# I- oxidation to I2 or IO3- [duplicate]

Is there any logical way to determine whether $$\ce{I-}$$ will oxidize to form $$\ce{I2}$$ or $$\ce{IO3-}$$ is a reaction ?

For example,
In reaction of $$\ce{KMnO4}$$ with $$\ce{I-}$$ in acidic medium $$\ce{I2}$$ will be formed, whereas in the reaction of $$\ce{KMnO4}$$ with $$\ce{I-}$$ in alkaline medium $$\ce{IO3-}$$ would be formed.

$$\ce{HNO3}$$ will oxidize $$\ce{I2}$$ to $$\ce{IO3-}$$.

Is it related to the oxidation power or the reaction conditions and medium or things out of my scope ? PS. I'm currently a high school student.

• My question was not really specific to KMnO4 however it does cover the general idea too Commented Apr 13, 2021 at 4:31
• Commented Apr 13, 2021 at 8:22

In alkaline solution, $$\ce{I2}$$ does not exist. It is transformed into $$\ce{IO3^-}$$ and $$\ce{I^-}$$ according to $$\ce{3I2 + 6OH^- -> IO3^- + 5I- + 3 H2O}$$ In acidic conditions the mixture $$\ce{IO3^- + 5I-}$$ reacts in the opposite direction : $$\ce{IO3^- + 5I- + 6 H^+ -> 3 I2 + 3 H2O}$$
• @Ashish. The two equations I gave you about iodine and its ions are only valid in dilute aqueous solution. $\ce{HNO3}$ is an acid if diluted in water. As a pure substance, it is a strong oxidizing substance. And it reacts with $\ce{I2}$ according to $\ce{3 I2 + 10 HNO3 -> 6 HIO3 + 10 NO + 2 H2O}$ Commented Mar 13, 2021 at 11:29