The standard reduction potential for `Cu^(2+)|Cu` is `+0.34V`. Calculate the reduction potential al `pH=14` for the above couple . `K_(sp)` of `Cu(OH)_(2)` is `1.0xx10^(-19)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Information
We have the following data:
- Standard reduction potential for the couple \( Cu^{2+} | Cu \): \( E^\circ = +0.34 \, V \)
- pH = 14
- \( K_{sp} \) of \( Cu(OH)_2 = 1.0 \times 10^{-19} \)

### Step 2: Calculate the Concentration of \( OH^- \) Ions
From the pH, we can calculate the concentration of \( OH^- \) ions using the formula:
\[ [OH^-] = 10^{-14 + pH} = 10^{-14 + 14} = 10^0 = 1 \, M \]

### Step 3: Calculate the Concentration of \( Cu^{2+} \) Ions Using \( K_{sp} \)
The solubility product expression for \( Cu(OH)_2 \) is given by:
\[ K_{sp} = [Cu^{2+}][OH^-]^2 \]
Substituting the known values:
\[ 1.0 \times 10^{-19} = [Cu^{2+}](1)^2 \]
Thus, we find:
\[ [Cu^{2+}] = 1.0 \times 10^{-19} \, M \]

### Step 4: Apply the Nernst Equation
The Nernst equation is given by:
\[ E = E^\circ - \frac{0.0591}{n} \log \left( \frac{1}{[Cu^{2+}]} \right) \]
Where \( n \) is the number of electrons transferred in the half-reaction. For the reduction of \( Cu^{2+} \) to \( Cu \), \( n = 2 \).

Substituting the values:
\[ E = 0.34 - \frac{0.0591}{2} \log \left( \frac{1}{1.0 \times 10^{-19}} \right) \]

### Step 5: Simplify the Logarithm
Calculating the logarithm:
\[ \log \left( \frac{1}{1.0 \times 10^{-19}} \right) = \log(10^{19}) = 19 \]

### Step 6: Substitute Back into the Nernst Equation
Now substituting back into the Nernst equation:
\[ E = 0.34 - \frac{0.0591}{2} \times 19 \]
Calculating the second term:
\[ \frac{0.0591}{2} \times 19 = 0.0591 \times 9.5 = 0.56145 \]
Thus:
\[ E = 0.34 - 0.56145 = -0.22145 \, V \]

### Final Answer
The reduction potential at \( pH = 14 \) for the couple \( Cu^{2+} | Cu \) is approximately:
\[ E \approx -0.022 \, V \]