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:
(a)0.1
(b)`10^(-13)`
(c)`10^(-15)`
(d)None of these

Options

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

Answer

To solve the problem, we need to find the solubility product constant (Ksp) of cadmium hydroxide (Cd(OH)₂) using the given electrochemical cell data. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the Anode and Cathode Reactions
The electrochemical cell is represented as:
\[ \text{Cd(s)} | \text{Cd(OH)}_2(s) | \text{NaOH(aq, 0.01M)} | \text{H}_2(g, 1 \text{ bar}) | \text{Pt(s)} \]

- **Anode Reaction (Oxidation)**:
\[ \text{Cd(s)} \rightarrow \text{Cd}^{2+}(aq) + 2e^- \]

- **Cathode Reaction (Reduction)**:
\[ 2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g) \]

### Step 2: Write the Overall Cell Reaction
Combining the anode and cathode reactions gives us the overall cell reaction:
\[ \text{Cd(s)} + 2\text{H}^+(aq) \rightarrow \text{Cd}^{2+}(aq) + \text{H}_2(g) \]

### Step 3: Use the Nernst Equation
The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q \]
Where:
- \( E_{\text{cell}} = 0.0 \, \text{V} \)
- \( E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \)
- \( n = 2 \) (number of electrons transferred)
- \( Q \) is the reaction quotient.

### Step 4: Calculate \( E^\circ_{\text{cell}} \)
Given \( E^\circ_{\text{Cd}^{2+}/\text{Cd}} = -0.39 \, \text{V} \) and \( E^\circ_{\text{H}^+/H_2} = 0 \, \text{V} \):
\[ E^\circ_{\text{cell}} = 0 - (-0.39) = 0.39 \, \text{V} \]

### Step 5: Determine the Concentration of \( \text{H}^+ \)
From the given concentration of NaOH (0.01 M), we can find the concentration of \( \text{OH}^- \):
\[ [\text{OH}^-] = 0.01 \, \text{M} \]
Using the water dissociation constant:
\[ K_w = [\text{H}^+][\text{OH}^-] = 10^{-14} \]
We can find \( [\text{H}^+] \):
\[ [\text{H}^+] = \frac{10^{-14}}{0.01} = 10^{-12} \, \text{M} \]

### Step 6: Substitute Values into the Nernst Equation
Substituting into the Nernst equation:
\[ 0.0 = 0.39 - \frac{0.0591}{2} \log \left( \frac{[\text{Cd}^{2+}]}{(10^{-12})^2} \right) \]
This simplifies to:
\[ 0.0 = 0.39 - 0.02955 \log \left( [\text{Cd}^{2+}] \times 10^{-24} \right) \]

### Step 7: Solve for \( [\text{Cd}^{2+}] \)
Rearranging gives:
\[ 0.02955 \log \left( [\text{Cd}^{2+}] \times 10^{-24} \right) = 0.39 \]
\[ \log \left( [\text{Cd}^{2+}] \times 10^{-24} \right) = \frac{0.39}{0.02955} \approx 13.2 \]
\[ [\text{Cd}^{2+}] \times 10^{-24} = 10^{13.2} \]
Thus,
\[ [\text{Cd}^{2+}] = 10^{13.2} \times 10^{24} = 10^{-11} \, \text{M} \]

### Step 8: Calculate Ksp of Cd(OH)₂
The dissociation of cadmium hydroxide is:
\[ \text{Cd(OH)}_2(s) \rightleftharpoons \text{Cd}^{2+}(aq) + 2\text{OH}^-(aq) \]
The solubility product expression is:
\[ K_{sp} = [\text{Cd}^{2+}][\text{OH}^-]^2 \]
Substituting the values:
\[ K_{sp} = (10^{-11})(10^{-2})^2 = 10^{-11} \times 10^{-4} = 10^{-15} \]

### Final Answer
The Ksp of Cd(OH)₂ is:
\[ \boxed{10^{-15}} \]