How does `Cu_2O` act as both oxidant and redcutant ? Explain with proper reactions showing the change of oxidation number in each case.

Explanation

In `Cu_2O` , copper is in +1 `(Cu^(+))` oxidation state and it undergoes disproportionation to form `Cu and Cu^(2+)` as :
`2Cu^(+)(aq)rarrCu(s)+Cu^(2+)(aq)`
Thus, `Cu^(+) ` or `Cu_2O` acts both as an oxidant as well as a reluctant.
For example,
(i) When heated in air , `Cu_2O` is oxidised to CuO.
`overset(+1" ")(Cu_2O)+1/2O_2rarroverset(+2" ")(2CuO)`
Thus, `Cu_2O` acts as a reluctant and reduces `O_2` to `O^(2-)`.
(ii) When heated with `Cu_2S` , it oxidises `S^(2-) " to " SO_2` and hence `Cu_2O` acts as an oxidant.
`overset(+1" ")(2Cu_2O)+overset(+1" ")(Cu_2S)rarroverset(0" ")(6Cu)+SO_2`