Sciencemadness Discussion Board - Making sodium iodide from iodine

Making sodium iodide from iodine

garage chemist - 31-3-2012 at 08:43

From several batches of aqueous acidic iodide-bearing liquor I have reclaimed the iodine by precipitation with hydrogen peroxide.
The iodine was filtered and washed with water, and the wet slurry-like iodine filter cake scraped out of the filter and stored in a jar.

Now I want to turn it into sodium iodide to be able to use it in syntheses.
The problem is that the iodine slurry contains a large and unknown amount of water, so I can't do any exact stochiometric reactions.
I thought about doing it as follows: Slowly add NaOH solution until the iodine is all dissolved and the solution no longer brown, then heat to a boil to disproportionate the hypoiodite completely to iodate and iodide.
Then add hydrazine solution while hot until there is no more nitrogen evolution to reduce the iodate to iodide, then evaporate everything to dryness to get my NaI.

I first thought of reacting the iodine itself with hydrazine to produce hydriodic acid and neutralize this afterwards, but then I read this:
http://www.chemicalforums.com/index.php?topic=20821.0
where someone reports that hydrazine increasingly disproportionates in strongly acidic HI solution forming ammonium iodide and nitrogen instead of reacting with the iodine, thus giving a large residue of NH4I upon distillation of the aqueous HI.

Thus I did an experiment: I added NaOH solution to an aqueous iodine suspension until everything was dissolved and the solution basic, heated until colorless, and then added a drop of hydrazine. There was immediate vigorous gas evolution.
So I have reason to assume that hydrazine is able to reduce iodate under basic conditions. Do you think I should do the whole batch (over 100g of iodine) based on this preliminary test?
Someone got a better idea of how to turn wet iodine into clean NaI?

There is a very old method that involves reacting iodine with iron filings under water, forming a solution of iron(II,III)iodide (something with the empirical formula Fe3I8) which is then precipitated with K2CO3 solution and the basic iron carbonate filtered off, with the filtrate being a solution of KI. Sounds messy and loss-ridden, what with the flocculent ferric carbonate, this would probably be a pain to filter.

The hydrazine method would be clean and without byproducts, that is, if it works.
There are patents describing the addition of iodine to an aqueous NaOH solution with added hydrazine to reduce the hypoiodite in situ, giving a solution of pure iodide. I can't work this way since I don't know how much iodine I have in total and how much water the wet iodine contains, thus I need a self-indicating method.

[Edited on 31-3-2012 by garage chemist]

S.C. Wack - 31-3-2012 at 09:27

Biltz² seems to be OK with the iron filings method.

Shake 7 or 8 g iron filings and 50 ml water in an Erlenmeyer flask with 25 g iodine added in small portions. Warm the mixture somewhat until all of the iodine has combined, and the color of the solution has become deep yellow (ferrous iodide); then pour off the liquid from the excess of iron. Add five grams more of iodine to the solution and heat until it is dissolved. Pour this solution into a boiling solution of 17 g potassium carbonate in 50 ml water. The mixture, which at first is very thick, becomes more fluid upon further heating, since the precipitate assumes a more compact form. A little of the solution when filtered must be perfectly colorless, and free from iron; if this is not the case, add a little more potassium carbonate to the boiling solution. Evaporate the filtrate to a small volume in a porcelain dish, filter again, and evaporate further in a beaker until crystals begin to separate. Then allow the solution to evaporate slowly by placing the beaker in a warm place (as on top of the hot closet). Drain the crystals in a funnel, wash them with a little cold water, and save the mother liquor for another crop of crystals. Yield, 25 to 35 g.

The potassium iodide when dissolved in a little water and acidified should not show any yellow color (free iodine), which would indicate the presence of iodate in the salt.

Nicodem - 31-3-2012 at 09:33

Quote: Originally posted by garage chemist

I thought about doing it as follows: Slowly add NaOH solution until the iodine is all dissolved and the solution no longer brown, then heat to a boil to disproportionate the hypoiodite completely to iodate and iodide.

