`E_(Cu^(2)//Cu)^(@) = 0.34 V`, what will be the reduction potential at pH=14 for same couple. Given `K_(sp)` of `Cu(OH)_(2) = 10^(-19)`

Options

A. `-0.22 V`
B. `-0.71 V`
C. `+0.22 V`
D. `+0.71 V`

Explanation

To find the reduction potential of the copper couple \((Cu^{2+}/Cu)\) at pH 14, we can follow these steps:

### Step 1: Determine the concentration of \(OH^-\) ions at pH 14
At pH 14, the concentration of hydrogen ions \([H^+]\) can be calculated as:
\[ [H^+] = 10^{-pH} = 10^{-14} \, \text{M} \]
Using the ionic product of water at 25°C, \(K_w = [H^+][OH^-] = 10^{-14}\):
\[ [OH^-] = \frac{K_w}{[H^+]} = \frac{10^{-14}}{10^{-14}} = 1 \, \text{M} \]

### Step 2: Use the solubility product \(K_{sp}\) of \(Cu(OH)_2\)
The solubility product \(K_{sp}\) for \(Cu(OH)_2\) is given as \(10^{-19}\). The dissociation of \(Cu(OH)_2\) can be represented as:
\[ Cu(OH)_2 (s) \rightleftharpoons Cu^{2+} (aq) + 2OH^- (aq) \]
The expression for \(K_{sp}\) is:
\[ K_{sp} = [Cu^{2+}][OH^-]^2 \]
Substituting the known concentration of \(OH^-\):
\[ 10^{-19} = [Cu^{2+}](1)^2 \]
Thus, we find:
\[ [Cu^{2+}] = 10^{-19} \, \text{M} \]

### Step 3: Apply the Nernst equation
The standard reduction potential \(E^\circ\) for the couple \((Cu^{2+}/Cu)\) is given as \(0.34 \, V\). The Nernst equation is:
\[ E = E^\circ - \frac{0.0591}{n} \log \left(\frac{1}{[Cu^{2+}]}\right) \]
where \(n\) is the number of electrons transferred (which is 2 for this reaction).

Substituting the values:
\[ E = 0.34 - \frac{0.0591}{2} \log \left(\frac{1}{10^{-19}}\right) \]
Calculating the logarithm:
\[ \log \left(\frac{1}{10^{-19}}\right) = \log(10^{19}) = 19 \]
Now substituting back into the equation:
\[ E = 0.34 - \frac{0.0591}{2} \times 19 \]
Calculating the term:
\[ E = 0.34 - 0.0591 \times 9.5 \]
\[ E = 0.34 - 0.56145 \]
\[ E = 0.34 - 0.56145 = -0.22145 \, V \]

### Final Answer
The reduction potential at pH 14 for the copper couple \((Cu^{2+}/Cu)\) is approximately:
\[ E \approx -0.22 \, V \]