4 ClO3- + 2 SbO+ + 2 H+ 4 ClO2 + Sb2O5 + H2O

Answer 1

4 ClO3- + 2 SbO+ + 2 H+ ==> 4 ClO2 + Sb2O5 + H2O

Cl on the left has oxidation number (O.N.) of +5

Cl on the right has O.N. of -4

Cl has gained electrons so it has been reduced. It (ClO3-)is thus the oxidizing agent.

Sb on the left has O.N. of +3

Sb on the right has O.N. of +5

Sb has lost electrons so it has been oxidized. It (SbO+) is thus the reducing agent.