At this point you could evaporate, mix wet with excess active charcoal and heat to a couple of hundreds °C to affect the reduction to NaI. Then you can leach out the NaI with water (or acetone?), filter, evaporate (perhaps also dissolve in acetone to remove the insoluble crap and evaporate again). I would try the reduction at a small scale first and with proper excess of carbon (you never know how energetic it can become). However, temperatures above 300 °C would probably be needed (but I know you have a tube furnace). Sodium iodate thermolyzes even alone to sodium iodide, but this decomposition might not be as clean as in the presence of a reducing reagent. The charcoal and the formed CO2 should also offer protection from oxygen.

(It's not that I think your idea of using hydrazine would not work. It just seems an overkill to use such an reagent.)

garage chemist - 31-3-2012 at 10:12

Thank you for the procedure and advice.
Unfortunately, the iodine slurry is not only wet, it also is inhomogenous, with water separating out of it during standing, so any weighings are useless and I don't know how much I'm using.

Thermal decomposition of iodate also sounds good. I have a little tabletop furnace in which I could heat the iodide-iodate mixture in porcelain crucibles.
I don't want to reduce with charcoal because it contains a lot of ash that introduces a variety of metal ions.
Does anyone know whether iodate forms periodate upon heating, or if it decomposes cleanly into iodide and oxygen?

I also once found that pure KI when melted and heated strongly in a test tube gives off considerable amounts of purple I2 vapor. I suspect that this is due to oxidation by aerial oxygen, forming iodine and potassium oxide which further reacts with the glass forming silicate. That's what I fear could happen with the iodide in porcelain crucibles too, and no, the oven doesn't allow usage of protective gas.

The hydrazine method still sounds best to me. I have plenty of hydrazine, and look at the stochiometry: 1 mole N2H4 (32g) reduces 2 moles I2 (508g)! I will only need a small amount of hydrazine.

S.C. Wack - 31-3-2012 at 10:32

It's clear from the Fe procedure that weighing anything is not really necessary.

watson.fawkes - 31-3-2012 at 11:11

Quote: Originally posted by garage chemist

I don't want to reduce with charcoal because it contains a lot of ash that introduces a variety of metal ions.

At the cost of more setup, you can reduce with carbon monoxide in gaseous form. Perhaps the easiest way is to put a porous plug of charcoal at the entrance of a tube furnace and then force air through the tube at a not-too-high rate. Vent exhaust gas well or flare it off.

peach - 31-3-2012 at 11:33

At one point (when I was extracting seaweed), I had a flask of hexane that iodine had been carried over into.

I wanted a quantitative for the iodine in the seaweed and needed to get the iodine back. I did this by recovering it as sodium iodide. To produce the iodide, I dropped small chunks of sodium into the cyclohexane and let it relux. Over time, the purple colour disappears and is replaced by a small pile of white salt sat at the bottom.

The choice of solvent reduces the rate of the reaction between the two, and means it's going to fall out rather than dissolve.

You could add some hexane (or similar) to that slurry, put it all into solution, filter any bits out, separate off the aqueous and then do the above.

Sublimating some iodine:

Something weird is going on here, the file is upright on my computer, but horizontal here...

The Iodine Iris (reminds me of Ulysses 31):

Expect mess:

[Edited on 31-3-2012 by peach]

AJKOER - 31-3-2012 at 12:44

I believe the following direct approach is best (which you mentioned). React I2 in a hot aqueous pure NaOH solution:

I2 + H2O <--> HI + HIO

NaOH + HI --> NaI + H2O

NaOH + HIO --> NaIO + H2O

NaIO --> 2/3 NaI + 1/3 NaIO3

So on net:

I2 + 2 NaOH --> H2O + 5/3 NaI + 1/3 NaIO3

or:

3 I2 + 6 NaOH --> 3 H2O + 5 NaI + NaIO3

(Source: http://en.wikipedia.org/wiki/Sodium_iodate)

The reaction requires 2 moles of NaOH for each mole of Iodine. Now, as there is a large solubility difference between NaI and NaIO3, upon cooling, most of the NaIO3 should separate out. I would then concentrate the solution (and/or add an external source of NaI) and cool the solution again to check for NaIO3. (End Of Synthesis).

----------------

To obtain Sodium iodine from the filtered NaIO3, dissolve in water and treat with Hydrogen sulfide. Proposed reaction (based on the reported reaction of HIO3 and H2S):

NaIO3 + 3 H2S --> 3 S (s) + NaI + 3 H2O

although I suspect filtering out the Sulfur is easier said than done. Note, with an unintentional excess of H2S:

2 NaI + H2S --> Na2S + 2 HI

so one would have to add NaOH to recover the NaI. Or add H2O2 to collect Iodine.

