Question

A galvanic cell consists of a standard hydrogen electrode and a copper electrode. Suppose that the copper electrode is immersed in a solution that is 0.1 M NaOH and that it is saturated with Cu(OH)2. Find the cell potential?

Answer 1

Cu(OH)2 Ksp = 5.92×10^–15

Cu (OH)2 -------------------------> Cu+2 + 2OH-

                                                     S        2S +0.1 = 0.1

Ksp = [Cu+2] [OH-]^2

5.92×10^–15   = [Cu+2] x (0.1)^2

[Cu+2] = 5.92×10^–13 M

for standard Hydrogen - electrode [H+] = 1.00 M

anode :

H2 ------------------> 2H+ + 2e-

cathode :

Cu+2 + 2e- ------------------> Cu

overall reaction :

H2   +        Cu⁺² -------------> 2H+      + Cu

1atm      (5.92×10^–13 M )       1M

for standard hydrogen electrode [H+] = 1.0M , P_H2 = 1 atm , Eo = 0.00V

E^o cell = 0.34 V

nernest equation :

E = Eo - 0.0591 / 2 * log [H+]^2 / [Cu+2]

    = 0.34 - 0.0591 / 2 x log [1 / 5.92×10^–13]

    = - 0.021 V

Ecell = - 0.021 V