Magpie
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oxidation of OH-
I have just found something surprising that I never knew before, ie, iodine will apparently oxidise the hydroxyl ion in aqueous solution.
I had recently read somewhere that "...hydroxide was to be added until the brown color of iodine disappeared." Since this was a surprise to me I went
right out to the lab and tried it. Sure enough that's what happens. Surmising that this must be a redox reaction I looked up the standard reduction
potentials for the two half-reactions. This showed that there is about a 0.1v driving force for the following reaction:
2I2 + 4OH- ---> O2 + 2H2O + 4I-
Have I made the correct analysis? Does anyone else find this surprising?
The single most important condition for a successful synthesis is good mixing - Nicodem
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kristofvagyok
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Never ever heard from that, the normal strong base and iodine reaction what I have known is the following:
3 I2 + 6 MOH → MIO3 + 5 MI + 3 H2O
where M is sodium, potassium, ect.
This reaction works perfectly and currently I have no idea how could the iodine turn the OH group to give oxygen as you have mentioned.
Did the reaction fizzed when you have tried it? Looks interesting, could you post a video?(:
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AJKOER
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Quote: Originally posted by Magpie  |
I had recently read somewhere that "...hydroxide was to be added until the brown color of iodine disappeared." Since this was a surprise to me I went
right out to the lab and tried it. Sure enough that's what happens. Surmising that this must be a redox reaction I looked up the standard reduction
potentials for the two half-reactions. This showed that there is about a 0.1v driving force for the following reaction:
2I2 + 4OH- ---> O2 + 2H2O + 4I-
Have I made the correct analysis? Does anyone else find this surprising? |
My immediate opinion is not surprising, but not completely correct (unless the context is clear) as I would not normally expect the above to be the
primary reaction products.
First, for example, Iodine and water (slow):
I2 + H2O <--> HOI + HI
Now, the major reaction depending on conditions (temperature, light, pH, impurities..) is usually the disproportionation reaction:
3 HOI --> HIO3 + 2 HI
and a secondary decomposition reaction:
2 HOI --> 2 HI + O2
which can be accelerated in the presence of organic material (like blood), light and certain metals (Ni, Cu,..), hence my comment relating to precise
experimental conditions.
Normally, in the presence of NaOH for example, I would expect the products in declining concentrations to be NaI, NaIO3 and some O2.
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Magpie
lab constructor
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| Quote: Originally posted by kristofvagyok | Never ever heard from that, the normal strong base and iodine reaction what I have known is the following:
3 I2 + 6 MOH → MIO3 + 5 MI + 3 H2O
where M is sodium, potassium, ect.
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I see that this is a disproportionation reaction. This is more consistent with my observations as I saw no bubbles of O2.
Thanks to you and AJKOER.
My experiment was not very quantitative. I placed a mL of 10% KI in a test tube then added a few mg of I2. This was way too much I2 so diluted an
aliquot to get a mL of light brown solution. Then I added a few drops of very old 10% NaOH. This turned it clear very quickly. I repeated this
looking closely for any bubble generation. I saw none.
The single most important condition for a successful synthesis is good mixing - Nicodem
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Magpie
lab constructor
Posts: 5939
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I was curious to compare the driving force for this reaction to the 0.13v I had calculated for the other reaction that evolved O2.
Using quite a few half reaction potentials from the table of standard potentials I came up with 1.49v. This definitely supports the
disproportionation reaction as predominant.
Is this a correct analysis?
The single most important condition for a successful synthesis is good mixing - Nicodem
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