Given the cell: `Cd(s)|Cd(OH)_2(s)|NaOH(aq,0.01M)|H_2(g,1"bar")|Pt(s)`
with `E_(cell)=0.0V."if"E_(Cd^(2+)|Cd)^(@)=-0.39V,"then"K_(sp)"of"Cd(OH_2)`is:

Options

A. 0.1
B. `10^(-13)`
C. `10^(-15)`
D. None of these

Answer

To solve the problem, we need to determine the solubility product constant (Ksp) of cadmium hydroxide (Cd(OH)₂) based on the given electrochemical cell and the provided data.

### Step-by-Step Solution:

1. **Identify the half-reactions**:
- The oxidation half-reaction involves cadmium (Cd) converting to cadmium ions (Cd²⁺):
\[ \text{Cd(s)} \rightarrow \text{Cd}^{2+}(aq) + 2e^- \]
- The reduction half-reaction involves hydrogen ions (H⁺) converting to hydrogen gas (H₂):
\[ 2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g) \]

2. **Overall cell reaction**:
Combining the half-reactions gives the overall cell reaction:
\[ \text{Cd(s)} + 2\text{H}^+(aq) \rightarrow \text{Cd}^{2+}(aq) + \text{H}_2(g) \]

3. **Use the Nernst equation**:
The Nernst equation relates the cell potential to the concentrations of the reactants and products:
\[ E_{cell} = E^\circ_{cell} - \frac{0.059}{n} \log Q \]
Where:
- \(E_{cell} = 0.0 \, V\)
- \(E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0 - (-0.39) = 0.39 \, V\)
- \(n = 2\) (number of electrons transferred)

4. **Calculate the reaction quotient (Q)**:
The reaction quotient \(Q\) for the overall reaction is given by:
\[ Q = \frac{[\text{Cd}^{2+}]}{[H^+]^2} \]
Given that the concentration of \(OH^-\) is \(0.01 \, M\), we can find the concentration of \(H^+\) using the ion product of water:
\[ K_w = [H^+][OH^-] = 10^{-14} \]
Therefore:
\[ [H^+] = \frac{10^{-14}}{0.01} = 10^{-12} \, M \]

5. **Substituting values into the Nernst equation**:
Rearranging the Nernst equation:
\[ 0 = 0.39 - \frac{0.059}{2} \log \left(\frac{[\text{Cd}^{2+}]}{(10^{-12})^2}\right) \]
Simplifying gives:
\[ 0.39 = \frac{0.059}{2} \log \left(\frac{[\text{Cd}^{2+}]}{10^{-24}}\right) \]

6. **Solving for \([\text{Cd}^{2+}]\)**:
Rearranging gives:
\[ \log \left([\text{Cd}^{2+}]\right) = \frac{0.39 \times 2}{0.059} + 24 \]
Calculating the left side:
\[ \log \left([\text{Cd}^{2+}]\right) = 13.23 + 24 = 37.23 \]
Therefore:
\[ [\text{Cd}^{2+}] = 10^{-37.23} \approx 5.88 \times 10^{-38} \, M \]

7. **Calculate Ksp**:
The dissociation of cadmium hydroxide is:
\[ \text{Cd(OH)}_2 \rightleftharpoons \text{Cd}^{2+} + 2\text{OH}^- \]
Thus:
\[ K_{sp} = [\text{Cd}^{2+}][\text{OH}^-]^2 \]
Substituting the values:
\[ K_{sp} = (5.88 \times 10^{-38})(0.01)^2 = 5.88 \times 10^{-38} \times 10^{-4} = 5.88 \times 10^{-42} \]

### Final Answer:
The Ksp of Cd(OH)₂ is approximately \(5.88 \times 10^{-42}\).