woelen
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iodate synthesis
As you may remember, some time ago I did quite some experimenting with making potassium bromate from potassium bromide. Now I did the same kind of
experiment with potassium iodide, making potassium iodate.
Making potassium iodate works very well, as well as making potassium bromate.
I took 10 grams of KI, and dissolved this in water, making a total volume of appr. 20 ml. I added 0.1 ml of a 20% solution of KOH.
I took a platinum anode and a titanium cathode. A current of 2 A was going through this liquid for 5 hours. After that period, the yield was just 1
gram (or even less, just a few tiny crystals at the bottom). Quite pathetic. The liquid become very hot, almost boiling.
I let cool down this liquid, and then dissolved 50 mg of K2Cr2O7 in 1 ml of water and added this to the solution. I then continued electrolysis, again
at 2 A, with the 1 gram of KIO3 still as solid at the bottom. After 2 hours, now I had a thick layer of very fine powder at the bottom. Yet 2 hours
later, the layer was even thicker. I was really pleased to see that good yield. The K2Cr2O7 is really helpful here.
I decanted the yellow liquid above the crystals, and I rinsed 5 times with 15 ml of cold distilled water. After the 5th rinse the rinsing liquid was
colorless. I let the crystals dry in air, above a warm radiator in a petri dish, loosely covered by a paper tissue (but not touched by) in order to
avoid dust contamination. The resulting crystals are very pale beige, almost white. But, when compared with a purely white solid, one can see that
they are off-white.
I added some of the crystals to 1 M H2SO4 and to my surprise, the liquid remains colorless, no formation of iodine. This is really surprising, this
means that by just rinsing 5 times, without recrystallizing, the solid KIO3 is totally free of iodide. Even tiny traces of iodide will lead to strong
brown coloration of a solution of an iodate, so this result is very neat.
I'm not 100% sure about the purity with respect to periodate. The acid-test only checks for iodide, not for periodate. Is there someone over here, who
knows a test, which can give any indication of how much periodate is in the iodate?
[Edited on 9-12-06 by woelen]
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Nerro
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I've looked up the solubility of KI and KIO3, they differ by a factor 4. This explains the purifying action of flushing it with water. How much of
your yield would you say you lost while rinsing?
On a sidenote, does anyone know why dichromate is such a good catalyst in these reactions? When making chlorate and bromate it is also useful. Does it
oxidize the I to +7? I it more succeptible to electrochemical oxidation once it has reacted with iodine than the iodine itself was?
[Edited on Sat/Dec/2006 by Nerro]
#261501 +(11351)- [X]
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woelen
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I used cold water from a fridge (somewhere between 0 and 10 C) and at that temperature the difference in solubility is MUCH more than a factor of 4.
It comes closer to a factor of 25. Have a look at the following table:
http://en.wikipedia.org/wiki/Solubility_table
I did not loose very much with the rinsing action. I noticed that KIO3 does dissolve in water, but very slowly. One needs to shake a lot and for a
long time to get it dissolved. KI on the other hand, is extremely soluble and also very quickly. So, I think you are right with your remark that this
explains the purifying action of the rinsing with water. But as an added remark, I am also surprised to see that inside the crystals hardly any iodide
is trapped.
The dichromate is not a catalyst to these reactions, it, however, prevents the back-reduction to iodide at the cathode. Without the dichromate, iodine
production is the same at the anode, but at the cathode only a little hydrogen is formed. Iodate and/or iodine are reduced back to iodide, hence the
lack of hydrogen production. With dichromate added (and just a very small amount is needed), there is a lot of hydrogen production at the cathode and
that is a good sign. That means little or no back-reduction of iodate and/or iodine. But why is this? I don't know, it just is a practical
observation.
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Nerro
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As to your remark on the possibility of KI getting trapped inside the KIO3 crystals, did the crystals dissolve in the H2SO4 1M? Did you shake and
leave it for some time? And if you did, did it discolour then?
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woelen
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My test for iodide-impurity was by adding a small spatula of my KIO3 to a few ml of 1 M H2SO4 and shaking, until all of it was dissolved. This results
in a clear colorless liquid. Addition of even a single tiny crystal of KI makes the liquid yellow/brown at once. From this I conclude that my KIO3
must be free of KI, otherwise I certainly would have noticed a yellow color in the solution. I did not even notice the faintest yellow, before adding
the KI.
[Edited on 10-12-06 by woelen]
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HeYBrO
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Sorry to bring up an old thread, but here is an interesting approach to potassium iodate, i can't seem to find any information on it by UTFSE.
https://www.youtube.com/watch?v=tpWb0j78uGU&list=UUa8oI2...
Does this reaction seem plausible?
