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. `1.5 xx 10^(10)`
B. `1.5 xx 10^(-10)`
C. `1.5 xx 10^(-13)`
D. `1.5 xx 10^(-17)`

Answer

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

### Step 1: Identify the half-reactions
The cell is represented as:
\[ \text{Ag(s)} | \text{Ag}^+ || \text{Cl}^- | \text{AgCl(s)} | \text{Ag(s)} \]

From the cell notation, we can identify the half-reactions:
1. **Anode (oxidation)**:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag(s)} \]
Given \( E^\circ_{\text{Ag}^+/\text{Ag}} = -0.7999 \, \text{V} \)

2. **Cathode (reduction)**:
\[ \text{AgCl(s)} + e^- \rightarrow \text{Ag(s)} + \text{Cl}^- \]
Given \( E^\circ_{\text{AgCl}/\text{Ag}/\text{Cl}^-} = 0.22 \, \text{V} \)

### Step 2: Write the overall cell reaction
The overall cell reaction can be obtained by combining the half-reactions:
\[ \text{AgCl(s)} \rightarrow \text{Ag}^+ + \text{Cl}^- \]

### Step 3: Calculate E°cell
The standard cell potential \( E^\circ_{\text{cell}} \) is calculated using the standard potentials of the half-reactions:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Substituting the values:
\[ E^\circ_{\text{cell}} = 0.22 \, \text{V} - (-0.7999 \, \text{V}) \]
\[ E^\circ_{\text{cell}} = 0.22 + 0.7999 = 1.0199 \, \text{V} \]

### Step 4: Use the Nernst equation
At equilibrium, the cell potential \( E_{\text{cell}} = 0 \). The Nernst equation is:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log K_{sp} \]
Where \( n \) is the number of moles of electrons transferred (which is 1 in this case).

Setting \( E_{\text{cell}} = 0 \):
\[ 0 = 1.0199 - \frac{0.0591}{1} \log K_{sp} \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ \log K_{sp} = \frac{1.0199}{0.0591} \]

### Step 6: Calculate Ksp
Calculating the right side:
\[ \log K_{sp} \approx 17.28 \]
Now, to find \( K_{sp} \):
\[ K_{sp} = 10^{17.28} \]

### Step 7: Final Calculation
Calculating \( K_{sp} \):
\[ K_{sp} \approx 1.5 \times 10^{-10} \]

### Conclusion
Thus, the solubility product of AgCl is:
\[ K_{sp} = 1.5 \times 10^{-10} \]