Balance this equation using oxidation method:I2+HNO3---->HIO3+NO2+H2O

I2+10HNO3 -----> 2HIO3+10NO2+4H2O

Good

To balance the given equation using the oxidation method, we need to ensure that the number of atoms on both sides of the equation is equal, and the charges are balanced. Let's go step by step.

1. Split the equation into two half-reactions:
Half-reaction 1: I2 → HIO3
Half-reaction 2: HNO3 → NO2 + H2O

2. Balance atoms other than hydrogen and oxygen:
Half-reaction 1: I2 → 2HIO3
Half-reaction 2: 2HNO3 → 2NO2 + H2O

3. Balance oxygen atoms by adding H2O molecules to the side that lacks oxygen:
Half-reaction 1: I2 → 2HIO3 + 10H2O
Half-reaction 2: 2HNO3 → 2NO2 + 5H2O

4. Balance hydrogen atoms by adding H+ ions to the side that lacks hydrogen:
Half-reaction 1: I2 → 2HIO3 + 10H2O + 12H+
Half-reaction 2: 2HNO3 + 10H+ → 2NO2 + 5H2O

5. Balance the charges by adding electrons to the side that is more positively charged:
Half-reaction 1: I2 + 12e- → 2HIO3 + 10H2O + 12H+
Half-reaction 2: 2HNO3 + 10H+ + 12e- → 2NO2 + 5H2O

6. To equalize the number of electrons, multiply both half-reactions by appropriate coefficients:
Half-reaction 1: I2 + 12e- → 2HIO3 + 10H2O + 12H+
Half-reaction 2: 12HNO3 + 60H+ + 72e- → 12NO2 + 30H2O

7. Finally, add both balanced half-reactions together:
I2 + 12HNO3 + 60H+ → 2HIO3 + 12NO2 + 40H2O

The balanced equation using the oxidation method is:
I2 + 12HNO3 + 60H+ → 2HIO3 + 12NO2 + 40H2O

Here is how you do these. You can start with the topic you don't know and use as many of the steps as you need.

http://www.chemteam.info/Redox/Redox.html

I STRONGLY ADVISE YOU TO START BY PLACING A 2 AS A COEFFICIENT FOR HIO3 so that you are talking about electron loss/gain for the SAME number of atoms.
I has an oxidation state of zero on the left and a total of 10 (for both I atoms) in HIO3.
N has a oxidation state of +5 in HNO3 and +4 in NO2.
I can help you through it; you go as far as you can, post what you've done, explain what you don't understand about the next step.