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AJKOER
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Here is an interesting citation on page 21 at https://www.google.com/url?sa=t&source=web&rct=j&... :
KBr + KClO3 ---) KBrO3 + KCl
So, if the reaction is valid and one can replace the bromide with iodide, than one may be able to produce KIO3 from KI and KClO3. I have the latter,
but not the iodide at the moment, so I may be able to try this out sometime.
-----------------
For those interested in exploring the underlying chemistry and a possible relationship to my suggested path above, please note on page 21 also, the
following reaction (which incorporates products produced in the electrolysis of a solution of KI and KCl):
3 I2 + 5 HOCl + H2O ---) 2 HIO3 + 5 HCl
which also results in the formation of iodate, albeit slowly based on the efficiency of the electrolysis procedure to form Iodine and Hypochlorous
acid (from Cl2 and water). Further, I would suspect for the electrolysis of a solution of KI only:
3 I2 + 5 HOI + H2O = 2 HIO3 + 5 HI
where the latter acids would react with any created KOH from the electrolysis to form KIO3 and KI. Now, those wondering on the role of KMnO4, to quote
from Wikipedia (see http://en.m.wikipedia.org/wiki/Chlorine_production ):
"Potassium permanganate can be used to generate chlorine gas when added to hydrochloric acid".
So, in the presence of any formed HCl (or HI), the KMnO4 would serve to generate more Cl2 and also with water more HOCl (or I2, and more HOI) to
assist in iodate formation by driving the above reaction to the right.
[Edited on 1-10-2014 by AJKOER]
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gdflp
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In the comments, the uploader did link to a reference. http://download.chemiestudent.de/protokol/ac/grund/annika/si... Unfortunately it's in german, which I don't speak, but I can understand the
chemical equations. I have all of the chemicals, so I might try this on the weekend. If I do, I'll post my results.
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gdflp
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Okay, I tried this yesterday and it definitely worked. I got a precipitate of MnO2 and the filtrate was clearish, at least, after I gravity filtered
it and lost more than half the yield due to my buchner funnel cracking I was
using ethanol denatured with methanol and it seemed to work fine. I boiled down the filtrate from 150ml to 25ml and cooled it in an ice bath. One
point I need to emphasize is to cool the solution slowly. I just threw mine straight into the ice bath and lost probably half the yield due to
formation of crystals which could pass through my medium grit filter paper, even after refiltering twice. I can't say anything about the yield
though, since I lost more than half of my yield to the funnel Product identity
after washing twice with denatured ethanol and once with distilled water was confirmed by adding a small amount of the product to concentrated
hydrochloric, and concentrated sulfuric acid. On addition of HCl, the solid fizzed and dissolved, turning the solution yellow and giving it a smell
of chlorine. On addition of H2SO4, a clear solution resulted indicating no iodide was present, though addition of a small amount of KI resulted in a
deep brown color instantly.
[Edited on 10-4-2014 by gdflp]
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HeYBrO
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Wow, awesome! I'm gonna give this ago once i procure some KI ! Anyone tried this with KCl?
Quote: |
implementation
10 g of potassium permanganate is dissolved in 250 ml of demineralized water. The solution is a concentrated potassium iodide solution. The mixture is
heated in a 500 ml three-necked flask equipped with a mantle heater for 30 minutes. Subsequently, the excess of permanganate is eliminated by the
addition of alcohol (decoloration of the solution). The solution is filtered with suction flask and suction filter. The filtrate thus obtained is
acidified with acetic acid and evaporated to the burner on a ceramic hob to the crystallization point. So that the crystallization is complete, the
solution after the first crystals have failed, cooled with ice. KIO3 the resulting crystals are filtered off and washed with ethanol.
observation
-It Had to be filtered several times since always came with brown stone in the filtrate.
- After adding acetic acid to give only a weak acidic solution, which was not acidified.

chemical
solubility

mp
KI
1000 g H2O at 20 ° C: 143 g
686 ° C
KMnO4
1000 g H2O at 20 ° C: 64 g
50 ° C
ETOH
Acetic acid dil.


- The evaporation was slow and was under constant stirring.
- When the first crystals were precipitated, the beaker was allowed to stand still on the ceramic hob for a few minutes, until the solution was cooled
with ice.
- It must be re-filtered several times to collect all KIO3 crystals in the filter cake. It was then rinsed with EtOH.
