Discussion Forum

The standard reduction potential for Cu²⁺ /Cu is + 0.34. Calculate the reduction potential at pH = 14 for the above couple. (K_sp Cu(OH)₂ = 1×10^-19

Answer:

Explanation:

When pH = 14 [H]⁺ = 10^-14 

and [OH^-] = 1M

K_sp = [Cu²⁺] [OH^-]² = 10^-19

.'. [Cu²⁺] = $\frac{10^{-19}}{[OH^{-]^{2}}}$ = 10^-19

The half cell reaction

Cu²⁺ + 2e^- → Cu

E = E⁰ - $\frac{0.059}{2}\log_{}{}\frac{1}{[Cu^{2+}]}$

= 0.34 - $\frac{0.059}{2}\log_{}{}\frac{1}{10^{-19}}$ = -0.22V