In the acidic medium, ${\mathrm{MnO}}_4^{-}$ convert to ${\mathrm{Mn}}^{2+}$ and ${\mathrm{I}}^{-}$ convert to ${\mathrm{I}}_2$.

$2{\mathrm{MnO}}_4^{-}+10{\mathrm{I}}^{-}+16{\mathrm{H}}^{+}\to 2{\mathrm{Mn}}^{2+}+5{\mathrm{I}}_2+8{\mathrm{H}}_2\mathrm{O}$

In neutral/faintly alkaline solution, ${\mathrm{MnO}}_4^{-}$ convert to ${\mathrm{MnO}}_2$ and ${\mathrm{I}}^{-}$ convert to ${\mathrm{IO}}_3^{-}$.

$2{\mathrm{MnO}}_4^{-}+{\mathrm{I}}^{-}+{\mathrm{H}}_2\mathrm{O}\to 2{\mathrm{MnO}}_2+2{\mathrm{OH}}^{-}+{\mathrm{IO}}_3^{-}$