Acidic solution balancing

Question

How to balance this equation?

IO3^- + I^- -> i2

Answer 1

IO₃^- + I^- -> I₂

The first thing you have to do is break this up into half-reactions. IO₃^- is reduced to I₂, I(+5) goes to I(0). In the other half reaction, I^- is oxidized to I₂.

Start with the oxidation half: I^- -> I₂

Add a coefficient to balance the iodine atoms: 2 I^- -> I₂

Now we must use electrons to balance the charge: 2 I^- -> I₂ + 2 e^-

Total charge of -2 on each size, total of 2 iodine atoms on each side, this is a balanced half-reaction.

Now reduction: IO₃^- -> I₂

Balance iodine atoms: 2 IO₃^- -> I₂

Now balance oxygens. Since we're in acidic aqueous solution, we can use water molecules to balance O:

2 IO₃^- -> I₂ + 6 H₂O

To balance H, remember water has H(+1) so we need H+ on the left:

12 H⁺ + 2 IO₃^- -> I₂ + 6 H₂O

But now we have a total charge of +10 on the left, neutral on the right, add 10 electrons to the left:

10 e^- + 12 H⁺ + 2 IO₃^- -> I₂ + 6 H₂O

Now we have to put the 2 half-reactions together.

2 I^- -> I₂ + 2 e^-

10 e^- + 12 H⁺ + 2 IO₃^- -> I₂ + 6 H₂O

Problem is that 2 electrons don't balance 10 electrons. We can fix this by multiplying the oxidation half by 5.

10 I^- -> 5 I₂ + 10 e^-

10 e^- + 12 H⁺ + 2 IO₃^- -> I₂ + 6 H₂O

Now we can add both sides of each reaction together to get the final reaction:

10 I^- + 10 e^- + 12 H⁺ + 2 IO₃^- -> 5 I₂ + 10 e^- + I₂ + 6 H₂O

Cancel the electrons from each side, and combine like terms:

10 I^- + 12 H⁺ + 2 IO₃^- -> 6 I₂ + 6 H₂O

All coefficients are multiples of 2, so divide:

5 I^- + 6 H⁺ + IO₃^- -> 3 I₂ + 3 H₂O

Looks balanced. Check it: 6 Iodine, 6 H, 3 O, on each side. Total charge 0 on each side.