The following cell is set up at `25^@C,`
`Ag(s)|Ag^(+)||Cl^(-)||AgCl(s)|Ag(s)`
Given that `E_(Ag^(+)//Ag)^(0)=0.7999V and E_(AgCl//Ag_((s))//Cl^(-))` 0.22 volt, the solubility product of AgCl is

Options

A. `1.8 xx 10^(10)`
B. `1.8 xx 10^(-10)`
C. `1.8 xx 10^(-13)`
D. `1.8 xx 10^(-17)`

Answer

To find the solubility product (Ksp) of AgCl from the given electrochemical cell, we can follow these steps:

### Step 1: Write the half-reactions
The half-reactions for the cell are:
1. Oxidation of silver (Ag):
\[ \text{Ag(s)} \rightarrow \text{Ag}^+ + e^- \]
2. Reduction of silver chloride (AgCl):
\[ \text{AgCl(s)} + e^- \rightarrow \text{Ag(s)} + \text{Cl}^- \]

### Step 2: Write the overall reaction
Combining the two half-reactions gives:
\[ \text{AgCl(s)} \rightleftharpoons \text{Ag}^+ + \text{Cl}^- \]

### Step 3: Write the expression for the solubility product (Ksp)
The solubility product (Ksp) for the equilibrium can be expressed as:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] \]

### Step 4: Use the Nernst equation
The Nernst equation is given by:
\[ E_{cell} = E^0_{cell} - \frac{0.0591}{n} \log Q \]
Where:
- \(E_{cell}\) is the cell potential.
- \(E^0_{cell}\) is the standard cell potential.
- \(n\) is the number of moles of electrons transferred (which is 1 in this case).
- \(Q\) is the reaction quotient, which is equal to \(K_{sp}\) for the dissolution of AgCl.

### Step 5: Calculate the standard cell potential (E^0_cell)
The standard cell potential can be calculated as:
\[ E^0_{cell} = E^0_{Ag^+/Ag} + E^0_{AgCl/Cl^-} \]
Substituting the given values:
\[ E^0_{cell} = 0.7999\,V + 0.22\,V = 0.5799\,V \]

### Step 6: Set up the Nernst equation
At equilibrium, \(E_{cell} = 0\):
\[ 0 = 0.5799 - \frac{0.0591}{1} \log K_{sp} \]

### Step 7: Solve for Ksp
Rearranging gives:
\[ \log K_{sp} = \frac{0.5799}{0.0591} \]
Calculating the right side:
\[ \log K_{sp} = 9.8122 \]
Now, converting from logarithmic form to exponential form:
\[ K_{sp} = 10^{-9.8122} \approx 1.54 \times 10^{-10} \]

### Conclusion
The solubility product of AgCl is:
\[ K_{sp} \approx 1.54 \times 10^{-10} \]