Cu\(_2\)S(s) + O\(_2\)(g) → 2CU(s) + SO\(_2\)(g)
What is the change in the oxidation number of copper in the reaction above?
Sulphur atom in Cu\(_2\)S is - 2. Thus, Cu\(_2\)S = 0
2Cu + S = 0
2Cu + (-2) = 0
2Cu = 2
\(\frac{2Cu}{2}\) = \(\frac{2}{2}\)
Cu = +1
On the RHS, Cu atom has an oxidation state of 0. Therefore, there is a change in oxidation number of Copper from +1 to 0
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