Question

"Write equations for the half-reactions that occur at the anode and cathode for the electrolysis of the following aqueous solution: CuBr2(aq)"

Possible oxidation reactions:
1. 2Br- -> Br2 + 2e- = -1.09
2. 2H2O -> O2 + 4H+ + 4e- = -1.23

Possible reduction reactions:
1. Cu2+ + 2e- -> Cu = 0.34
2. 2H2O + 2e- -> H2 + 2OH- = -0.83


I thought that, since the first oxidation (Br) and first reduction (Cu2+) were more positive, they would occur first, but it keeps saying my answer is wrong. I'm worried it may just be some sort of formatting thing.

This is my current answer:
Cu2+(aq) + 2e- -> Cu(s), 2Br-(aq) -> Br2(l) + 2e-

Answer 1

Cu^2+ comes out before H^+ at the negative electrode.
But I think the 1.23 occurs before 1.09.