Question:
Balance the following reaction: $$\ce{CuS + SO4^{2-} -> CuO + SO2}$$
BalanceBelow, the following reaction:number mentioned in the parenthesis is the oxidation number of sulfur.
$\ce{CuS + SO4^{2-} -> CuO + SO2}$
My Efforts: Number mentioned in the brackets oxidation number of S
II have found out that the oxidation half-reactions and reduction half-reactions are as follows:
$\ce{CuS^{(-2)}-> ^{(4)}SO2}$ Oxidation half reaction
$\ce{^{(6)}SO4^{2-} -> ^{(4)}SO2}$ Reduction half reaction\begin{align} \ce{CuS^{(-2)} &-> ^{(4)}SO2}&& \text{Oxidation half reaction}\\ \ce{^{(6)}SO4^{2-} &-> ^{(4)}SO2}&& \text{Reduction half reaction} \end{align}
Problem:
In this, what about $\ce{CuO}$ which in the product side.? I have also learned that we need to balance the spectator ionions before charges if the spectator ion is other than oxygen and hydrogen.
Here, there is no $\ce{Cu}$ in reduction half reaction. So, I think there is something is wrong in my reduction half-reaction.