l0k1 - 20-11-2012 at 11:44
Lugol's Iodine is a solution containing 10% KI and 5% I2 by weight in distilled water. It happily goes in any place you might otherwise use tincture
but is obviously more concentrated.
An electrochemical reaction involving KI somehow has been rattling about in my brain and I'm wondering what would happen if you put current through
lugol's solution neutralised with something like citric acid to snarf up the generated KOH and keep the iodine from protonating again.
Doing a bit of stochiometry it looks like somewhere between 40 and 70ml of your average distilled vinegar per litre of lugol's provides counterions
for the potassium hydroxide that forms in the solution preventing it from forming the iodide. Assuming that the solution continues to conduct at
around 7% concentration of salts, I don't see why you won't get (albeit perhaps slower) oxidation of iodide to iodine. When done just with an oxidiser
in acidic conditions.
Stronger acid is required when chemically oxidising to I2 because the electron comes from the iodate intermediate, but in the electrochemical reaction
the oxidation occurs all at once at the surface of the electrode, taking the electrons away without the intermediate formation of the iodate.
Since it's such a simple thing to do, a 9v battery, graphite pencil electrodes and the appropriate mix of lugol's and distilled vinegar should prove
the concept by turning into a translucent red solution with precipitate and lots of H2 bubbling off (shotglass size).
elementcollector1 - 20-11-2012 at 11:54
Why bother with electrolysis? Chemical methods are almost always faster and easier.
l0k1 - 20-11-2012 at 12:06
to eliminate the need to have a strong solution of HCl. time is not an issue.
A micro-scale test with a 9v battery with 50 drops of 15% lugol's and at first 6 drops distilled vinegar, later increased to about 20 , which did not
visibly reduce the generation of O2 + I2 on one side and the H2 on the other. Tiny crystals of I2 form on the surface mainly due to the adhering
bubbles of oxygen and hydrogen but once the reaction is complete the solution should be nearly pale yellow/red. Not having done the workup it might
not be as facile as I think it will be but I figure I can let it go until well beyond much of the water being decomposed and solid potassium acetate
starts to precipitate, but that will wash out with a little splash of water.
[Edited on 20-11-2012 by l0k1]
a layer of fine crystals builds up on the bottom of the solution and the solution colour is slowly turning from red to brown. the reaction speed is
not far off comparable to peroxide and acid.
[Edited on 20-11-2012 by l0k1]
the solution turned entirely to yellow/brown and lots and lots of fine crystals floating around slowly settling. one of the electrodes was all matte
black and pitted, so i think the gas coming from it must have been carbon dioxide, although i wouldn't be certain some amount of fine carbon powder
has flaked off in the process.
[Edited on 20-11-2012 by l0k1]
a larger test was performed with a 15V 5A AC/DC converter and the oxidation of the carbon was far more vigorous, the reaction is probably best done in
an ice bath because it generates quite a bit of heat (not nearly enough to make the iodine irritating past a metre away). This current was run for
about an hour on about 50ml of lugol's and is being let to settle. It will be allowed to settle and further electrolysis after the solid iodine fully
settles and is decanted off and washed and filtered. i don't think the oxidation was complete in an hour and a lot of the graphite was eroded (pretty
sure into CO2 because of bubbling on the same electrode the iodine forms around). Aside from the matter of the eroding electrode, the reaction system
seems quite sound and knowing there isn't an oxidising acid being generated or diffused is very reassuring.
so it is slower, depending on how much power you can put into it, but no requirement to have a strong HCl solution.
[Edited on 21-11-2012 by l0k1]
PHILOU Zrealone - 21-11-2012 at 02:39
You may wel get a mixed Kolbe electrolysis products...
2 CH3-CO2(-) --> CH3-CH3(g) + 2 CO2(g) + 2 e(-)
2 I(-) --> I2(s/g) + 2 e(-)
CH3-CO2(-) + I(-) --> CH3-I(g) + 2 e(-)
Citric acid will be normaly completely destroyed to CO2 and H2O. Alfa hydroxyacids does not survive wel electrolysis.
[Edited on 21-11-2012 by PHILOU Zrealone]
l0k1 - 21-11-2012 at 02:51
i am pretty sure ethane is unlikely to be produced. how would one detect if MeI is being made? considering the electrode that is eroding the gas from
the electrode the I2 forms, it seems more likely the carbon is oxidised than anything else, and on the other electrode is hydrogen... i don't know.
[Edited on 21-11-2012 by l0k1]
well. pardon me: https://en.wikipedia.org/wiki/Ethane yes potassium acetate produces ethane. and this forms through the intermediate of methyl CH3 radicals which
obviously would form methyliodide if conditions were acidic and HI acid levels were high enough. except they aren't. I don't think CH3- radicals bind
to unconcentrated HI ions nor do they bind to I2.
however it is quite a problem that the acetic acid turns into ethane, this means that gradually the pH will rise and there will be hydrolysis of I2 to
KI.
[Edited on 21-11-2012 by l0k1]
well, after some attempts at workup it appears the I2 is thoroughly contaminated with graphite flakes.
something i had considered was a split cell with NaOH in the one that would be reducing to NaOH anyway. the combined cell thus described above does
not efficiently oxidise iodides to iodine. it does to a small degree but really not worth it.
[Edited on 21-11-2012 by l0k1]
*puts aside the electrodes for after a more intensive study, and practical as well, has been done*
i think the only practical electrochem reaction for oxidising iodides would have to be the NaCl cell with the Cl2 bubbled through KI solution. I still
don't know what the best electrode material is but presumably some noble metal plating is probably involved. a friend of mine has rhodium coins, i
recall him searching for shot and wire and it wasn't onerously expensive.
[Edited on 21-11-2012 by l0k1]
IrC - 21-11-2012 at 15:05
"well, after some attempts at workup it appears the I2 is thoroughly contaminated with graphite flakes"
Wouldn't you be able to use sublimation to purify the I2? Some type of distillation setup in a sealed environment where pure I2 collected in a cooler
location?