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Author: Subject: Electrolysis of Sodium Iodide to get Iodine
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[*] posted on 10-8-2021 at 17:59
Electrolysis of Sodium Iodide to get Iodine


I know about the chemical method. I just want to entertain this one as well. The only proper discussion of this I can find says that it can be used to make sodium iodate.

I would start with adding a NaOH solution to turn the Iodine into NaI and removing the povidone. I will then take that solution (Hopefully only containing NaI and possibly excess NaOH) and electrolyze it, with I2 formation on the cathode(could be wrong I never remember this) and H2 on the anode. I would hope that this would form I2 without forming an excess of iodates, but I suspect it would. I would then be able to filter off the Iodine powder/crystals and have a pure product.

I will not be able to conduct this test for several days as I would prefer to do everything outside and it will most likely rain. However any suggestions would be helpful. I have a 12V power supply and a variable power supply. I have used the 12V one for electrolysis before, so I am considering starting with using that one, but I don't know how sensitive to voltage the reaction would be.




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[*] posted on 10-8-2021 at 18:36


I tried electrolyzing a KI solution, but my electrodes kept eroding away. I tried it with both nickel and graphite, and they each failed on the iodine side after ~12 hours. So you'll want something better for your electrodes at least.


[Edited on 8/11/2021 by Metallophile]
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[*] posted on 10-8-2021 at 19:04


Whereas iodine normally has limited solubility in water, in the presence of iodides it readily dissolves forming triiodide anion.



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[*] posted on 10-8-2021 at 19:07


I was thinking about using copper for the anode and either another copper one for the cathode or some graphite. With graphite not working I don't have my hopes up that much, but I will try anyway. I would want to try with a short run of maybe 3-4 hours to test the durability of the electrodes. If neither of those works I could just use copper and then put the Iodine/Cu(OH)x mixture in a different solvent to dissolve the I2, then evaporate the solvent.



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[*] posted on 10-8-2021 at 19:08


What would triiodides do if further electrolyzed? Could that produce I2 or likely not?



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[*] posted on 10-8-2021 at 19:43


Iodine in alkaline solution disproportionate in to iodate and iodide.



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[*] posted on 10-8-2021 at 20:10


Quote: Originally posted by Bedlasky  
Iodine in alkaline solution disproportionate in to iodate and iodide.


And making it acidic creates triiodides iirc. I'll see what happens if after I make the NaI, I make it slightly acidic(maybe using acetic acid or another acid). Then I could see what the triiodide does under electrolysis. Although everything I can think of doesn't really affect a triiodide at all. Possibly if it goes to 2/3 completion all of the iodine will be in triiodide and then it could fall out of solution? Then though, with bad electrodes there would be a ton more contamination. Although, doesn't adding acid make triiodide?

[Edited on 8-11-2021 by opfromthestart]




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[*] posted on 10-8-2021 at 20:59


Yes, in acidic solution reaction goes the opposite way.

[Edited on 11-8-2021 by Bedlasky]




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[*] posted on 11-8-2021 at 07:23


Instead of reacting the initial iodine with NaOH, I am considering using Na2S2O3 instead, since

2OH- + I2 -> I- + OI- + H2O

Where the hypoiodite turns into iodate pretty soon. This is what I would like to avoid. With thiosulfate, it reacts as

2 S2O3 2- + I2 -> S4O6 2- + 2 I- (According to slightly sketchy sources)

This is closer to the initial reaction that I would want as a start to the electrolysis. However, I could not find any resources to what happens when this tetrathionate ion is electrolyzed. It could possibly do something like

S4O6 2- + 2 OH- -> 2S + 2 SO3 2- + H2O (Completely made this up, this doesn't even need electrolysis so idk)

Which could cause sulfide gas to form, which I would want to avoid. I am unsure how to neutralize this ion except for oxidation, which would most likely also oxidize the I- back into I2, which I don't exactly want. Any other methods of removing the thiosulfate after it has reacted, or confirmation or correction on the behavior of thiosulfate and tetrathionate under electrolysis, would be helpful.




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[*] posted on 11-8-2021 at 08:26


Actually, would metabisulfite react in a similar way, or is thiosulfate necessary?



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[*] posted on 11-8-2021 at 11:22


So, I did an experiment with sodium metabisulfite stump remover I currently have.

I measured 10mL of water and put it in a beaker. I then added 1g of Na2S2O5. I swirled the beaker until it all dissolved. I took a pipette and drew up some iodine povidone solution. I dropped a few drops into the solution and they cleared after some time. I checked the pH and it seemed to be <6, but they aren't very good. I am getting better ones soon.

I'm not quite sure what reaction is going on as any reaction that would create NaI would seem to also create sulfuric acid, which I believe would be very unfavorable.

Edit:

I redid the experiment to get more quantitative results. I prepared the initial solution the same way. After adding 4mL of iodine solution, the solution turned a light yellow/green. After adding 10mL more, no further change of color occurred.

