# Balancing disproportionation reaction of hypoiodous acid

In aqueous solution, the acid $$\ce{HIO}$$ disproportionates according to the following equation where m, n, p and q are simple whole numbers in their lowest ratios.

$$\ce{mHIO → nI2 + pHIO3 + qH2O}$$

This equation can be balanced using oxidation numbers.

In $$\ce{HIO}$$ and $$\ce{HIO3}$$, the oxidation state of iodine is +1 and +5 respectively. Therefore in $$\ce{HIO}$$, iodine lost 4 electrons to iodine atoms, therefore n is 2 and finally p is 1. However, when the electrons are received by the iodine atoms, they should form iodide ions instead of molecules. How would you tackle this question?

• They are not and they should not. // Do not you think HIO is strong enough oxidizer to prevent creation of easily oxidized I- ? Sep 4, 2021 at 18:00
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Sep 4, 2021 at 19:50

The disproportionation rection of $$\ce{HIO}$$ cannot produce $$\ce{HIO3}$$ and $$\ce{HI}$$ (or $$\ce{I^-}$$ ions) in acidic solution. The reason is that $$\ce{I^-}$$ and $$\ce{IO3^-}$$ ions do react with one another, if $$\ce{H^+}$$ ions are present, to give : $$\ce{IO3^- + 5 I^- + 6 H^+ -> 3 I2 + 3 H2O}$$ If no ion $$\ce{H^+}$$ is present, the previous reaction does not occur, and the solution contains simultaneously the ions $$\ce{I-}$$ and $$\ce{IO3^-}$$