-The Filter cake was scraped from the filter into a crystallizing dish. -The Crystals were dried in the crystallizing dish. calculations
Weighing KMnO4: 10.05 g
Weighing CI: 5.05 g
+ → conc. KI solution weighing H2O: 35.39 g
1000g H2O >> 143 g KI
x g H2O >> 5 g KI
x = 34.97 g
Weighing crystallizing: 89.30 g
Weighing Kristallisiersschale + KIO3 crystals: 91.67 g amount of KIO3: 2,37g
yield calculation
M (KI) = 165.91 g / mol M (KIO3) = 214 g / mol
Based on AI
Of 1 mol of KI produced 1 mol of KIO3
165.91 g → 1 mol 5.05 g → 0.03 mol
From 0.03 mol KI KIO3 arise 0.03 mol

theoretical yield
0.03 mol * 214 g / mole = 6.51 g
Useful yield
2.37 g
(2.37 g / 6.51 g) * 100 = 36.4% = yield evaluation / Questions
Were obtained 2,37g potassium iodate. This corresponds to a yield of 36.4%. Some loss is einzuberechnen during filtration and the scraping of the
filter. During the final filtration product could be reached in the mother liquor and not have been completely scraped from the filter. In addition,
secondary reactions occur, consume potassium, eg arises KOH.
The addition of ethanol to the reaction mixture results in a redox reaction with potassium permanganate and the formation of CO2 and H2, as well as
manganese dioxide, which causes a discoloration of the solution.
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From Aforementioned link.
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gdflp
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I don't believe that permanganate is a strong enough oxidizer to oxidize chloride, or even bromide, but the chemicals aren't that expensive so I'd
give it a shot.
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HeYBrO
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Hmm, It might just oxidise it to chlorine. I don't have any KCl so i can't try it.
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blargish
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Would anyone be able to confirm that KI can react with KClO3 to produce KIO3 in aqueous solution? My initial instinct was that
I2 would be formed and be precipitated out of solution; however, when looking at the redox equation in neutral/basic conditions, it seems
that enough OH- is created via
ClO3- + 6 I- + 3 H2O ==> Cl- + 3 I2 + 6 OH-
to ensure that all of the iodine reacts to form iodate and iodide via
3 I2 + 6 OH- ==> 5 I- + IO3- + 3H2O
This should allow the net reaction to occur, where
ClO3- + I- ==> Cl- + IO3-.
Does this actually occur?
BLaRgISH
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Metacelsus
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It seems unlikely. What would the mechanism be (would it go through chlorine oxides)?
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blargish
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Is it necessary for that to be the case? That is what I was unsure of.
I think that looking at the reduction potentials of the reaction would say whether or not potassium chlorate would be a strong enough oxidizer to
oxidize iodide straight to iodate and in turn be reduced to chloride. However, I am unable to find the reduction potential for iodate to iodide,
whilst the reduction potential for chlorate to chloride in basic conditions is 0.64V as per wikipedia.
BLaRgISH
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AJKOER
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Quote: Originally posted by blargish ![](images/xpblue/lastpost.gif) | Would anyone be able to confirm that KI can react with KClO3 to produce KIO3 in aqueous solution? My initial instinct was that
I2 would be formed and be precipitated out of solution; however, when looking at the redox equation in neutral/basic conditions, it seems
that enough OH- is created via
ClO3- + 6 I- + 3 H2O ==> Cl- + 3 I2 + 6 OH-
to ensure that all of the iodine reacts to form iodate and iodide via
3 I2 + 6 OH- ==> 5 I- + IO3- + 3H2O
This should allow the net reaction to occur, where
ClO3- + I- ==> Cl- + IO3-.
Does this actually occur? |
Your first reaction appears reasonable forming free iodine given the often cited reaction between an iodide and a hypochlorite:
NaClO + 2 KI + H2O ---> I2 + NaCl + 2 KOH
Then, there is this reference I just found citing a 1920 article published in the Journal of the American Chemical Society (the link address is http://pubs.acs.org/appl/literatum/publisher/achs/journals/c... ) which claims that the action of Iodine on a chlorate liberates Chlorine and
produces an iodate. In the current situation, this would be:
I2 + 2 KClO3 ---> 2 KIO3 + Cl2
Interestingly, the authors comment are actually quite favorable on this path to an iodate, or, on the particular point of the article, Iodic acid
(HIO3).
[Edit] Now, in an aqueous reaction in a closed vessel, the final products can still be KIO3 and KCl, but the reaction appears to be more complex than
I at first suspected. Also, one may be able to breakup the preparation into steps starting with any of a variety of methods to first produce Iodine
followed by the action of a chlorate to create the iodate.
I also found a discussion of the so called Chlorate variation of the experimental iodine clock sequence (see http://en.m.wikipedia.org/wiki/Iodine_clock_reaction ) that focuses on the action of a chlorate on an iodide in an acidic environment. Again,
iodate is one of the final products.
[Edited on 1-11-2014 by AJKOER]
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AJKOER
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For safety reasons, I will rewrite the author's equation, as his text is very clear on the employment of "an acidified solution" of the chlorate to
avoid the very unsafe generation of ClO2 by the application of say, concentrated H2SO4, to a dry chlorate salt:
I2 + 2 KClO3(aq) --Acid--> 2 KIO3 + Cl2
It is unsafe because the formed ClO2 gas is not diluted with an inert gas, as would be the case of say employing an excess of Oxalic acid which would
add CO2. In such a dry salt scenario using Sulfuric acid, any Chlorine dioxide formed could explode and splatter the acid.
