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 need to calculate the reduction potential of the copper couple, \( Cu^{2+} | Cu \), at a pH of 14 using the given standard reduction potential and the solubility product \( K_{sp} \) of \( Cu(OH)_2 \).

### Step-by-Step Solution:

1. **Identify the Reaction and Given Data:**
- The standard reduction potential for the reaction \( Cu^{2+} + 2e^- \rightarrow Cu \) is given as \( E^\circ = +0.34 \, V \).
- The \( K_{sp} \) of \( Cu(OH)_2 \) is \( 1.0 \times 10^{-19} \).
- The pH is given as 14.

2. **Calculate the Concentration of Hydroxide Ions:**
- At pH 14, the concentration of hydroxide ions \( [OH^-] \) can be calculated using the formula:
\[ [OH^-] = 10^{-14 + pH} = 10^{-14 + 14} = 10^0 = 1 \, M \]

3. **Write the Dissolution Reaction for \( Cu(OH)_2 \):**
- The dissolution of copper hydroxide can be represented as:
\[ Cu(OH)_2 (s) \rightleftharpoons Cu^{2+} (aq) + 2OH^- (aq) \]

4. **Set Up the Expression for \( K_{sp} \):**
- The solubility product \( K_{sp} \) is given by:
\[ K_{sp} = [Cu^{2+}][OH^-]^2 \]
- Substituting the known values:
\[ 1.0 \times 10^{-19} = [Cu^{2+}](1)^2 \]
- Therefore, the concentration of \( Cu^{2+} \) is:
\[ [Cu^{2+}] = 1.0 \times 10^{-19} \, M \]

5. **Use the Nernst Equation:**
- The Nernst equation is given by:
\[ E = E^\circ - \frac{0.0591}{n} \log \left( \frac{1}{[Cu^{2+}]} \right) \]
- Here, \( n = 2 \) (the number of electrons transferred).
- Substituting the values:
\[ E = 0.34 - \frac{0.0591}{2} \log \left( \frac{1}{1.0 \times 10^{-19}} \right) \]

6. **Calculate the Logarithm:**
- The logarithm can be simplified:
\[ \log \left( \frac{1}{1.0 \times 10^{-19}} \right) = \log(10^{19}) = 19 \]

7. **Substitute Back into the Nernst Equation:**
- Now substituting the logarithm back into the equation:
\[ E = 0.34 - \frac{0.0591}{2} \times 19 \]
- Calculate \( \frac{0.0591 \times 19}{2} \):
\[ \frac{0.0591 \times 19}{2} = 0.56145 \]

8. **Final Calculation of \( E \):**
- Now calculate \( E \):
\[ E = 0.34 - 0.56145 = -0.22145 \, V \approx -0.22 \, V \]

### Final Answer:
The reduction potential at \( pH = 14 \) for the couple \( Cu^{2+} | Cu \) is approximately \( -0.22 \, V \).