I also added some baking soda to the final result of both products. They both bubbled, so I assume there was some acidic bisulfite in the reaction. After neutralization, I added about 10mL of the iodine solution and the color still vanished every time, so the reaction does not seem to depend much on pH.

[Edited on 8-11-2021 by opfromthestart]

[Edited on 8-11-2021 by opfromthestart]




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[*] posted on 11-8-2021 at 12:06


I'm guessing the reaction is somewhat of the form

2HSO3- + I2 + 2H2O -> 2I- + 2HSO4- + 2H+

But I am not completely sure as I have not used sulfites pretty much at all, and the formation of either of those acids seems unlikely. Maybe using a stoichiometric amount of baking soda beforehand to form carbonic acid, and then CO2 gas, would help the reaction progress while acting as a form of buffer with its own solubility.

[Edited on 8-11-2021 by opfromthestart]




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[*] posted on 11-8-2021 at 12:39


Doesn't this work like with bromide? Where you electrolize just enough to get the right ratio iodide/iodate before you add a strong acid so the iodate oxidizes the iodide? Woelen has a nice description on his website.
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[*] posted on 11-8-2021 at 13:36


Almost, but I would like to have the formation of the iodine itself not be the result of an oxidizing chemical, but the electrolysis itself. I will still look at the website since it will probably follow many of the same steps up to that point. However, given my goal, I would like to prevent the formation of iodate as much as possible, or to reduce the iodate into iodide once it forms. But I think this would end up either reducing the iodate into iodine or reduce my actual iodine product back into iodide, both of which wouldn't be desirable. My goal is to have the oxidation from I- to I2 be purely from electrolysis.



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[*] posted on 11-8-2021 at 13:50


Quote: Originally posted by Tsjerk  
Woelen has a nice description on his website.


If you are referring to https://woelen.homescience.net/science/chem/exps/KBrO3_synth...
then this is not what I am going for.




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[*] posted on 11-8-2021 at 14:36


http://www.sciencemadness.org/talk/viewthread.php?tid=7180

But I don't understand one thing. You want to make iodine from iodine povidone. So why you need electrolysis? You dissolved it in NaOH to destroy povidone. After filtration you have clean solution of NaI/NaIO3. So only thing you need to do is add acid and voila, your iodine is reformed and precipitate from solution.




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[*] posted on 11-8-2021 at 14:46


I don't need electrolysis, but I just think it would be cool so its the method I want to try out. I am considering not using NaOH to dissolve it, and using metabisulfite to turn the iodine into NaI. Then I would want to get rid of the polymer, then do electrolysis to try to get iodine that way.

My idea was that, given that you can electrolyze metal salts to metals, why couldn't you do the same for elemental anions. So I want to see if a solution of only NaI would be able to produce I2, hopefully as crystals but probably either way.

Since the reaction that makes iodates seems to be accelerated at higher temperatures, would running it in an ice bath reduce the amount of iodate formed?

[Edited on 8-12-2021 by opfromthestart]




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[*] posted on 11-8-2021 at 18:04


NaOH doesn't break down pvp very well, it complexes with it!

KOH, on the other hand, will separate pvp from povidone iodine solutions.



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[*] posted on 11-8-2021 at 20:00


Quote: Originally posted by happyfooddance  
NaOH doesn't break down pvp very well, it complexes with it!



So then, could I add the metabisulfite until all of the iodine should be as NaI, then add NaOH to pull the PVP out of solution, filter, then neutralize the base and attempt electrolysis then? I believe then only NaI and Na2SO4 should be in solution at that point, and I believe that the sulfate will not react under electrolysis, so it should be purely with the I- ion, either that or OH-, but hopefully it will react more with the iodine.




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[*] posted on 11-8-2021 at 22:39


Not all pvp is created equal; however, in my experience when using NaOH the iodine has remained complexed with the pvp.

But when I've used KOH, a thick white goop falls out of soln. The clear liquid that remained was a solution of KI, and when acidified and oxidized with aqueous HCl and H2O2, elemental iodine falls out of solution and can be easily filtered or decanted.
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[*] posted on 12-8-2021 at 04:14


I do not have any KOH on hand, but if I attempt it with NaOH and it doesn't work correctly, I will use that.

I have previously made a small amount of iodine using the same process as https://www.youtube.com/watch?v=FNf8PSda7iI. I suspect the yellow color in the gooey precipitate is what you are referring to as the iodine still in complex. It still seems that enough of the iodine is no longer in complex that it can react under the oxidation reaction and have a decent amount of product. Using NaOH instead of KOH should affect yield but probably wouldn't affect the reaction after that point, just change the reaction to a different cation, affects solubility slightly but should work fine under electrolysis.