[Edited on 31-3-2012 by AJKOER]

Endimion17 - 31-3-2012 at 13:11

I think you can't get sodium iodide out. Have you checked the solubility of related salts? It's extremely soluble, and everything else will precipitate before it does. NaI will at the end, being dried together with the impurities. Maybe some solvent extraction, but I can't think of any appropriate, alkali metal iodide specific one.

I remember discussing this with a colleague, and reaching a conclusion that only a direct synthesis has sense, such as the refluxing peach mentioned.

(Or was it potassium iodide? Anyway, I distinctly remember this problem with alkali metal iodides and the solubility issue.)

AndersHoveland - 31-3-2012 at 21:55

Perhaps heating (perhaps 80 °C) with urea and NaOH could reduce iodine to iodide. Urea would be an ideal reducing agent because it is cheap, readily obtainable as fertilizer, it would not boil out of hot solution like NH4OH, and its oxidation products (mostly CO2 and N2) can easily be removed from solution.

Chlorine and urea often just form chlorourea, but with the NaOH keeping the pH higher, and the heat to decompose intermediate chloramines, the urea should all be oxidized.

The reaction of chlorine with ammonium hydroxide is complex, and can result in different reaction products depending on the reactant ratios and pH. But one of the reactions is:
8NH3 + 3Cl2 --> N2 + 6NH4Cl

Obviously this has implications for reducing iodine.

[Edited on 1-4-2012 by AndersHoveland]

turd - 1-4-2012 at 01:53

Quote: Originally posted by peach

Something weird is going on here, the file is upright on my computer, but horizontal here...

Probably your camera saved the way you held it while taking the picture in the EXIF metadata of the JPEG. Your desktop software pays attention to this EXIF-tag, whereas the forum software does not.
The wonders of modern technology!

garage chemist - 1-4-2012 at 11:32

I subjected the whole batch of iodine to the NaOH/hydrazine treatment today. There were some interesting and strange things going on.

First I put the iodine slurry into a 500ml flask and prepared a solution of 40g NaOH in 160g water, carefully weighed out to be 200g in total.
I then added the NaOH solution until the brown color had disappeared, and heated to nearly boiling. Testing the pH I found it to be at 14, so i added some extra iodine from an old selfmade sample I had standing around. When enough had been added to turn the solution weakly brown/purple again, I found that the pH was at 12 and it did not change at all when I added more iodine. So I added some more NaOH to make the brown/purple color disappear (the solution was yellow then), and the pH was at 14 again.
At this point the unused NaOH solution was weighed back, and from the reacted amount of NaOH the hydrazine requirement was calculated and weighed out as 100% N2H5OH (just 4,45g!) which I had available.
This was added dropwise to the boiling hot solution and there was strong fizzing from the nitrogen evolution. Every drop of hydrazine made the solution turn colorless for a few seconds and the yellow color always reappeared when the hydrazine addition was interrupted. After most of the hydrazine had been added, the solution stayed colorless and the fizzing slowly died down. A little excess was added and the pH was 14 at this point. Now I added some more iodine to lower the pH, which again produced fizzing and I managed to bring it to exactly 7.
The only thing that remains to be done is rotovapping the solution down to dryness. The excess hydrazine protects the iodide from aerial oxidation and keeps the solution colorless.

What I've learned from this is that the disproportionation of iodine only occurs at pH above 12, and that the reaction between NaOH and iodine is not as nicely self-indicating as I thought it would be. If enough NaOH is added to make the brown color disappear then a considerable excess of NaOH is present.
The final pH adjustment can be conveniently done with iodine after the hydrazine addition.

blogfast25 - 1-4-2012 at 12:02

Nice work. Still not sure why you're hung up on NaI instead of KI. The latter is easier to crystallise.

I bet your iodide bearing liquor came from Grignards, right? Nice bit of recycling!

[Edited on 1-4-2012 by blogfast25]

garage chemist - 1-4-2012 at 12:09

It wasn't from Grignards, it was from several rather unsuccessful attempts at ethyl iodide preparation from H3PO4, KI and EtOH. These only gave very low yields of product and most of the iodide stayed in the distillation residue.

