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The half equation for the reduction of the permanganate would be
$$\ce{5e- + 8H+ + MnO4- -> Mn^2+ + 4H2O}$$
But with regards to the iodide, I am very confused. I have seen some sources say that $\ce{I-}$ is oxidised to $\ce{IO3-}$ in this reaction, and others saying that it is oxidised to $\ce{I2}$.
How stable is $\ce{IO3-}$ in the presence of $\ce{I-}$ under acidic conditions?
It would end up being in an equilibrium with very few $\ce{IO3-}$ anions and lots of iodine, due to the following equilibrium:
$$\ce{2 IO3- + 12 H3O+ + 10 I- <=> 6 I2 + 18 H2O}$$
The problem here is that in acidic medium iodide ($\ce{I-}$) is oxidized to iodine ($\ce{I2}$). But in neutral or faintly alkaline medium, it is oxidized to iodate ($\ce{IO3-}$).
Also in neutral or alkaline medium the ionic half equation of reduction of permanganate would change to:
$$\ce{MnO4- + 2 H2O + 3e- -> MnO2 + 4 OH-}$$
