`Ag|Ag^(+), KI||Ag I| Ag` emf is E, then `K_(sp)` of AgI is given as:

Options

A. `K_(sp)=(nF)/(2.303RT) logE^(@)`
B. ln `K_(sp)=Nf[(deltaE^(@))/(delta)-E^(@)]`
C. ln `K_(sp)=(nF)/(E^(@))`
D. log `K_(sp)=(nFE^(@))/(2.303RT)`

Answer

To find the solubility product constant (Ksp) of AgI given the emf (E) of the cell Ag|Ag⁺, KI||AgI|Ag, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Cell Reaction**:
The cell consists of two half-cells: Ag|Ag⁺ and AgI|Ag. The Ag|Ag⁺ half-cell involves the reduction of silver ions (Ag⁺) to solid silver (Ag). The AgI|Ag half-cell involves the dissolution of solid AgI into Ag⁺ and I⁻ ions.

2. **Write the Nernst Equation**:
The Nernst equation relates the cell potential (E) to the standard cell potential (E°) and the concentrations of the reactants and products:
\[ E = E° - \frac{RT}{nF} \ln Q \]
where:
- \(E\) is the cell potential,
- \(E°\) is the standard cell potential,
- \(R\) is the universal gas constant,
- \(T\) is the temperature in Kelvin,
- \(n\) is the number of moles of electrons transferred,
- \(F\) is Faraday's constant,
- \(Q\) is the reaction quotient.

3. **Determine the Reaction Quotient (Q)**:
For the dissolution of AgI:
\[ \text{AgI (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{I}^- (aq) \]
The reaction quotient \(Q\) at equilibrium is given by:
\[ Q = \frac{[\text{Ag}^+][\text{I}^-]}{1} = [\text{Ag}^+][\text{I}^-] \]
At equilibrium, \(Q = K_{sp}\).

4. **Set Up the Equation at Equilibrium**:
At equilibrium, the cell potential \(E\) becomes 0 (since no net reaction occurs). Thus, we can rearrange the Nernst equation:
\[ 0 = E° - \frac{RT}{nF} \ln K_{sp} \]
Rearranging gives:
\[ E° = \frac{RT}{nF} \ln K_{sp} \]

5. **Solve for Ksp**:
Rearranging the equation to solve for \(K_{sp}\):
\[ \ln K_{sp} = \frac{nFE°}{RT} \]
Taking the exponential of both sides gives:
\[ K_{sp} = e^{\frac{nFE°}{RT}} \]

6. **Substituting Values**:
If we know the values of \(n\), \(F\), \(R\), \(T\), and \(E°\), we can substitute them into the equation to find \(K_{sp}\).

### Final Expression:
Thus, the solubility product constant \(K_{sp}\) of AgI can be expressed as:
\[ K_{sp} = e^{\frac{nFE°}{RT}} \]