Balancing a Redox Reaction in Acidic Solution

[Bound]

I can balance this equation using the half reaction method, but I want to learn to use the oxidation number method with this particular question to "broaden my abilities".

Homework Statement

ICl → Cl^- + IO₃ + I₂

The Attempt at a Solution

I: +1 → +6: lost 5e^-
I: +1 → 0: gained 1e^-
(I'm unsure of the next step)

Using the half reaction method, I know the balanced equation to be:
12ICl + 6H₂O → 5I₂ + 12Cl^- +2IO₃ + 12H⁺

 

[AGNuke]

First of all, I am not sure if IO₃ is actually an oxide of Iodine, but if the question is hypothetical, I will do as asked, but for your information, it might be a disproportionation reaction in which the reaction goes as follows :
$ICl \rightarrow I_2 + IO_3^- + Cl^-$

Note that I formed Iodide Radical, which really exists. But still, I'd take IO₃ in the account.

Change in ON to form I₂ = 1
Change in ON to form IO₃ = 5

Write the LHS as ICl + ICl. Let 1st ICl being oxidised to IO₃ and 2nd ICl being reduced to I₂. We know their ON change. Now "Cross Balance" the ON difference as coefficients of reactants. Thus LHS becomes

ICl + 5 ICl. (ON diff. of 1st is 5, crossed to 2nd. Likewise, 2nd has ON diff of 1, crossed to 1st)

Now form the respective products from respective reactant stoichiometrically. Form 5/2 I₂ from 5 ICl and 1 IO₃ from 1 ICl. Now I am sure you can balance the Cl, H and O. That's Oxidation Number method.