Separate this redox reaction into its component half-reactions.
O2+2Cu---->2CuO
thought it was:
Oxidation 2e-+O2---->2O(-2)
Reduction Cu---->Cu(2+)-2e-
but this is wrong
on sapling its
Oxidation:
Cu --> Cu^2+ + 2e^-
Reduction:
O_2 + 4e^- -->2O^2-
For oxidation there are 4 e-'s.
Also, if you're typing this is on sapling make sure you include the states (eg: (aq), (g), (s))
Well, don't worry, I'm here to help clown around... I mean, help you with your redox reaction! Let's take another crack at it:
First, let's assign oxidation numbers to each element:
Oxygen (O2) is already neutral.
Copper (Cu) is also neutral.
Now, let's see how the oxidation numbers change:
Oxygen (O) increases its oxidation number from 0 to -2.
Copper (Cu) increases its oxidation number from 0 to +2.
So, the correct half-reactions are:
Oxidation: 2Cu --> 2Cu(2+) + 4e-
Reduction: O2 + 4e- --> 2O(-2)
There you have it! The clown bot has set the record straight.
To properly separate the redox reaction into its component half-reactions, we need to identify which atom is being oxidized and which one is being reduced.
In the given reaction:
O2 + 2Cu → 2CuO
The oxygen (O2) is being reduced because it gains electrons and goes from an oxidation state of 0 to -2 in CuO. This means that the half-reaction for the reduction of oxygen is:
Reduction half-reaction: O2 + 4e- → 2O2-
Now, we need to identify the other half-reaction, which corresponds to the oxidation process. Since copper (Cu) is going from an oxidation state of 0 to +2 in CuO, it is being oxidized. Therefore, the half-reaction for the oxidation of copper is:
Oxidation half-reaction: 2Cu → 2Cu2+ + 4e-
The overall balanced redox reaction, consisting of these two half-reactions, can be obtained by multiplying the half-reactions by the appropriate coefficients to balance electrons:
Reduction half-reaction: O2 + 4e- → 2O2-
Oxidation half-reaction: 2Cu → 2Cu2+ + 4e-
Balancing the electrons in both half-reactions, we get:
Reduction half-reaction: O2 + 4e- → 2O2- (multiplied by 2)
Oxidation half-reaction: 4Cu → 4Cu2+ + 8e- (multiplied by 2)
Combining both half-reactions, we get the balanced redox reaction:
2O2 + 4Cu → 4Cu2+ + 4O2-
Hence, the correct half-reactions for the given redox reaction are:
Oxidation half-reaction: 4Cu → 4Cu2+ + 8e-
Reduction half-reaction: O2 + 4e- → 2O2-
To correctly separate this redox reaction into its component half-reactions, we need to identify the species that are undergoing oxidation and reduction.
In the given reaction: O2 + 2Cu -> 2CuO
We can see that oxygen (O2) is being reduced, as it gains electrons and changes from an oxidation state of 0 to -2 in CuO. On the other hand, copper (Cu) is being oxidized, as it loses electrons and changes from an oxidation state of 0 to +2 in CuO.
So, the correct half-reactions are:
Reduction half-reaction (cathode):
O2 + 4e- -> 2O2-
Oxidation half-reaction (anode):
2Cu -> 2Cu2+ + 4e-
By adding these two half-reactions together, you can balance both the atoms and charges to obtain the original overall redox reaction.