Why does ozone disproportionate on reaction with potassium iodide?

In this redox reaction, iodide ion is oxidized to iodine and ozone reduces to hydroxide ion.

$$\ce{2KI + O3 + H2O -> I2 + O2 + 2KOH}$$

This seems like a redox reaction since iodide ion is getting oxidized and ozone is getting reduced. Then why is $O_{2}$ formed in this reaction?

• To me, this hardly qualifies as disproportionation. Jul 31 '17 at 15:34
• @Ivan Neretin yeah you are right. This isn't disproportionation since ozone is only getting reduced and not oxidized. But the question still remains. Why is oxygen being formed? Jul 31 '17 at 15:40
• Well, $\ce{O2}$ is much less active than $\ce{O3}$, so the reduction just stops right there for kinetic reasons. If you apply a lot of $\ce{KI}$ and wait for some days, all oxygen will get reduced. Jul 31 '17 at 15:43
• An intermediate step is still a step. If you reduce $\ce{O3}$, then one atom out of three gets reduced and the rest transform to the more stable $\ce{O2}$. If you don't reduce it, then there is no reaction and nothing to speak about. Jul 31 '17 at 15:55
• @Eloise I think you have that backward. We use electric discharge and strict conditions of temperature and pressure to convert oxygen to ozone. To reduce ozone to oxygen all you need is time; it self-disproportionates. If you don't have time, there are several catalysts that will make the disproportionation go faster.
– zwol
Jul 31 '17 at 17:23

The reaction equation does not really give information about the full mechanism, but one possibility is that the ozone is first reduced to ozonide ion, which is known in compounds with heavier alkali metals like potassium,; then the ozonide ion disprportionates to give diatomic oxygen:

$\ce{O3}+\ce{e-}\rightarrow \ce{O3-}$

$\ce{O3-}+\ce{e-}+\ce{H2O}\rightarrow 2 \ce{OH-}+\ce{O2}$

Note that the reducing agent is not specified as iodide ion. Any mild reducing agent can act similarly.

As @Ivan said in the comment:

An intermediate step is still a step. If you reduce $\ce{O3}$, then one atom out of three gets reduced and the rest transform to the more stable $\ce{O2}$. If you don't reduce it, then there is no reaction and nothing to speak about.

This is the case. In an intermediate step, ozone gets reduced to diatomic oxygen and a monoatomic oxygen, also called nascent oxygen. Generally, this nascent oxygen is unstable so it reacts/gets used up to oxidize potassium iodide to iodine.

(Source)

Diatomic oxygen is stable and does not get used up in the reaction leaving it untouched during the reaction.