The electrochemical cell shown below is a concentration cell. `M|M^(2+)(` saturated solution of sparingly soluble salt, `MX_(2))||M^(2+)(0.001 mol dm^(-3))|M`
The `emf` of the cell depends on the difference in the concentration of `M^(2+)` ions at the two electrodes. The `emf` of the cell at `298` is `0.059V`.
The solubility product `(K_(sp),mol^(3) dm^(-9))` of `MX_(2)` at 298 based on the information available the given concentration cell is `(` Take `2.303xxRxx298//F=0.059V)`

Options

A. `1xx10^(-15)`
B. `4xx10^(-15)`
C. `1xx10^(-12)`
D. `4xx10^(-12)`

Explanation

To solve the problem, we need to determine the solubility product (Ksp) of the sparingly soluble salt MX₂ using the information provided about the electrochemical cell.

### Step-by-Step Solution:

1. **Understand the Cell Configuration**:
The electrochemical cell is a concentration cell with the following configuration:
- Anode: M | M²⁺ (saturated solution of MX₂)
- Cathode: M²⁺ (0.001 mol/dm³) | M

2. **Use the Nernst Equation**:
The Nernst equation for a concentration cell can be expressed as:
\[ E_{\text{cell}} = E^\circ - \frac{RT}{nF} \ln \left( \frac{[M^{2+}]_{\text{cathode}}}{[M^{2+}]_{\text{anode}}} \right) \]
Since this is a concentration cell, \(E^\circ = 0\). Thus, the equation simplifies to:
\[ E_{\text{cell}} = -\frac{RT}{nF} \ln \left( \frac{[M^{2+}]_{\text{cathode}}}{[M^{2+}]_{\text{anode}}} \right) \]

3. **Substitute Known Values**:
Given that \(E_{\text{cell}} = 0.059 \, V\), \(n = 2\) (since 2 electrons are involved in the half-reaction), and using the provided relation \( \frac{2.303 \cdot R \cdot 298}{F} = 0.059 \):
\[ 0.059 = -\frac{0.059}{2} \log \left( \frac{0.001}{[M^{2+}]_{\text{anode}}} \right) \]

4. **Rearranging the Equation**:
Rearranging gives:
\[ -2 = \log \left( \frac{0.001}{[M^{2+}]_{\text{anode}}} \right) \]
This implies:
\[ \frac{0.001}{[M^{2+}]_{\text{anode}}} = 10^{-2} \]
Therefore:
\[ [M^{2+}]_{\text{anode}} = 0.001 \times 10^{2} = 0.1 \, \text{mol/dm}^3 \]

5. **Calculate Solubility (s)**:
For the salt MX₂, the dissociation can be represented as:
\[ MX_2 \rightleftharpoons M^{2+} + 2X^{-} \]
If the solubility of MX₂ is \(s\), then:
- The concentration of \(M^{2+}\) ions will be \(s\)
- The concentration of \(X^{-}\) ions will be \(2s\)

The solubility product (Ksp) is given by:
\[ K_{sp} = [M^{2+}][X^{-}]^2 = s(2s)^2 = 4s^3 \]

6. **Substituting the Value of s**:
We have found that:
\[ [M^{2+}]_{\text{anode}} = 0.1 \, \text{mol/dm}^3 \Rightarrow s = 0.1 \]
Therefore:
\[ K_{sp} = 4(0.1)^3 = 4 \times 0.001 = 0.004 \, \text{mol}^3/\text{dm}^9 = 4 \times 10^{-15} \, \text{mol}^3/\text{dm}^9 \]

### Final Answer:
The solubility product \(K_{sp}\) of MX₂ at 298 K is:
\[ K_{sp} = 4 \times 10^{-15} \, \text{mol}^3/\text{dm}^9 \]