CoolFool - 28-6-2017 at 21:39
I read on wikipedia that KMnO4 reacts with KI to form K2MnO4. But if the reaction takes place in an aqueous solution would the K2MnO4 decompose back
to KMnO4 MnO2 and KOH. In the end would i be able to extract the Iodine?
JJay - 28-6-2017 at 22:12
I don't think that will work out well. Even assuming that the reaction goes cleanly, the MnO2 would precipitate along with the iodine and complicate
purification.
nezza - 29-6-2017 at 01:47
In alkaline solution iodide does reduce permanganate to manganate. If permanganate is added dropwise to an alkaline solution of iodide the purple
manganate is reduced to green manganate as a solution. No MnO2 is precipitated at low concentrations. I presume the iodide is oxidised to hypoiodite
or iodate. If the green manganate solution is acidified iodine is formed as a brown solution if excess iodide is present and the manganate is reduced
to soluble Manganese(II). Manganate is only stable in alkaline solution and in the absence of other contaminants disproportionates in acid to
permanganate and Mn2+ or MnO2 depending on the pH.
chornedsnorkack - 29-6-2017 at 07:22
From the Pourbaix diagrams, iodine is stable at pH <5. At higher pH it dismutes, to iodine and iodate.
Manganate is stable at pH>13.
So:
at pH<5:
2KMnO4+8HI->2KI+3I2+2MnO2+4H2O
at 5<pH<13:
2KMnO4+KI+H2O->KIO3+2MnO2+2KOH
at pH>13:
6KMnO4+KI+6KOH->6K2MnO4+KIO3+3H2O
Also, iodine tincture is a solution of iodine in ethanol. But ethanol can react with permanganate, and also is miscible with water.
[Edited on 29-6-2017 by chornedsnorkack]
ninhydric1 - 29-6-2017 at 11:33
Actually, some brands of iodine tincture are a solution of iodine, potassium iodide, and ethanol due to iodine being on the DEA list I chemicals.
clearly_not_atara - 29-6-2017 at 12:19
In other words, it sounds like you'll get iodate as the primary product. Iodic acid itself is quite stable, relative to other oxyhalic acids (save
perchloric acid).
For iodine I'm guessing the simplest method would use 3% H2O2 in dilute phosphoric acid. I think this works... just ensure [H3PO4] > [I-].
Triiodide production is a problem.
[Edited on 29-6-2017 by clearly_not_atara]
AJKOER - 3-7-2017 at 05:01
The primary compounds by weight are water and iodide. Note, the action of ozone on iodide (in excess) first creates IO- and not IO3- (see abstract
discussion available at https://www.researchgate.net/publication/231291875_Oxidation...). As such, I would expect in acidified iodide solutions to which is added some
KMnO4 (creating HOI) some I2 formation.
This agrees with chornedsnorkack and nezza comments above.
Reaction:
I- (aq) + HOI (aq) + H+ (aq) → I2 (aq) + H2O (ℓ)
Note, with H2O2:
2 I- (aq) + H2O2 (aq) + 2 H+ (aq) → I2 (aq) + 2H2O (ℓ)
[Edited on 3-7-2017 by AJKOER]