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[*] posted on 13-8-2021 at 11:20


Experiment 2:

Intent:
To confirm the molar ratio in the reaction between the metabisulfite and iodine. I suspect it should be one mole metabisulfite to two moles of I2. I also suspect that the reaction would cause the solution to become acidic. This is why sodium bicarbonate is added.

Experiment steps:
Add 125mL of iodine povidone solution to a beaker. Add 1.3g NaHCO3 to make the solution basic.
Fill a beaker with 10mL water and add .5g sodium metabisulfite. Stir until all is dissolved.
Slowly add metabisulfite solution to the iodine solution. Mark the point when most of the red/yellow color disappears.

Results:
Solution bubbled a small amount(or it might have just been from stirring, it was small) while solution was being added. After the initial 10mL solution was added, the solution was a deep red, but started to be translucent/transparent. Made another 10mL of metabisulfite solution and continued adding it.
After a total of 12mL of solution had been added, the solution turned a transparent yellow color, similar to urine(?). Another 1.3g NaHCO3 was added after 13mL was added, as I had forgot to earlier. It started to bubble very visibly and the color of the solution changed to a light yellow. The rest of the metabisulfite solution was added, with no significant color change. The pH was 7 after this process.

Conclusion:
Well, I realized halfway through that it would obviously become acidic due to the formation of HSO3- ions in solution. Maybe I should try neutralizing it beforehand to just see the reaction between SO3 2- and iodine. I believe the color is due to the povidone still being in solution, and possibly holding some of the iodine with it.

I think adding KOH to remove the povidone, then adding metabisulfite to both make the solution acidic and get rid of any iodates would probably work to get a relatively pure solution of mixed iodides, along with sulfates. Once this would be taken to completion, I would attempt electrolysis on the resulting solution. I would probably add excess metabisulfite to allow for the solution to be acidic, and just have to run the reaction longer to account for the solution destroying the initial iodine that would be formed.

When I redo this experiment with the already neutralized sulfite, I will start with the 20mL of solution to mitigate the problem I faced in this experiment.

Other:
From calculations, I believe that the electrolysis would occur at 1.45V. However, the synthesis of iodate would occur at .55V. I believe running this reaction at low temperatures, as in in an ice bath, would possibly mitigate this reaction, as the formation of chlorates seems to be accelerated at high temperatures. The formation of I2 from iodide also appears to create hydroxide in
$$2I^- + 2H_2O \to I_2 + H_2 + 2OH^-$$
so periodic addition of dilute sulfuric acid may be needed. I am not sure how this would affect copper electrodes but I assume it would not be good. I may try graphite/charcoal electrodes and see how that works, but I would want to fine tune this part of the reaction first.

Is there a good test for if iodates are in solution? I want to make sure they are actually being reduced.




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[*] posted on 13-8-2021 at 14:25


Are you still running at 12V? I don't think you will ever be able to do any electrolysis on a preparatory scale with that. Try to find a 5v adapter. 12V will destroy any electrode.

Making bromine by electrolysis of bromide to bromate (or chloride to chlorate) works without destroying the electrodes (barely) because the bromine reacts to bromate. Any halogen around will destroy your electrodes, having it react with the formed hydroxide saves them.
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[*] posted on 13-8-2021 at 15:25


I have a variable voltage power supply, so I will try running it at about 2v with that. Would halogens destroy the electrodes even if they aren't metal? I was planning on using a graphite rod for one/both of the electrodes.



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[*] posted on 18-8-2021 at 08:28


Experiment 3:

Intent:
Uhhhhhh just see if adding NaOH will make the PVP separate out properly, and seeing the effects of adding metabisulfite to the basic solution.

Steps:
Add 1.5g NaOH to 30mL water.
Add 125mL I-PVP to a beaker.
Add the NaOH solution slowly to the iodine until the solution is clear.
Filter off any PVP precipitate.
Make a solution of 10mL water, 1g Na2S2O5.
Add this to the filtered solution.

Results:
Well, I definitely should have actually calculated how much hydroxide I needed since I royally messed that up. I added the initial 1.5g, then another 3g, then another 1g. By this time I had enough solution to almost fill my 250mL beaker so I put it on a hotplate to evaporate some of the solution. It was yellowish when put on the hot plate. After some time, the color got lighter but it clouded up a bit. The pH was tested at this time and it was 12. As it evaporated, clearish globules started to form. pH was still around 12.
After evaporating to 75mL, two layers formed: a watery top layer and a thick bottom layer. I believe this bottom layer to be the PVP, or a product of it. Both layers are pretty transparent. I filtered off the top layer through 3 coffee filters. When stirred, the bottom layer become opaque.

I made a solution of 10mL water, 1.2g Na2S2O5. This was added to the filtered solution. No change in color was detected. Another solution of metabisulfite was prepared. This was added to the remaining PVP solution. No change was observed here either.

To be continued.

[Edited on 8-18-2021 by opfromthestart]




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