# Balancing a Redox Reaction in Acidic Solution

1. Oct 21, 2012

### 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".

1. The problem statement, all variables and given/known data
ICl → Cl- + IO3 + I2

3. 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+

2. Oct 22, 2012

### 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.