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

Solution

Chemistry ma’am · 4 steps

At pH 14, Kₛₚ tells you how little Cu²⁺ is left, then put that into Nernst.

1. Find the hydroxide. At pH = 14, pOH = 0, so [OH⁻] = 1 M.
2. Free copper ions. Cu(OH)₂ is saturated, so Kₛₚ = [Cu²⁺][OH⁻]². With [OH⁻] = 1 M, [Cu²⁺] = 1 × 10⁻²⁰ M.

   E = E° - 0.059/2 log 1/[Cu²⁺] = 0.34 - 0.0295 × 20

3. Put in numbers. For Cu²⁺ + 2e⁻ → Cu, n = 2. Then log(10²⁰) = 20, and 0.0295 × 20 = 0.59 V.
4. So. E = 0.34 - 0.59 = -0.25 V = -25 × 10⁻² V, so x = 25.

The common slip: 84. Taking n = 1 instead of 2 doubles the correction to 1.18 V and gives E = -0.84 V, so 84, not 25.

The idea to keep. When a sparingly soluble salt sits on the electrode, use Kₛₚ to find the free metal ion concentration. That concentration is what goes into the Nernst equation.

Answer from the NTA final answer key. Solution by PrepWiser.