Answer:
\[C{{u}^{+}}\] undergoes disproportionation to form \[C{{u}^{2+}}\] and Cu.
\[2\,C{{u}^{+}}(aq)\xrightarrow{\,}\,C{{u}^{2+}}(aq)+Cu(s)\]
Thus, \[C{{u}^{+}}\] or \[C{{u}_{2}}O\] acts both as an oxidant as well as a reductant
(i) When heated in air,\[C{{u}_{2}}O\] is oxidised to\[CuO,\]
\[\overset{+1}{\mathop{C{{u}_{2}}}}\,\overset{-2}{\mathop{O}}\,\,+1/2\overset{0}{\mathop{{{O}_{2}}}}\,\xrightarrow{\,}\,2\overset{+2}{\mathop{Cu}}\,\,\overset{-2}{\mathop{O}}\,\]
i.e., \[C{{u}_{2}}O\] acts as a reductant and reduces \[{{O}_{2}}\] to \[{{O}^{2-}}\].
(ii) When heated with \[C{{u}_{2}}S,\] it oxidises \[{{S}^{2-}}\] to \[S{{O}_{2}}\] and hence\[C{{u}_{2}}O\] acts as an oxidant
\[2\overset{+1}{\mathop{C{{u}_{2}}}}\,\overset{-2}{\mathop{O}}\,+\overset{+1}{\mathop{C}}\,{{u}_{2}}\overset{-2}{\mathop{S}}\,\xrightarrow{\,}\,6\overset{0}{\mathop{Cu}}\,+\overset{+4}{\mathop{S{{O}_{2}}}}\,\]
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