Question

1)Find oxidised and reduced substance ,oxidising and reducing agent.
(i)CuO + H2 ----> Cu + H2O
(ii)CuO + C ----> Cu + Co
(iii)MnO2 + 4HCl ----> MnCl2 + Cl2 + 2H2O
(iv) 2H2S + SO2 ----> 2H2O + 3S
(v)Zn(s) + H2SO4(g) ----> ZnSO4(aq) +H2(g)​

Answer 1

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CuSo4 = Cu + So(1) + O_4

Answer 2

Explanation:

Answer

                             MnO2+4HCl⟶MnCl2+2H2O+Cl2

Oxidation state:   +4               −1             +2                            0

Increase in oxidation state = Oxidation ⇒

As chlorine increases from −1⟶0⇒ (Oxidation of HCl) 

Decrease in oxidation state = Reduction ⇒ As manganese decreases from +4→+2⇒ (Reduction of MnO2)

⇒MnO2 is reduced, HCl is oxidized, MnO2 is oxidizing agent, HCl is reducing agent.

Hope it helps !!!