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Question:

The reduction potential at pH = 14 for the Cu2+/Cu couples is: [Given, EoCu2+/Cu = 0.34V; KspCu(OH)2 = 1 × 10⁻¹⁹]

-0.34 V

0.22 V

0.34 V

-0.22 V

Solution:

Given, pH = 14, ∴ pOH = 0 and [OH⁻] = 1 M
[Cu²⁺][OH⁻]² = Ksp = 1 × 10⁻¹⁹
[Cu²⁺] = 1 × 10⁻¹⁹ M
For the half reaction, Cu²⁺ + 2e⁻ → Cu
EoCu2+/Cu = EoCu2+/Cu - 0.0591/2 log1/[Cu²⁺] = 0.34 - 0.0591/2 log10¹⁹ = -0.22 V