There's one good reason to have NaI instead of KI: the former is very soluble in acetone and can be used to turn alkyl chlorides into iodides since the byproduct NaCl continuously precipitates from the acetone solution. See the preparation of diiodomethane from dichloromethane on Rhodium! This doesn't work with KI.

peach - 1-4-2012 at 14:15

How long did you leave that to reflux GC?

When I tried it, with red phosphorus, it yielded but quite a bit less than the theoretical. I had seen mention of leaving it to stir under reflux overnight, whereas I think I left it for three hours. I also applied excessive washing of the distillate I expect.

[Edited on 1-4-2012 by peach]

DJF90 - 1-4-2012 at 15:33

Sodium iodide forms an adduct with acetone which is used in the purification of the latter. Being that the adduct crystallises out from the acetone solution (under the correct conditions), I suspect any crude NaI can be purified by dissolving in boiling acetone (check Amarego under "acetone" for details, NaI 100g in 400g acetone), hot filtration to remove junk and then cooling to -8*C in a salt-ice bath produces crystals of the 1:1 adduct NaI-Me2CO. Filtering and then warming of this product allows distillation of pure acetone (leaving behind pure NaI).

If I were in your position GC, I'd save the hydrazine and go the Fe route mentioned above. Couple that with this purification method and I suspect any trace metals from the Fe powder will be minimal in your product.

simba - 1-4-2012 at 17:04

NaOH isn't the best way to do what you want, use thiosulfate instead. Thiosulfate will react instantly with iodine and doesn't generate any iodate.

Just mix the iodine crystals with powdered thiosulfate, wash it with sparingly with dry ethanol, and evaporate the ethanol, preferably with vacuum assist, since it reacts quite easily with air and convert back to iodine.

AJKOER - 1-4-2012 at 18:13

Thiosulfates react with halogens differently, per Wikipedia:

2 S2O3 2− (aq) + I2 (aq) → S4O6 2− (aq) + 2 I− (aq)

S2O3 2− (aq) + 4 Br2 (aq) + 5 H2O(l) → 2 SO4 2− (aq) + 8 Br− (aq) + 10 H+ (aq)

S2O3 2− (aq) + 4 Cl2 (aq) + 5 H2O (l) → 2 SO4 2− (aq) + 8 Cl− (aq) + 10 H+ (aq)

Now in photography it is know to fix black and white negatives after the developing, ammonium thiosulfate as a fixing salt is three to four times faster than Sodium thiosulfate (see http://en.wikipedia.org/wiki/Ammonium_thiosulfate). This suggests the possible alternative use of (NH4)2S2O3 instead of NaS2O3 (followed by the addition of NaOH and heat to remove the ammonia). However, there is still the issue of Tetrathionate, which is a product of the oxidation of thiosulfate, S2O32−, by iodine, I2:

2S2O32− + I2 → S4O62− + 2I−
(Source: http://en.wikipedia.org/wiki/S4o6 )

All of which suggests an alternative route to me of first, aqueous HI, followed by addition of NaOH (you may wish to checkout a prior Sciencemadness thread on routes to aqueous HI).

[Edited on 2-4-2012 by AJKOER]

Nathaniel - 5-4-2012 at 08:12

http://www.versuchschemie.de/topic,10904,-Iodwasserstoffs%E4...

You just need sulfur, iron and HCl....nothing complex. After running H2S through iodine suspension and obtaining HI, you can run it into NaOH solution (don't make it too basic, measure pH during bubbling in HI and add NaOH later if needed to keep the concentration of sodium and iodide ions the same)

FeS + 2HCl --> FeCl₂ + H₂S
H₂S + I₂ --> S + 2HI
HI + NaOH --> NaI + H₂O

The apparatus is a bit more complex (but again nothing special...If you don't have everything it's possible to improvise)

I think the reactions and the pictures are enough for anyone to figure out what the point is even if they don't know German

AJKOER - 5-4-2012 at 09:06

Actually, per my prior thread reference ( http://www.sciencemadness.org/talk/viewthread.php?action=pri...
) to quote myself, which suggests a slight modification from Nathaniel:

"Actually, I was thinking one could add a little Aluminum Sulfide, Al2S3, to Iodine suspended in water.

