CuS is oxidized by molecular oxygen to produce gaseous sulfur trioxide and solid copper (II) oxide. Write 2 balanced half reactions
CuS + 2O2 ==> CuO + SO3
(S^2- ==> S^6+ + 8e)
(O2 + 4e--> 2O^2-)
You want to make the electron gain/loss equal so multiply eqn 1 by 1 and eqn 2 by 2 like this, then add them together.
1*(S^2- ==> S^6+ + 8e)
2*(O2 + 4e-->2O^2-)_
________________________________________S^2- + 2O2 ==> s^6+ + 4O^2- and rearrange to
S^2- + 2O2 ==> SO3 + O^2- Note that this doesn't contain the cu(II) ion but you can add that if you wish.
I did it right thank you for confirmation!
To balance the half-reactions, we first need to identify the oxidation state of each element involved in the reaction.
CuS: The oxidation state of Cu (copper) is +2, and the oxidation state of S (sulfur) is -2.
O2: Oxygen is diatomic and exists as O2. The oxidation state of oxygen is 0.
Based on the given information, the two balanced half-reactions for this reaction can be written as follows:
Oxidation half-reaction: CuS -> CuO
Reduction half-reaction: O2 -> SO3
Now, let's balance each half-reaction step-by-step:
For the oxidation half-reaction:
CuS -> CuO
To balance the sulfur (S) atoms, add a coefficient of 3 in front of CuS:
3CuS -> CuO
For the reduction half-reaction:
O2 -> SO3
To balance the oxygen (O) atoms, add a coefficient of 3/2 in front of O2:
3/2 O2 -> SO3
Now, let's balance the charges by adding electrons (e-) to each half-reaction:
Oxidation half-reaction:
3CuS -> 3CuO + 6e-
Reduction half-reaction:
3/2 O2 + 6e- -> SO3
The electrons involved are equal in both half-reactions. To equalize the number of electrons, multiply the oxidation half-reaction by 2 and the reduction half-reaction by 3:
6CuS -> 6CuO + 12e-
9/2 O2 + 18e- -> 3SO3
Now, sum up both half-reactions:
6CuS + 9/2 O2 -> 6CuO + 3SO3