In the balanced chemical reaction,
IO−3+a I−+b H+⟶c H2O+d I2
a, b, c and d respectively correspond to
5, 6, 3, 3
IO−3+al−+bH+⟶CH2O+dl2
Step 1: I−1⟶I2(oxidation)
IO−3⟶I3(reduction)
Step 2: 2IO−3+12H+⟶I2+6H2O
Step 3: 2IO−3+12H++10e⟶I2+6H2O
[2I−⟶I2+2e]×5
Step 5: IO−3+5I−+6H+⟶3I2+3H2O
On comparing, a=5, b=6, c=3, d=3