Chemistry Question:

Question

The standard reduction potential for Cu²⁺, the voltage measured between a standard hydrogen electrode and a Cu electrode immersed in a solution where [Cu²⁺] = 1.00 M, is 0.340 V. The positive voltage indicates that the following reaction is spontaneous. H₂(g) + Cu²⁺(aq) → 2H⁺(aq) + Cu(s) What would be the Cu²⁺ reduction potential if the standard hydrogen electrode was kept the same, but copper electrode were instead immersed in a solution of 0.100 M NaOH that was saturated with Cu(OH)₂? (The Ksp of Cu(OH)₂ is 2.20 × 10⁻²⁰.)

Answer 1

This is an interesting question requiring the use of both the Ksp equation and the Nernst equation. First, we'll have to find the actual concentration of Cu²⁺ ions in solution, and then use the Nernst equation to find the voltage.

Cu(OH)₂(s) ==> Cu²⁺(aq) + 2OH^-(aq)

Ksp = 2.20x10^-20 = [Cu²⁺][OH^-]² = [Cu²⁺][0.100]² (use 0.100 for OH^- since it is a common ion from NaOH)

2.20x10^-20 = [Cu²⁺][0.01]²

[Cu²⁺] = 2.2x10^-18 M

Nernst equation:

Ecell = Eºcell - RT/nF ln Q

Assuming 25ºC (298K), the Nernst is simplified to

E_cell = Eº_cell - 0.0592 / n log Q where Q = [H⁺]² / [Cu²⁺] and n = 2

E_cell = 0.340 - 0.0296 log (1² / 2.2x10^-18]

E_cell = 0.340 - 0.0296 log 4.55x10¹⁷

E_cell = 0.340 - 0.523

E_cell = -0.183 V

Be sure to check the math