Justify that the reaction

$bold 2 bold Cu subscript bold 2 bold O bold left parenthesis bold s bold right parenthesis bold space bold plus bold space bold Cu subscript bold 2 bold S bold left parenthesis bold s bold right parenthesis bold space bold space bold rightwards arrow bold space bold space bold 6 bold Cu bold left parenthesis bold s bold right parenthesis bold space bold plus bold space bold SO subscript bold 2 bold left parenthesis bold g bold right parenthesis$
is a redox reaction. Identify the species oxidised/ reduced, which acts as an oxidant and which acts as a reductant.

Solution

Writing O.N. of each atom above its symbol,

space space space stack bold 2 bold Cu subscript bold 2 with bold plus bold 1 on top stack bold O bold left parenthesis bold s bold right parenthesis with bold minus bold 2 on top bold space bold space bold plus bold space stack bold Cu subscript bold 2 with bold plus bold 1 on top stack bold S bold left parenthesis bold s bold right parenthesis with bold minus bold 2 on top bold space bold space bold space bold space bold space bold rightwards arrow bold space bold space bold space bold space stack bold 6 bold Cu bold left parenthesis bold s bold right parenthesis with bold 0 on top bold space bold plus bold space bold S with bold plus bold 4 on top stack bold O subscript bold 2 bold left parenthesis bold g bold right parenthesis with bold minus bold 2 on top

Since O.N. of copper decreases from +1 (in Cu₂O) or Cu₂S to 0 in Cu, therefore Cu is reduced. Since the O.N. of S increases from -2 (in Cu₂S) to +4 (in SO₂), therefore S is oxidised. Hence, the above reaction is a redox reaction.
Further, O in Cu₂O helps S in Cu₂S to increase its O.N. from -2 to +4, therefore Cu₂O is an oxidant, and S in Cu₂S helps Cu both in Cu₂O and Cu₂S to decrease its O.N. of +1 to zero in Cu metal, therefore, Cu₂S is the reductant.