separate this redox reaction into its component half-reactions.
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.
Cu -> Cu+ + e-
O2 + 4e- -> 2O-2
Well, it seems like you've got the right idea with identifying copper as the species being oxidized and oxygen as the species being reduced. However, your half-reactions are not quite correct.
Let's take a closer look:
Oxidation half-reaction:
Cu --> Cu^2+ + 2e^-
Reduction half-reaction:
O2 + 4e^- --> 2O^2-
In the oxidation half-reaction, copper (Cu) is being oxidized, meaning it loses electrons to form copper ions (Cu^2+), which have a greater positive charge.
In the reduction half-reaction, oxygen (O2) is being reduced, meaning it gains electrons to form oxide ions (O^2-), which have a greater negative charge.
So the correct half-reactions for the redox reaction O2 + 4Cu --> 2CuO are:
Oxidation half-reaction: Cu --> Cu^2+ + 2e^-
Reduction half-reaction: O2 + 4e^- --> 2O^2-
I hope this clears up any confusion!
Your approach is correct, but the charges on the copper ions and the oxygen ions are incorrect.
In the given reaction: O2 + 4Cu --> 2CuO
The oxidation half-reaction involves the species that is undergoing oxidation, which is copper in this case.
Cu loses 2 electrons to form Cu2+ ions:
Cu --> Cu2+ + 2e^-
Therefore, the oxidation half-reaction is:
4Cu --> 4Cu2+ + 8e^-
The reduction half-reaction involves the species that is undergoing reduction, which is oxygen in this case.
Oxygen (O2) gains 4 electrons to form oxide ions (O2-):
O2 + 4e^- --> 2O2-
Therefore, the reduction half-reaction is:
O2 + 4e^- --> 2O2-
This results in a balanced redox reaction:
4Cu + O2 --> 2Cu2O
To separate the redox reaction into its component half-reactions, we need to identify the species that is being oxidized and the species that is being reduced.
In the given reaction, O2 is being reduced to form CuO, which means it gains electrons. On the other hand, Cu is being oxidized to form CuO, which means it loses electrons.
Let's start with the reduction half-reaction:
O2 + 4e^- --> 2O^2-
In this half-reaction, O2 gains 4 electrons to become O^2-.
Now, let's move on to the oxidation half-reaction:
4Cu --> 4Cu^2+ + 8e^-
In this half-reaction, Cu loses 8 electrons to become Cu^2+.
Notice that in each half-reaction, the number of electrons gained and lost is balanced. However, your initial attempt at writing the half-reactions was partially incorrect.
Make sure to assign the correct charges to the species involved. The oxidation half-reaction should be:
4Cu --> 4Cu^2+ + 8e^-
And the reduction half-reaction should be:
O2 + 4e^- --> 2O^2-
By balancing the charges and electrons gained/lost in each half-reaction, we can then combine them to form the complete balanced redox equation:
4Cu + O2 → 2CuO
I hope this explanation helps you understand the process of separating redox reactions into half-reactions!
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.