${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}\frac{{\mathrm{Cl}}_2}{\text{ step }-1}⟶{\mathrm{CH}}_3\mathrm{CHO}\frac{3{\mathrm{Cl}}_2}{\text{ step-2 }}\to$Cl₃CCHO. In the above reaction, the role
of Cl₂ in step-1 and step-2 respectively

Since loss of  hydrogen in  step-1 it  is Oxidation . However step -II Involves replacement of by Cl, we can say that it is chlorination reaction.