Balance this equation:

I2+HNO3 = HIO3+NO2+H2O

Redox. I gains +5 (it is oxidized)

N goes from +5, to +4, so it takes 5N for each I

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

To balance the equation I2 + HNO3 = HIO3 + NO2 + H2O, we need to ensure that the number of each type of atom is equal on both sides of the equation.

1. Start by counting the number of atoms for each element on both sides of the equation:
On the left side:
Iodine (I): 2
Nitrogen (N): 1
Hydrogen (H): 1
Oxygen (O): 3

On the right side:
Iodine (I): 1
Nitrogen (N): 1
Hydrogen (H): 1
Oxygen (O): 5

2. Since there are more oxygen atoms on the right side than the left side, let's balance the oxygen atoms first.

Add a coefficient of 2 in front of HIO3 on the right side:
I2 + HNO3 = 2HIO3 + NO2 + H2O

3. Now, recheck the count of each atom:
On the left side:
Iodine (I): 2
Nitrogen (N): 1
Hydrogen (H): 1
Oxygen (O): 3

On the right side:
Iodine (I): 2
Nitrogen (N): 2
Hydrogen (H): 2
Oxygen (O): 9

4. The iodine atoms are balanced, but the nitrogen and hydrogen atoms still need balancing.

Add a coefficient of 4 in front of NO2 on the right side and a coefficient of 2 in front of HNO3 on the left side:
2I2 + 2HNO3 = 2HIO3 + 4NO2 + H2O

5. Recheck the count of each atom again:
On the left side:
Iodine (I): 4
Nitrogen (N): 2
Hydrogen (H): 2
Oxygen (O): 6

On the right side:
Iodine (I): 4
Nitrogen (N): 4
Hydrogen (H): 2
Oxygen (O): 14

6. Now the equation is balanced with respect to the number of atoms on each side.

The balanced equation is:
2I2 + 2HNO3 = 2HIO3 + 4NO2 + H2O