The standard electrode of a metal ion `(Ag|Ag^(o+))` and metal `-` insoluble salt anion `(Ag|AgCl|Cl^(c-))` are related as
(a)`E^(c-)._(Ag^(o+)|Ag)=E^(c-)._(Cl^(c-)|AgCl|Ag)+(RT)/(F)ln K_(sp)`
(b)`E^(c-)._(Cl^(c-)|AgCl|Ag)=E^(c-)._(Ag^(o+)\Ag)+(RT)/(F)lnK_(sp)`
(c)`E^(c-)._(Ag^(o+)|Ag)=E^(c-)._(Cl^(c-)|AgCl|Ag)+(RT)/(F)ln.([Cl^(c-)])/(K_(sp))`
(d) `E^(c-)._(Cl^(c-)|AgCl|Ag)=E^(c-)._(Ag^(o+)|Ag)+(RT)/(F)ln.(K_(sp))/([Cl^(c-)])`

Options

A. `E^(c-)._(Ag^(o+)|Ag)=E^(c-)._(Cl^(c-)|AgCl|Ag)+(RT)/(F)ln K_(sp)`
B. `E^(c-)._(Cl^(c-)|AgCl|Ag)=E^(c-)._(Ag^(o+)\Ag)+(RT)/(F)lnK_(sp)`
C. `E^(c-)._(Ag^(o+)|Ag)=E^(c-)._(Cl^(c-)|AgCl|Ag)+(RT)/(F)ln.([Cl^(c-)])/(K_(sp))`
D. `E^(c-)._(Cl^(c-)|AgCl|Ag)=E^(c-)._(Ag^(o+)|Ag)+(RT)/(F)ln.(K_(sp))/([Cl^(c-)])`

Explanation

To solve the problem, we need to analyze the relationship between the standard electrode potentials of the given half-cells and the solubility product constant (Ksp) for the silver chloride (AgCl) system.

### Step-by-Step Solution:

1. **Identify the Half-Reactions**:
- For the reduction of silver chloride:
\[ \text{AgCl (s)} + e^- \rightarrow \text{Ag (s)} + \text{Cl}^-(aq) \]
- For the oxidation of silver:
\[ \text{Ag (s)} \rightarrow \text{Ag}^+(aq) + e^- \]

2. **Write the Overall Reaction**:
The overall reaction combining both half-reactions is:
\[ \text{AgCl (s)} \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq) \]

3. **Equilibrium Condition**:
At equilibrium, the cell potential \(E_{cell}\) is zero, which implies:
\[ Q_{cell} = K_{sp} \]
where \(Q_{cell}\) is the reaction quotient.

4. **Using the Nernst Equation**:
The Nernst equation relates the standard cell potential to the concentrations of the products and reactants:
\[ E_{cell} = E^\circ_{cell} - \frac{RT}{F} \ln Q \]
Since \(E_{cell} = 0\) at equilibrium, we can rearrange the equation:
\[ 0 = E^\circ_{cell} - \frac{RT}{F} \ln K_{sp} \]
This leads to:
\[ E^\circ_{cell} = \frac{RT}{F} \ln K_{sp} \]

5. **Expressing \(E^\circ_{cell}\)**:
The standard cell potential can be expressed in terms of the individual half-cell potentials:
\[ E^\circ_{cell} = E^\circ_{Ag^+|Ag} - E^\circ_{AgCl|Ag} \]

6. **Rearranging the Equation**:
Rearranging gives us:
\[ E^\circ_{AgCl|Ag} = E^\circ_{Ag^+|Ag} - \frac{RT}{F} \ln K_{sp} \]

7. **Final Formulation**:
This can be rewritten as:
\[ E^\circ_{Cl^-|AgCl|Ag} = E^\circ_{Ag^+|Ag} + \frac{RT}{F} \ln K_{sp} \]

### Conclusion:
From the analysis, we find that the correct relationship is:
\[ E^\circ_{Cl^-|AgCl|Ag} = E^\circ_{Ag^+|Ag} + \frac{RT}{F} \ln K_{sp} \]
This corresponds to option (b):
\[ E^{c-}_{Cl^{-}|AgCl|Ag} = E^{c-}_{Ag^{+}|Ag} + \frac{RT}{F} \ln K_{sp} \]