Question

At 298 K, the standard reduction potential for Cu²⁺/Cu electrode is 0.34 V. 
Given: Ksp Cu(OH)2 = 1 × 10^-20 
Take \(\frac{2.303RT}{F}\)= 0.059V 
The reduction potential at pH = 14 for the above couple is (-)x x 10^-2 V. 
The value of x is_____.

Answer 1

Correct Answer: 25

1. Dissociation of Cu(OH)\(_2\): \[ \text{Cu(OH)}_2 (\text{s}) \rightleftharpoons \text{Cu}^{2+} (\text{aq}) + 2\text{OH}^- (\text{aq}). \] \[ K_\text{sp} = [\text{Cu}^{2+}] [\text{OH}^-]^2. \] 2. At pH = 14: \[ \text{pH} = 14 \implies \text{pOH} = 0 \implies [\text{OH}^-] = 1 \, \text{M}. \] Substituting in the solubility product expression: \[ [\text{Cu}^{2+}] = \frac{K_\text{sp}}{[\text{OH}^-]^2} = \frac{1 \times 10^{-20}}{1^2} = 1 \times 10^{-20} \, \text{M}. \] 3. Reduction Half Reaction: \[ \text{Cu}^{2+} (\text{aq}) + 2\text{e}^- \rightarrow \text{Cu} (\text{s}). \] Using the Nernst equation: \[ E = E^\circ - \frac{0.059}{n} \log_{10} \left( \frac{1}{[\text{Cu}^{2+}]} \right). \] Here: - \(E^\circ = 0.034 \, \text{V}\), - \(n = 2\), - \([\text{Cu}^{2+}] = 1 \times 10^{-20}\). 4. Substitute Values in Nernst Equation: \[ E = 0.034 - \frac{0.059}{2} \log_{10} \left( \frac{1}{1 \times 10^{-20}} \right). \] Simplify the logarithmic term: \[ \log_{10} \left( \frac{1}{1 \times 10^{-20}} \right) = \log_{10} (10^{20}) = 20. \] Substituting back: \[ E = 0.034 - \frac{0.059}{2} \times 20. \] \[ E = 0.034 - 0.59. \] \[ E = -0.25 \, \text{V}. \] 5. Express in Required Format: The reduction potential is \(-x \times 10^{-2}\), where \(x = 25\). 
Final Answer: \( \boxed{25} \)