Al2S3 + 6 H2O --> 2 Al(OH)3 + 3 H2S

H2S + I2 + H2O --> 2 HI + S + H2O "

To quote from Wikipedia on Al2S3: "The material is sensitive to moisture, hydrolyzing to hydrated aluminium oxides/hydroxides.[1] This can begin when the sulfide is exposed to the atmosphere. The hydrolysis reaction generates gaseous hydrogen sulfide (H2S)."

This would be followed by the addition of NaOH to form NaI per Nathaniel.
--------------------------

However, they are even better methods latter revealed to prepare HI (and then NaI). To quote from Simba on Blogfast25's suggested "bisulphite" method:

"Indeed, it sounds too good to be true. I don't know if it is a good method for producing HI because I haven't tried it, but it does generate HI after all.

I2 + NaHSO3 + H2O → 2 HI + NaHSO4 "
---------------------

Also AndersHoveland's suggestion which turns out actually to be a commercial preparation:

"The industrial preparation of HI involves the reaction of I2 with hydrazine, which also yields nitrogen gas.[5]

2 I2 + N2H4 → 4 HI + N2

When performed in water, the HI must be distilled."
-----------------------

Please ignore my other suggested synthesis in the cited thread as the above paths are possible superior with the possible exception relating to ones personal availability of required reagents.

DJF90 - 5-4-2012 at 11:14

The problem with

Quote:

However, they are even better methods latter revealed to prepare HI (and then NaI). To quote from Simba on Blogfast25's suggested "bisulphite" method:

"Indeed, it sounds too good to be true. I don't know if it is a good method for producing HI because I haven't tried it, but it does generate HI after all.

I2 + NaHSO3 + H2O → 2 HI + NaHSO4 "

is that the pKa of HI is about 7 units lower than H2SO4 (-9 vs -2 approx.) and bisulfate and so the equilibrium will lie as NaI + H2SO4. This is problematic upon concentration (and heating) as the iodide reduces the sulfuric acid, forming Iodine, SO2 and H2S in varying quantities. Chilling the solution directly after formation may lead to crystallisation of sodium bisulfate or of sodium iodide, depending on the solubility of the two in aq. HI and aq. H2SO4 respectively.

AJKOER - 5-4-2012 at 12:47

DJF90:

A point of clarification, the ScienceMadness tread I alluded is not restricted to aqueous HI.

My understanding of the reaction:

I2 + NaHSO3 + H2O → 2 HI (g) + NaHSO4

My supposition is supported by borohydride on another forum (Opiophile.org) which came up in a google search on this particular reaction. The topic discussed is the synthesis of desomorphine (link per your discretion), to quote:

"As for HI, well I suggest that the very much OTC sodium metabisulfite is a good source. Correct me if I'm wrong on this, but:

Na2S2O5 (sodium metabisulfite) + H2O + I2 ---> 2HI + 2NaHSO3 (sodium bisulfite).
NaHSO3 (sodium bisulfite) + H2O + I2 ----> 2HI + NaHSO4 (sodium bisulfate).

The reactions are reversible so usually an acid is added to push the reaction forward BUT as long as the HI is driven off as gas, the reaction is driven forward. Seems ideal for producing the 50-100ml of 57% HI needed for the experimentor in this field...."

Subject: "Thought on desomorphine synthesis", Link:
http://forum.opiophile.org/archive/index.php/t-35096.html?s=...

barley81 - 17-1-2013 at 16:50

Brauer says that ammonium iodide can be made by reacting ammonia, H2O2, and iodine:
"Powdered iodine (100 g.) is reacted with 280 ml. of 10% ammonia water (i.e., double the stoichiometric quantity) and 600 ml. of 3% H2O2 (i. e., 33% excess). The I2 dissolves and O2 is evolved. In some cases, further H202 solution must be added until the reaction mixture becomes pure yellow. The solution is evaporated on a steam bath."
(p289)

You can then add sodium carbonate or potassium carbonate (or hydroxide) and boil to obtain NaI / KI.

weiming1998 - 17-1-2013 at 17:38

An alkaline solution of iodine will oxidise alcohols due to hypoiodite formation, in which the hypoiodite is reduced to iodide (http://www.chemguide.co.uk/organicprops/alcohols/iodoform.ht...). If the product of the reaction is an aldehyde/ketone containing a methyl group, iodoform will be produced as a precipitate. So adding sodium hydroxide to iodine in methanol (or any alcohol that will not produce aldehyde/ketones with a methyl group) should do the trick, forming iodide and the oxidized organic products. The alcohol/oxidation products should have a relatively high volatility, so that it can be evaporated from the solution without contaminating the iodide.

