How to do this redox reaction question?

There are 3 O atoms in each O₃ molecule. In the reaction shown, 2 of the oxygen atoms are converted to an oxygen molecule and the rest one is converted to an H₂O molecule. The conversion from O₃ to O₂ is not redox, because there is no change in oxidation number. However, the conversion of O₃ to H₂O in an acidic medium is reduction, because the oxidation number of O decreases from 0 (in O₃) to -2 (in H₂O). The half equation for reduction is :
O₃(g) + 2H⁺(aq) + 2e⁻ → O₂(g) + H₂O(l)

On the other hand, I⁻ ions are oxidized. The half equation for oxidation is :
2I⁻(aq) → I₂(aq) + 2e⁻

Combining the above 2 half equations, one would get the overall equation.

Look at the iodine. It is an ion, not an element. It goes from -1 on the left to 0 on the right, so it is oxidized.

One oxygen goes from 0 on the left to -2 on the right in the H20, so it is reduced.

O₃ + 2H⁺ + 2e⁻ → O₂ + H₂O