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AJKOER
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Quote: Originally posted by AndersHoveland  |
Actually, the equation is:
(3)CaO + (3)I2 --> Ca(IO3)2 + (2)CaI2
(in water obviously, first hypoiodite is formed but this is unstable and rapidly disproportionates)
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AndersHoveland, I believe we both rush this reaction. Your oxygen doesn't balance. However, you inspired a re-look and assuming the disproportionation
and further, a hydrolysis of the Calcium salts, I believe I have an interesting result. To start I present some rescaled reaction to permit visual net
balancing:
6 I2 + 6 H2O <----> 10 HI + 2 HIO3
In presence of CaO, the reaction, however, moves to the right as the HI and HIO3 react with the CaO:
5 CaO (s) + 10 HI <---> 5 H2O + 5 CaI2
CaO (s) + 2 HIO3 <---> H2O + Ca(IO3)2
But, I have written the reactions as reversible assuming possible hydrolysis. The net reaction so far is:
6 I2 + 6 H2O + 6 CaO (s) ----> [5 H2O + 5 CaI2] + [H2O+ Ca(IO3)2]
Upon replacing the quantities in [] with their left side equation equivalents above:
6 I2 + 6 H2O + 6 CaO (s) ----> [5 CaO (s) + 10 HI] + [CaO (s) + 2 HIO3]
Noting that CaO is repeated on both sides and dividing by two:
3 I2 + 3 H2O --CaO Catalyst--> 5 HI + HIO3
In other words, the basic CaO pushes the Iodine water disproportionation to the right acting as a catalyst. The final product agrees with my cited
reference and the water is, in effect, removed.
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AndersHoveland
Hazard to Other Members, due to repeated speculation and posting of untested highly dangerous procedures!
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| Quote: |
Your oxygen doesn't balance.
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Quite right. The equation in the last post has now been corrected.
This really is not a complicated or obscure reaction. All halogens (with with exception of fluorine) react with hydroxyl ions to form
hypohalites XO[-], which tend to be unstable to varying degrees and disproportionate to halides X[-] and halates, containing the
anion of the general formula XO3[-], where X is a halogen.
(2)OH[-] + Cl2 --> Cl[-] + ClO[-] + H2O
(3)ClO[-] --> ClO3[-] + (2)Cl[-]
Hypochlorites disproportionate in boiling solution, or on exposure to light. Hypobromites decompose after several minutes at room temperature.
Hypoiodite is less stable, and quickly disproportionates. http://pubs.acs.org/doi/abs/10.1021/ac60213a058
Iodic acid, HIO3, oxidizes [at room temperature] hydroiodic acid, HI, to form iodine and water. source: http://pubs.acs.org/doi/abs/10.1021/j150061a001
I am not saying there is not some slight equilibrium in the other direction. The equilibrium will shift, giving different reaction products depending
on whether the pH is acidic or alkaline. By alternatively adding acid, then base, then acid again, you could continue to change iodine solution back
and forth between iodate/iodide.
[Edited on 22-12-2011 by AndersHoveland]
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AJKOER
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OK, on my referenced reaction of certain metal oxides (which per the cited source include those of Ca, Fe and Zn) with Iodine and water to reputedly
form aqueous HI (Hydriodic acid), AndersHoveland is implying that with the oxide of Calcium only CaI2 and Ca(IO3)2 are formed, whereas and I am
postulating some hydrolysis so as to account for the author's statement of HI formation. Our equations are now in agreement, but I go a step further.
To gain some insight, I investigated what, if any, reported reactions are incurred between Iodine water and Iron oxide, and also Zinc oxide. I found
another dated source (Hand book of chemistry, Volume 5, By Leopold Gmelin) with more details on the reaction between Iron Oxide and I2 + H2O:
On page 249: "2. Aqueous hydriodic acid prepared from 126 parts of iodine, when gently heated with excess of recently precipitated hydrate of ferric
oxide, dissolves a quantity corresponding to 62-56 parts of the anhydrous oxide. The solution smells of iodine and exhibits the reactions mentioned
under (1); hence it would appear that part of the ferric oxide and the hydriodic acid decompose each other, yielding protiodide of iron, free iodine,
and water. (Kerner.) "
With respect to Zinc, per page 28: "Aqueous Iodide of Zinc or Hydriodate of Zinc-Oxide.—Iodide of zinc deliquesces in the air. The same solution is
formed by keeping zinc and iodine immersed in water till the liquid loses its colour.—Colourless, somewhat acid liquid, containing 24 per cent. of
zinc oxide to 76 per cent. of hydriodic acid. (Gay-Lussac.) The highly concentrated solution [overcharged with oxide?] when diluted with water,
deposits hydrated zinc-oxide free from iodine. (Rammelsberg.)" My take on this is that air is slowly oxidizing the Zn to ZnO, which then creates the
hydriodic acid.
