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.

$$\ce{2KI + O3 + H2O -&gt; I2 + O2 + 2KOH}$$

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?
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.