The standard reduction potential of a silver chloride electrode (metal -sparingly soluble salt electrode) is `0.209V` and for silver electrode is `0.80V`. If the moles of `AgCl` that can dissolve in `10L` of a `0.01M NaCl` solution is represented as `10^(-2)` then find the value of `z`.

Explanation

To solve the problem, we need to find the value of \( z \) in the expression \( 10^{-z} \) that represents the moles of AgCl that can dissolve in 10 L of a 0.01 M NaCl solution. We will use the given standard reduction potentials and the solubility product (Ksp) of silver chloride (AgCl).

### Step-by-Step Solution:

1. **Write the Nernst Equation**:
The Nernst equation for the silver chloride electrode can be written as:
\[ E_{AgCl/Ag} = E^\circ_{Ag} + \frac{0.0591}{n} \log K_{sp} \]
Where:
- \( E_{AgCl/Ag} \) is the standard reduction potential of the AgCl electrode (0.209 V).
- \( E^\circ_{Ag} \) is the standard reduction potential of the silver electrode (0.80 V).
- \( n \) is the number of electrons transferred (1 for AgCl).

2. **Substitute the Values**:
Substitute the known values into the Nernst equation:
\[ 0.209 = 0.80 + \frac{0.0591}{1} \log K_{sp} \]

3. **Rearrange to Solve for \( \log K_{sp} \)**:
Rearranging the equation gives:
\[ 0.209 - 0.80 = 0.0591 \log K_{sp} \]
\[ -0.591 = 0.0591 \log K_{sp} \]
Now, divide both sides by 0.0591:
\[ \log K_{sp} = \frac{-0.591}{0.0591} \approx -10 \]

4. **Calculate \( K_{sp} \)**:
From the logarithmic relationship, we find:
\[ K_{sp} = 10^{-10} \]

5. **Set Up the Solubility Expression**:
Let \( X \) be the solubility of AgCl in the 0.01 M NaCl solution. The solubility product expression for AgCl is:
\[ K_{sp} = [Ag^+][Cl^-] = X(X + 0.01) \]
Since NaCl provides 0.01 M Cl\(^-\), we can substitute:
\[ 10^{-10} = X(X + 0.01) \]

6. **Solve the Quadratic Equation**:
This can be approximated as:
\[ 10^{-10} = X^2 + 0.01X \]
Ignoring the \( 0.01X \) term for small \( X \):
\[ 10^{-10} \approx X^2 \]
Thus:
\[ X \approx 10^{-5} \]

7. **Calculate Moles of AgCl in 10 L**:
The moles of AgCl that can dissolve in 10 L is:
\[ \text{Moles of AgCl} = X \times 10 = 10^{-5} \times 10 = 10^{-4} \]

8. **Express in Terms of \( 10^{-z} \)**:
We have:
\[ 10^{-4} = 10^{-z} \]
Therefore, \( z = 4 \).

### Final Answer:
The value of \( z \) is \( 4 \).