separate this redox reaction into its component half-reactions.

Question

O2 + 4Cu-->2CuO

Oxidation half-reaction: ?
Reduction half-reaction: ?

I believe that the Oxidation is copper and that reduction is oxygen. So I tried to write it as

4Cu^+ --> 4Cu + 4e^-
O2^2- +4e^- ---> 2O

but it is wrong and I don't understand why. Please help.

Answer 1

The half reactions are
Cu(s) ==> Cu^2+ + 2e (you had 1+ for Cu)
O2(g) + 4e ==> 2O^2- (you can see how this differs from what you had).
The Cu is the oxidation half; O2 is the reduction half.

Answer 2

Cu -> Cu+ + e-

O2 + 4e- -> 2O-2