Cl02⟶2Cl+5O3−+10e−
To balance O and H, we first find out side deficient in ' O′ atoms. Cl2+6H2O⟶2ClO3+10e−
Then find out side deficient in H and add H2O, then add equal number of OH on opposite side. Cl2+6H2O⟶2ClO3+12H2O
Adding OH− Cl2+12OH−⟶2ClO3+6H2O+10e−