The standard potential of `Cl^(c-)|AgCl|Ag` half cell is related to that of `Ag^(o+)|Ag` though the expression.
`E^(c-)._(Ag^(o+)|Ag)=E^(c-)._(Cl^(c-)|AgCl|A)+(RT)/(F)ln K_(sp)(AgCl)`
True/False

Explanation

To determine whether the statement regarding the standard potential of the half-cell `Cl^(-)|AgCl|Ag` in relation to `Ag^(+) | Ag` is true or false, we can analyze the given expression step by step.

### Step-by-Step Solution:

1. **Understanding the Half-Cells**:
- The half-cell `Cl^(-)|AgCl|Ag` involves the reduction of chloride ions and the formation of silver chloride (AgCl).
- The half-cell `Ag^(+) | Ag` involves the reduction of silver ions to silver metal.

2. **Standard Potentials**:
- The standard potential for the half-cell `Ag^(+) | Ag` is denoted as `E°(Ag)`.
- The standard potential for the half-cell `Cl^(-)|AgCl|Ag` is denoted as `E°(Cl)`.

3. **Nernst Equation**:
- The Nernst equation relates the cell potential to the standard potential and the concentration of the reactants/products.
- For a general cell reaction, the Nernst equation is given by:
\[ E = E° - \frac{RT}{nF} \ln Q \]
- Here, \( Q \) is the reaction quotient, \( n \) is the number of electrons transferred, \( R \) is the universal gas constant, and \( F \) is Faraday's constant.

4. **Equilibrium Condition**:
- At equilibrium, the cell potential \( E \) becomes zero. Thus, we can set up the equation:
\[ 0 = E°(Cl) - E°(Ag) - \frac{RT}{nF} \ln K_{sp}(AgCl) \]
- Rearranging gives:
\[ E°(Cl) = E°(Ag) + \frac{RT}{nF} \ln K_{sp}(AgCl) \]

5. **Identifying the Expression**:
- The expression provided in the question is:
\[ E°(Ag) = E°(Cl) + \frac{RT}{F} \ln K_{sp}(AgCl) \]
- This expression is not correctly rearranged as per the equilibrium condition derived above.

6. **Conclusion**:
- The statement in the question is **False** because the correct relationship should have \( E°(Cl) \) on the left side, not \( E°(Ag) \).

### Final Answer:
The statement is **False**.