Step 1**: Therefore, Statement 1 is true.

### Step 2: Understanding Statement 2
**Statement 2**: For this concentration cell, \( E_{\text{cell}} = \frac{RT}{nF} \ln \left( \frac{[\text{Cl}^-]_{\text{LHS}}}{[\text{Cl}^-]_{\text{RHS}}} \right) \).

- The Nernst equation relates the cell potential to the concentrations of the reactants and products. For a concentration cell, the equation can be expressed as:
\[ E_{\text{cell}} = E_{\text{cell}}^\circ - \frac{RT}{nF} \ln \left( \frac{[\text{Products}]}{[\text{Reactants}]} \right) \]
- Since \( E_{\text{cell}}^\circ = 0 \), the equation simplifies to:
\[ E_{\text{cell}} = -\frac{RT}{nF} \ln \left( \frac{[\text{Cl}^-]_{\text{RHS}}}{[\text{Cl}^-]_{\text{LHS}}} \right) \]
- By inverting the fraction inside the logarithm, we can rewrite it as:
\[ E_{\text{cell}} = \frac{RT}{nF} \ln \left( \frac{[\text{Cl}^-]_{\text{LHS}}}{[\text{Cl}^-]_{\text{RHS}}} \right) \]

**Conclusion for Step 2**: Thus, Statement 2 is also true.

### Step 3: Assessing the Relationship Between the Statements
- Both statements are true, but Statement 2 does not serve as a correct explanation for Statement 1. Statement 1 simply states that the standard cell potential is zero, while Statement 2 provides a formula for calculating the cell potential based on concentrations.

### Final Answer
Both statements are true, but Statement 2 is not the correct explanation for Statement 1.