Would excess ethanol work? Because when the ethanol is not in excess, iodoform precipitates, but an excess might reduce all the hypoiodites to iodide before it could react with the formed acetaldehyde.

phlogiston - 18-1-2013 at 08:09

Why not directly react elemental sodium with your iodine?
To do this in a slow, controlled (boring!) fashion, you could slowly add a dilute solution of iodine in an organic solvent to elemental sodium.

No need for expensive hydrazine and complicated disproportionationg reacts, and probably near 100% yield.

elementcollector1 - 18-1-2013 at 08:24

Quote: Originally posted by phlogiston

Why not directly react elemental sodium with your iodine?
To do this in a slow, controlled (boring!) fashion, you could slowly add a dilute solution of iodine in an organic solvent to elemental sodium.

No need for expensive hydrazine and complicated disproportionationg reacts, and probably near 100% yield.

If one has elemental sodium...

weiming1998 - 18-1-2013 at 16:46

Quote: Originally posted by phlogiston

Sodium is probably harder to acquire in appreciable quantities than hydrazine (which you can make yourself). Even if you do have some sodium, why waste large amounts (grams) reducing iodine when you can reduce iodine with a more common reducing agent and even NaOH+heat?

AJKOER - 20-1-2013 at 19:58

OK, here is an easy inexpensive route that may work just fine.

1. Dissolve NaOH into a small amount of cold H2O2 (creating a mildly concentrated to dilute solution) in a deep vessel as in the next step oxygen gas will form.

2. Add Iodine. Keep the solution cold and stir. With time, all the Iodine should be converted into just NaI.

Logic:

2 NaOH + I2 --> NaI + NaIO + H2O

2 NaIO + 2 H2O2 --> 2 NaI + 2 H2O + O2 (g)

and by dissolving the NaOH in H2O2, as any NaIO is created, it should decomposed into iodide and oxygen. This is based on the well known fast reaction of NaClO and H2O2 to generate O2 (which I have performed on many occasions):

2 NaClO + 2 H2O2 --> 2 NaCl + 2 H2O + O2 (g)

This reaction could also speed up the rate at which Iodine normally dissolves in NaOH.

Avoid an excess of Iodine (will lower the pH and create a more concentrated Iodine solution), warm solutions, strong light, and Iron and other heavy metals (best glass vessel) as all of these factors will foster faster disproportionation into Iodate (which we want to minimize).

Now , why not a very dilute H2O2/NaOH solution? I am assuming one wants to avoid a similar reaction of dilute Chlorine water and H2O2 as reported in Watt's Dictionary of Chemistry, which has a tendency of forming HOCl (here HOI).

[Edited on 21-1-2013 by AJKOER]

weiming1998 - 21-1-2013 at 03:18

Quote: Originally posted by AJKOER

OK, here is an easy inexpensive route that may work just fine.

1. Dissolve NaOH into a small amount of cold H2O2 (creating a mildly concentrated to dilute solution) in a deep vessel as in the next step oxygen gas will form.

2. Add Iodine. Keep the solution cold and stir. With time, all the Iodine should be converted into just NaI.

Logic:

2 NaOH + I2 --> NaI + NaIO + H2O

2 NaIO + 2 H2O2 --> 2 NaI + 2 H2O + O2 (g)

and by dissolving the NaOH in H2O2, as any NaIO is created, it should decomposed into iodide and oxygen. This is based on the well known fast reaction of NaClO and H2O2 to generate O2 (which I have performed on many occasions):

2 NaClO + 2 H2O2 --> 2 NaCl + 2 H2O + O2 (g)

This reaction could also speed up the rate at which Iodine normally dissolves in NaOH.

Avoid an excess of Iodine (will lower the pH and create a more concentrated Iodine solution), warm solutions, strong light, and Iron and other heavy metals (best glass vessel) as all of these factors will foster faster disproportionation into Iodate (which we want to minimize).

Now , why not a very dilute H2O2/NaOH solution? I am assuming one wants to avoid a similar reaction of dilute Chlorine water and H2O2 as reported in Watt's Dictionary of Chemistry, which has a tendency of forming HOCl (here HOI).

