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elementcollector1
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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...
Elements Collected:52/87
Latest Acquired: Cl
Next in Line: Nd
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weiming1998
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| 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. |
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?
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AJKOER
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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]
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weiming1998
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| 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.
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phlogiston
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| Quote: | | 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? |
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.
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"If a rocket goes up, who cares where it comes down, that's not my concern said Wernher von Braun" - Tom Lehrer |
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AJKOER
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| 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.
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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]
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ScienceSquirrel
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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?
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barley81
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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.
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AJKOER
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| 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).
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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)
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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]
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