[Edit] As clearly both the above and the Chlorate version of the Iodine clock reaction occur in an acidified reaction (but, apparently, at a near
neutral pH per the discussion below), I will cite and expand on the reaction. To quote from the Wikipedia referenced article:
"An experimental iodine clock sequence has also been established for a system consisting of iodine potassium-iodide, sodium chlorate and perchloric
acid that takes place through the following reactions.[3]......
Chlorate ion oxidizes iodide ion to hypoiodous acid and chlorous acid in slow and rate-determining step:
ClO3− + I− + 2 H+ → HOI + HClO2
Chlorate consumption is accelerated by reaction of hypoiodous acid to iodous acid and more chlorous acid:
ClO3− + HOI + H+ → HIO2 + HClO2
More autocatalysis when newly generated iodous acid also converts chlorate in the fastest reaction step:
ClO3− + HIO2 → IO3− + HClO2
In this clock the induction period is the time it takes for autocatalytic process to start after which the concentration of free iodine falls rapidly
as observed by UV/VIS spectroscopy."
At this point, my take on the implied net reaction would be:
3 ClO3− + I− + 3 H+ → IO3- + 3 HClO2
In my opinion, the reaction could continue with following corresponding reactions:
3 ClO2− + I− + 3 H+ → IO3- + 3 HClO
3 ClO− + I− + 3 H+ → IO3- + 3 HCl
For a final net reaction of:
3 ClO3− + 3 I− + 3 H+ → 3 IO3- + 3 HCl
Or, on rescaling:
ClO3− + I− + H+ → IO3- + HCl
demonstrating that the two cited preparation paths are basically related as in the presence of any formed iodate, an excess of iodide and acid, per
the same Wikipedia reference, we have the reaction:
IO3− + 5 I− + 6 H+ → 3 I2 + 3 H2O
[Edit] To quote from Mellor:
"Iodine separates from an acidified soln. of potassium iodide: 2Cl02+10HI=2HCl+4H20+5I2; in neutral soln.: 6Cl02+10KI=4KI03 +6KCl+3I2; and in the
bicarbonate soln.: 2Cl02+2KI=2KCl02+I2, whereby 80 per cent, of the chlorine dioxide is converted into the chlorite."
With respect to ClO2, a possibly interesting point would be to consider the cited reaction:
3 ClO3− + I− + 3 H+ → IO3- + 3 HClO2
at its half way point, the reaction mix would consist of:
3/2 HClO3 + 1/2 I− + 1/2 IO3- + 3/2 HClO2
As HClO3 + HClO2 = ClO2 + H2O, this can be thougth of as the action of action ClO2 on an aqueous iodide (which is Cheddite Cheese prior comment on the
possible reaction mechanism), but in not such excess so that the solution becomes acidic (per Mellor, who also references the employment of the KI
salt).
[Edited on 3-11-2014 by AJKOER]
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blargish
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Quote: Originally posted by AKJOER ![](images/xpblue/lastpost.gif) |
For a final net reaction of:
3 ClO3− + 3 I− + 3 H+ → 3 IO3- + 3 HCl
Or, on rescaling:
ClO3− + I− + H+ → IO3- + HCl
[Edited on 2-11-2014 by AJKOER] |
If you are correct, then once again the net reaction is merely
ClO3- + I- ==> IO3- + Cl-
However, as you show, the process seems to be a whole lot more complicated.
Now in regards to your previous mentioning of the reaction between chlorate and iodine to form iodate and liberate chlorine, I also found a video on
youtube where one synthesized potassium iodate from potassium chlorate and iodine. (https://m.youtube.com/watch?v=RJBmNYKyb7s). However, he describes the formation of an iodate/iodic acid double salt in acidic conditions
(KH(IO3)2), which then converts to iodate in basic conditions. He cites the following 2 reactions but does not give a source.
2ClO3- + H+ + I2 = H(IO3)2- + Cl2
3H2O + 5ClO3- + 3I2 = 5Cl- + 3H(IO3)2- + 3H+
If the formation of KH(IO3)2 is true, it may complicate the picture of the reaction
I'll be getting decent quantities of potassium chlorate and iodide soon and will be able to conduct a few experiments for myself.
BLaRgISH
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AJKOER
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Blargish:
See my Mellor comment I just inserted.
With respect to possible double salts, my experience is that they are only reliably and quickly form when working with high viscosity solutions (that
is, highly concentrated).
They often early in their formation (before hardening) can be decomposed by dilution and warming.
[Edited on 2-11-2014 by AJKOER]
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