Pure water is saturated with pure solid `AgCl`, a silver electrode is placed in the solution and the potential is measured against normal calomel electrode at `25^(@)C`. This experiment is then repeated with a saturated solution of `Agl`. If the difference in potential in the two cases is `0.177V`. What is the ratio of solubility product (solubility) of `AgCl` and Agl at the temperature of the experiment?

Options

A. `10^(3)`
B. `10^(6)`
C. `10^(2)`
D. `10^(2)`

Explanation

To solve the problem, we need to find the ratio of the solubility products (Ksp) of AgCl and AgI based on the given potential difference when a silver electrode is placed in saturated solutions of both salts. Here are the steps to arrive at the solution:

### Step-by-Step Solution:

1. **Understanding the Dissociation of AgCl and AgI:**
- The dissociation of AgCl in water can be represented as:
\[ \text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) \]
- The dissociation of AgI in water can be represented as:
\[ \text{AgI (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{I}^- (aq) \]

2. **Setting Up the Nernst Equation:**
- For AgCl, the Nernst equation for the silver electrode can be written as:
\[ E_1 = E^\circ - \frac{0.0591}{1} \log \left( \frac{1}{C_1} \right) \]
- For AgI, the Nernst equation can be written as:
\[ E_2 = E^\circ - \frac{0.0591}{1} \log \left( \frac{1}{C_2} \right) \]
- Here, \(C_1\) is the concentration of \(\text{Ag}^+\) from AgCl and \(C_2\) is the concentration of \(\text{Ag}^+\) from AgI.

3. **Finding the Potential Difference:**
- The difference in potential between the two experiments is given as:
\[ E_1 - E_2 = 0.177 \, V \]
- Substituting the expressions for \(E_1\) and \(E_2\):
\[ \left( E^\circ - \frac{0.0591}{1} \log \left( \frac{1}{C_1} \right) \right) - \left( E^\circ - \frac{0.0591}{1} \log \left( \frac{1}{C_2} \right) \right) = 0.177 \]
- This simplifies to:
\[ -0.0591 \log \left( \frac{1}{C_1} \right) + 0.0591 \log \left( \frac{1}{C_2} \right) = 0.177 \]
- Rearranging gives:
\[ 0.0591 \left( \log C_2 - \log C_1 \right) = 0.177 \]

4. **Solving for the Concentration Ratio:**
- This can be rewritten using the properties of logarithms:
\[ 0.0591 \log \left( \frac{C_2}{C_1} \right) = 0.177 \]
- Dividing both sides by 0.0591:
\[ \log \left( \frac{C_2}{C_1} \right) = \frac{0.177}{0.0591} \approx 3 \]
- Converting from logarithmic form gives:
\[ \frac{C_2}{C_1} = 10^3 = 1000 \]
- Therefore, we can express the ratio of concentrations as:
\[ \frac{C_1}{C_2} = \frac{1}{1000} \]

5. **Relating Concentration to Solubility Product:**
- The solubility product \(K_{sp}\) for each salt can be expressed as:
\[ K_{sp}(\text{AgCl}) = [\text{Ag}^+][\text{Cl}^-] = C_1^2 \]
\[ K_{sp}(\text{AgI}) = [\text{Ag}^+][\text{I}^-] = C_2^2 \]
- Since \(C_1\) and \(C_2\) are the concentrations of \(\text{Ag}^+\) from AgCl and AgI respectively, we can relate the solubility products:
\[ \frac{K_{sp}(\text{AgCl})}{K_{sp}(\text{AgI})} = \frac{C_1^2}{C_2^2} = \left( \frac{C_1}{C_2} \right)^2 = \left( \frac{1}{1000} \right)^2 = \frac{1}{1000000} \]

### Final Result:
The ratio of the solubility products of AgCl and AgI is:
\[ \frac{K_{sp}(\text{AgCl})}{K_{sp}(\text{AgI})} = 10^{-6} \]