[pzona69]

I'm trying to oxidize KI using H2O2 (aq), and for my products (trying to work it out on paper), I'm getting HI and KO.

KO seems like some kind of weird thing I shouldn't be getting, and I'm not even sure H2O2 (aq) would actually oxidize KI. I just wanted to know what the formula would look like.

[enahs]

This is a typical lab showing the production of Oxygen gas (remember Potassium Iodide is soluble in water).


KI -> K⁺_(aq) + I^-_(aq)

H₂O_{2 (aq)}+ I^-_(aq) -> H₂O_(l) + IO^-_(aq)
IO^-_(aq) + H₂O_{2 (aq)} -> I^-_(aq) + H₂O_(l) + O_2(g)


Sometimes you can see a slight yellow color due to a side reaction, given the correct conditions of:
2H⁺_(aq) + IO^-_(aq) + 2I^- -> I₃^-_(aq) + H₂O_(l)

[pzona69]

My ultimate goal here is actually to get pure (relatively pure anyway) I₂. I couldn't just dissolve the KI in water and separate it that way (?), so I figured I would need to oxidize it first to get HI, and then go from there. Please tell me if I'm wrong, or if there's an easier way.

[enahs]

My ultimate goal here is actually to get pure (relatively pure anyway) I₂. I couldn't just dissolve the KI in water and separate it that way (?), so I figured I would need to oxidize it first to get HI, and then go from there. Please tell me if I'm wrong, or if there's an easier way.

I am at work, so I do not have access to my library as well as various book-marks, or time to search. But I remember, and Wiki confirms (or I confirm wiki!) that I₂ can be produced by oxidizing iodides with Chlorine gas (Cl₂), and also via Manganese Dioxide in acid solution.

I would start with those, looking around on the internet and trying to find more precise reaction conditions, etc.

Wiki on Iodine production, and other stuff

[pzona69]

That's very helpful, thank you.

I'm working with a mix of KI with some (not sure exactly how much) NH₄I.
Would the oxidation of ammonium iodide be the same as the potassium iodide? I'm thinking it would be similar, but I just want to make sure.

[enahs]

Ammonium Iodide is easily oxidized to I₂ just by the presence of air (O₂ in air to me more precise). I do not see it causing a problem unless you have a lot of extra NH₄⁺ in the solution, causing various side reactions and generally making things more complicated.

You have probably already oxidized out all the I from the Ammonium Iodide.

[AWK]

All soluble inorganic iodides can be easily converted to iodine by oxidation in acidic, basic or neutral media.
Reaction with H2O2, even diluted, goes immediately.
2KI + H2O2 = 2KOH + I2.
It is better to acidify reaction because KOH can react further with iodine., eg:
2KOH + I2 = KOI + KI + H2O

[IITian]

If you have good ventilation, you can bubble Cl2(g) through a solution of KI.

[Kryolith]

The reaction of potassium iodide and halogen free copper sulfate yields very pure iodine.

[Mr. Jack]

In my experience, a small amount of H2SO4 and H2O2 will oxidize nearly all the KI to I2.  It's very effective and gives 90+% theoretical yield.

[AWK]

The reaction of potassium iodide and halogen free copper sulfate yields very pure iodine.

But with yield only 50 %

[pzona69]

The reaction of potassium iodide and halogen free copper sulfate yields very pure iodine.

Under what type of conditions does the reaction occur? Are there any catalysts I should know about?

[Kryolith]

Under what type of conditions does the reaction occur? Are there any catalysts I should know about?

Normal aqueous conditions, no catalysts. Iodine will precipitate.

But with yield only 50 %

Of course you have the same quantity of CuI as a "byproduct". But pure iodine in lab scale quantities (correct formulation?) is expensive

[AWK]

Of course you have the same quantity of CuI as a "byproduct". But pure iodine in lab scale quantities (correct formulation?) is expensive

Not the same quantity - only an equivalent quantity, ie
2 moles CuI per mole I₂ and yield calculated for iodine is only 50 %

Reaction of KI with H₂O₂ gives yield of I₂ equal to 100%.
KI and I2 are relatively expensive hence such improvement of yield is profitable