What is the potential of this cell ( in V) at 25 oC if the copper electrode is placed in a solution in which [Cu2+] = 5.1×10-10 M? I found this answer to be 0.065 which is correct. The second part: If the copper electrode of the above cell is placed in a solution of 0.270 M NaOH that is saturated with Cu(OH)2, what is the potential of the cell (in V, measured in the same direction as in question 5) at 25 oC? Ksp = 1.6×10-19. I'm using the Nernst equatuion. Here's how I think it goes: The NaOH is soluble so it dissolves but the Cu(OH)2 isn't. It's Ksp is very small but it will dissociate a little. So the dissociation equation I got is: Cu(OH)2 ---> Cu2+ + 2OH- Ksp = [Cu2+][OH-]^2 = [x][2x]^2 x = 3.42e-7 To find Ecell I said Cu gets oxidized and OH reduced and my Ecell = -0.74 Plugging into the Nernst equations: E=-0.74-(0.0592/4)*log[Cu2+][OH-]^2 E=-0.64 This is wrong. Can someone explain to me please. Thanks!