How can we distinguish b/w the oxidizing and reducing agent in the disproportionation reaction reaction.Cu2O + Cu2s ----> Cu+SO2.

Cu⁺ undergoes disprportionation to form Cu²⁺ and Cu. Thus Cu+ and Cu₂O acts as an oxidant as well as reductant.

When heated with Cu₂s,Cu₂O oxidises S^2- to SO₂ in which oxidation of S is +4.

2Cu⁺² O + Cu₂S^2- = 6Cu⁰  + SO⁺⁴₂

Thus, ?Cu2O acts as an oxidant.

Approved

Sorry, this is not a dispropationation reaction. Its a redox one, i just wrote it in a hurry. 

Well acc to ur answer,  Cu 1+ is showing disproptionation to 2+ and 0 . How this possible ??? Explain.

Its is getting reduced from 1+ to 0 while S^2- wi oxidized from -2 to +4 oxidation state.  Well as such Cu2O as well Cu+ should act as the oxidizing agent and S^2- should be the reducing agent.

But in the book, Cu₂S  is the reducing agent and Cu2O is the oxidising one.

I just wanna clear my doubt.