[Edited on 21-1-2013 by AJKOER]

I don't think this will work. Even if it does, you will need a lot of H2O2. The reason is that iodide salts are used to catalyse the decomposition of H2O2 in reactions like the Elephant's Toothpaste. This is because even under neutral/basic conditions, H2O2 will still oxidise iodide. So what would happen is that the iodine would disproportionate to an equimolar mixture of IO- and I- ions. The H2O2 will then reduce some IO-, but also oxidise some I- to IO-. By the time all the H2O2 has been converted to H2O and O2, there is no (or very little) net decrease in the amount of IO- in solution.

phlogiston - 21-1-2013 at 12:44

Quote:

Because your sodium iodide will most likely be quite contaminated with NaOH. This may or may not be a problem for his application.
I guess it depends on what he has available. BTW sodium, too, can be made at home as several people on this forum have shown.

AJKOER - 21-1-2013 at 15:26

Quote: Originally posted by weiming1998

.....
I don't think this will work. Even if it does, you will need a lot of H2O2. The reason is that iodide salts are used to catalyse the decomposition of H2O2 in reactions like the Elephant's Toothpaste. This is because even under neutral/basic conditions, H2O2 will still oxidise iodide. So what would happen is that the iodine would disproportionate to an equimolar mixture of IO- and I- ions. The H2O2 will then reduce some IO-, but also oxidise some I- to IO-. By the time all the H2O2 has been converted to H2O and O2, there is no (or very little) net decrease in the amount of IO- in solution.

Weiming:

Thanks for the review, but per this source (https://uwaterloo.ca/chem13news/sites/ca.chem13news/files/up... ) to quote:

"The NaI solution is made slightly basic in order to prevent a side reaction involving the oxidation of I- to I2 by H2O2."

So, to the extent that I2 and NaOH produce NaI, it apparently survives in a higher pH environment with H2O2. I am assuming that most of the NaIO is reduced to NaI and that the loss to iodate is acceptable.

However, interestingly per this reference (http://www.ingentaconnect.com/content/nrc/cjc/2001/00000079/... ), the question of intermediaries in the I2 and H2O2 with varying pH appears to be a source of controversy. To quote:

" Although there have been a number of studies of the reduction of I2 [by H2O2], there exists a great degree of controversy regarding the intermediates involved, the effect of buffers, and the general rate law (1–9). Because the rates and the mechanism of this reaction are important in predicting the pH dependence of iodine behaviour in reactor containment building after a postulated reactor accident, we have undertaken a kinetic study of I2 reduction by H2O2 in aqueous solution over a pH range of 6–9. The experiments were performed using stopped-flow instrumentation and monitoring the decay of I–3 spectrophotometrically."

[Edited on 21-1-2013 by AJKOER]

ScienceSquirrel - 21-1-2013 at 17:23

Can I respectfully suggest that AJKOER carries out an experiment with a solution of potassium or sodium iodide, potassium or sodium hydroxide and hydrogen peroxide to demonstrate hydrogen peroxide acting as a reducing agent towards iodide?

barley81 - 21-1-2013 at 20:04

Iodine CAN be reduced by hydrogen peroxide in basic solution. I said earlier that this works with ammonia. I think ammonia is especially suited since the pH of 10% ammonia is not high enough to effect disproportionation of I2 to iodate/iodide, but the ammonia can also react with the HI and keep the pH from dropping low enough for the oxidation of I- back to I2.

AJKOER - 22-1-2013 at 06:52

Quote: Originally posted by AJKOER

OK, I actually can across a reference on my proposed reaction (see http://www.allreactions.com/index.php/group-1a/natrium/sodiu... ). Here are a few of the listed reactions of interest:

2 NaOH (hot) + I2 + H2O2 =2 NaI + O2↑ + 2H2O

2 NаОН (cold) + I2 + H2S(gas) = 2 NaI + S↓ + 2H2O.

2 NaOH (diluted) + FeI2 = 2NaI + Fe(OH)2 ↓ (in the atmosphere of N2)

NaIO3 + H2O + 2 Fe = NaI + 2 FeO(OH)↓ (boiling, in air).