In conclusion, apparently the presence of incomplete, decomposition or hydrolysis reactions account for the apparent discrepancy.
Source Link:
http://books.google.com/books?id=OgI5AAAAMAAJ&pg=PA28&am...
As a sidebar on the reaction of Tin and Iodine in relation to aqueous HI, here is an extract from this new source, pages 82 to 83:
" Tin And Iodine.
A. Photiodide Of Tin, or Stannous Iodide.—Iodostannous Acid. 1. When tin-filings are heated with a twofold quantity of iodine, a brown-red,
translucent compound is formed, which yields a dingy orangeyellow powder, and fuses very easily (according to Sir H. Davy, it volatilizes when heated
more strongly); the combination is attended with noise and ignition. (Gay-Lussac, Rammelsberg, Pogg, 48, 169.) According to Henry (Phil. Trans. 1845,
363), tin heated with twice its weight of iodine, yields two compounds, the protiodide SnI, and the bin iodide SnI2, which may be separated by
sublimation, tho latter volatilizing at 180° C. while the former remains fixed at a red heat. 2. A dilute solution of protochloride of tin mixed
with a very slight excess of iodide of potassium soon deposits a large quantity of yellowis-red crystalline tufts. If the mixture was warm, the iodide
of tin is deposited, on cooling, in fine yellowish-red needles. (Boullay, Ann. Chim. Phys. 34, 372.) Gay-Lussac likewise obtained orange-yellow silky
crystals, by treating the iodide of tin (1) with a small quantity of water, and pouring off the liquid from the separated hydrate of stannous oxide—
that liquid containing stannous oxide with excess of hydriodic acid. According to Gay-Lussac, these crystals consist of acid hydriodate of stannous
oxide, but, according to Boullay's analysis, they are formed of protiodide of tin.
Protiodide of tin, when heated in the air in contact with stannic oxide, leaves a substance, which, when treated with water, is resolved "into stannic
oxide and hydriodic acid which dissolves. (Boullay.) Tho iodide prepared by (I) is easily decomposed by water—especially with the aid of heat, and
the more completely in proportion as tho quantity of water is greater—yielding hydrated stannous oxide which is precipitated, and aqueous hydriodic
acid containing a trace of tin. (Gay-Lussac, Rammelsberg.) Dissolves in water sparingly, but without decomposition; fuses readily and forms a
dark-red mass, having a crystalline texture; yields a powder of a red colour and considerable lustre, like red lead. (Henry.) 1 The iodide (2)
dissolves in water without decomposition, and more abundantly in warm than in cold water; it is likewise soluble in hydrochlorate of stannous oxide.
Hence it is not precipitated from a solution of that compound by small quantities of iodide of potassium. (Boullay.) Iodide of tin combines with the
more basic metallic iodides. (Boullay.)"
[Edited on 23-12-2011 by AJKOER]
[Edited on 23-12-2011 by AJKOER]
[Edited on 23-12-2011 by AJKOER]
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AJKOER
Radically Dubious
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Having dismissed the idea of creating HI (aqueous or otherwise) using H2O2 and Iodine, it seems only fair to come across an old reference in Mellor,
page 939 (link below) on a discussion of reactions with H2O2 and various compounds that would give the complete opposite possibility of even
non-aqueous HI formation. To quote:
(Context is upon adding H2O2)
"Iodine, in the presence of alkali carbonates, is transformed into hydrogen iodide."
No chemical equation given. My guess:
3 I2 + 3 H2O2 + Na2CO3 -----> 4 HI (g) + 2 NaIO3 + H2O + CO2 (g)
where the reaction sequence (speculation) is:
3 I2 + 3 H2O2 ----> 6 HOI --> 4 HI (g) + 2 HIO3
and: Na2CO3 + 2 HIO3 --> 2 NaIO3 + H2O + CO2 (g)
There may also be some additional reactions with the escaping HI gas like the formation of some Sodium Iodide and Iodine vapors from the reaction of
HI on HIO3.
Source: "A comprehensive treatise on inorganic and theoretical chemistry" by Joseph William Mellor", Page 939
LINK:
http://books.google.com/ebooks/reader?printsec=frontcover&am...
Note, this suggested preparation route is similar to my previous alluded method from a different source citing the presence of the oxides of Na, Ca,
Ba, Mg, K and Sr in the reaction of Iodine and H2O, where the water was said to be decomposed (with the possible production of hydrogen iodide?).
Here, we are adding CO2 to the oxides to form alkaline carbonates and replacing the water with H2O2.
[Edited on 27-12-2011 by AJKOER]
[Edited on 27-12-2011 by AJKOER]
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