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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.

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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 : $$\ce{IO3^- + 5I- + 6 H^+ -> 3 I2 + 3 H2O}$$

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  • $\begingroup$ Oh didnt think of that. This helps a lot but how do you explain this theory with HNO3 + I2 reaction !? Where HNO3 is an acid ? $\endgroup$
    – Ashish
    Commented Mar 13, 2021 at 10:04
  • $\begingroup$ @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 $\ce{3 I2 + 10 HNO3 -> 6 HIO3 + 10 NO + 2 H2O}$ $\endgroup$
    – Maurice
    Commented Mar 13, 2021 at 11:29

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