2 NaI = 2Na + I2 (t>1400° С )

8 NaI (solid) + 9 H2SO4 (conc.) = 4 I2↓ + H2S↑ + 4Н2О + 8 NaHSO4 (30—50° С )

2 NaI (solid) + 4 HNO3 (conc.) = I2↓ + 2NO2↑ + 2H2O + 2NaNO3 (boiling, impurity of НIO3).

6 NaI + 2 H2O + O2 → 4NaOH + 2 Na[I(I)2] (normal temp.,on the light)

4 NaI + 4 НСl (diluted.) + O2 = 2I2↓ + 4 NaCl + 2 H2O (normal temp.,on the light).

2 NaI (cold) + E2 = 2NaE + I2↓ (E = Cl, Br)

Nal (hot) + 3 H2O + 3Cl2 = NaIO3 + 6HCl "
------------------------------------------------------------------------------

Now there could be a potential issue with my suggested synthesis as my initial opinion was the following reaction:

[2 NaOH + H2O2] (Dilute, Cold?) + I2 --> 2 NaI + O2↑ + 2H2O

to forestall, I thought, any disproportionation into Iodate. Note, this is clearly an issue as with hot solid NaI added to Chlorine water, or is it hot aqueous NaI combined with Chlorine water (issues of solution's pH, concentration, temperature and liberation/action of free Chlorine vs. HOCl...), as per the last cited reaction above, Iodate is formed. Also, apparently replacing H2O2 with H2S gas should be performed in the cold! Nevertheless, the author states one should add hot NaOH (solid or aqueous?) to H2O2 + I2.

In any event, any unwanted Iodate can apparently be removed by boiling with Iron in an open vessel and filtering out as noted above (the fourth listed reaction).

[EDIT] Per the NaOH reference page (see http://www.allreactions.com/index.php/group-1a/natrium/sodiu... ), I now suspect that it is in fact hot solid NaOH that is added to Iodine. The author apparently distinguishes between hot concentrated and just hot solid. Reactions of interest to quote:

2 NaOH(hot) + I2 + H2O2 =2 NaI + O2 ↑ + 2H2O

2 NaOH (conc., cold) + E2 = NaEO + NaE + H2O (E = Cl, Br, I)

6 NaOH (conc.,hot) + 3E2 = NaEO3 + 5NaE + 3H2O. (E = Cl, Br, I)

6 NaOH + 3 Br2 + 2(NH3-H2O) --> (time) 6 NaBr + N2 ↑ + 8 H2O (normal temp)

4 NaOH + 6NO = 4 NaNO2 + N2 + 2H2O (300—400°С )

2 NaOH (cold) + NO + NO2 = 2NaNO2 + H2O

4 NaOH (hot) + 4 NO2 + O2 = 4NaNO3 + 2H2O

2 NaOH (conc., hot) + 3 H2O + Al2O3 = 2 Na[Al(OH)4]

NaOH (conc.) + Al(OH)3 = Na[Al(OH)4].

2 NaOH (conc. 60%) + H2O + ZnO = Na2[Zn(OH)4] (90°С )

2 NaOH (conc.) + Zn(OH)2 = Na2[Zn(OH)4] (normal temp.)

2 NaOH (diluted, conc.) + Zn + 2 SO2 = Na2S2O4 + Zn(OH)2↓.

2 NaOH + 2 H2O + З Н2О2 (conc.) = Na2O2-2H2O2-4H2O↓ (0°С )

Na2O2-2H2O2-4H2O = Na2O2 + 2 H2O2 + 4 H2O (normal temp., over conc. H2SO4)
-------------------------------------------------------------------------------

My revised suggested synthesis would be to add dry I2 to dry hot NaOH in a closed chamber, shake as some of the Iodine will have sublimed and finish off by slowly adding H2O2. In my opinion, the reaction sequence is:

NaOH (dry, hot) + I2 <--> HOI + NaI

For another source see http://books.google.com/books?id=5Hf-uK58OZIC&pg=SA3-PA2... which parallels the reaction of HOCl and NaCl thst I cited (with references) in a recent thread:

NaCl + HOCl <--> NaOH + Cl2

Then, with more NaOH:

NaOH + HOI --> NaOI + H2O

The final addition of aqueous H2O2 decomposes the NaOI:

2 NaOI + 2 H2O2 --> 2 NaI + 2 H2O + O2 (g)

[Edited on 22-1-2013 by AJKOER]

[Edited on 22-1-2013 by AJKOER]