Consider the electrode `Ag|AgCl(s),Cl^(-)(0.1M),i.e.,` silver electrode in contact with` 0.1M KCl` solution saturated with `AgCl` . If it is combined with the electrode `Ag|Ag^(o+)(0.1M)` to form a complete cell, the `EMF` would be `(K_(sp)of AgCl=10^(-10)` at `25^(@)C)`

Options

A. `0.799`
B. `-.6363`
C. `0.59`
D. `0.472`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the cell notation
We have two electrodes:
1. Anode: \( \text{Ag} | \text{AgCl}(s), \text{Cl}^-(0.1M) \)
2. Cathode: \( \text{Ag} | \text{Ag}^+(0.1M) \)

The cell notation can be written as:
\[ \text{Ag} | \text{AgCl}(s), \text{Cl}^-(0.1M) || \text{Ag}^+(0.1M) | \text{Ag} \]

### Step 2: Write the expression for \( K_{sp} \) of \( \text{AgCl} \)
The solubility product constant \( K_{sp} \) for \( \text{AgCl} \) is given by:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] \]
Given \( K_{sp} = 10^{-10} \) and \( [\text{Cl}^-] = 0.1 \, M \), we can find \( [\text{Ag}^+] \).

### Step 3: Calculate the concentration of \( \text{Ag}^+ \)
Using the \( K_{sp} \) expression:
\[ 10^{-10} = [\text{Ag}^+][0.1] \]
Rearranging gives:
\[ [\text{Ag}^+] = \frac{10^{-10}}{0.1} = 10^{-9} \, M \]

### Step 4: Calculate the EMF of the cell
For a concentration cell, the EMF can be calculated using the Nernst equation:
\[ E_{cell} = E^\circ - \frac{0.059}{n} \log \frac{[\text{Ag}^+]_{anode}}{[\text{Ag}^+]_{cathode}} \]
Here, \( E^\circ = 0 \) (since both electrodes are the same metal), and \( n = 1 \).

Substituting the concentrations:
- \( [\text{Ag}^+]_{anode} = 10^{-9} \, M \)
- \( [\text{Ag}^+]_{cathode} = 0.1 \, M \)

### Step 5: Substitute values into the Nernst equation
\[ E_{cell} = 0 - \frac{0.059}{1} \log \frac{10^{-9}}{0.1} \]
This simplifies to:
\[ E_{cell} = -0.059 \log(10^{-9} \times 10) \]
\[ E_{cell} = -0.059 \log(10^{-8}) \]
Using the property of logarithms:
\[ E_{cell} = -0.059 \times (-8) = 0.472 \, V \]

### Final Answer
The EMF of the cell is:
\[ E_{cell} = 0.472 \, V \]