Sciencemadness Discussion Board - New Way to Prepare Hydroiodic acid (HI)?

New Way to Prepare Hydroiodic acid (HI)?

AJKOER - 20-12-2011 at 18:06

The new untried method for preparing HI parallels the reaction of Cl2 and H2O:

Cl2 + H2O <==> HCl + HClO

A similar reaction is reported to occur with Iodine:

I2 + H2O <==> HI + HIO

Now, upon adding excess H2O2 to Chlorine water:

HCl + HClO + H2O2 --> 2 HCl + H2O + O2 (g)

So, assuming a similar reaction (?) with iodine:

I2 + H2O2 + H2O ---> 2 HI + H2O + O2 (g)

However, it is known that H2O2 will oxidize an excess of HI:

2HI + H2O2 ---> I2 + 2 H2O

See: http://www.kbcc.cuny.edu/academicDepartments/PHYSCI/PL/chm12...

This interestingly theoretical parallels one of proposed reaction with HCl/H2O2 etching solution (namely Cl2 + H2O).

Searching also found the following reaction:

"The second step is the oxidation of of iodine by hydrogen peroxide:

5 H2O2 (aq) + I2 (aq) --> 2 HIO3 (aq) + 4 H2O (l)"

Source: http://chemed.chem.purdue.edu/demos/main_pages/22.15.html

This suggests that excess H2O2 will dissolve and oxidize the I2 (or HI to I2) to HIO3. But, barring excess H2O2 or excess HI, will just mostly HI be created as required? A source that supports this view is: "Iodine(+1) reduction by hydrogen peroxide" by G. Schmitz. To quote: "The mechanism is complicated and the reducing action of hydrogen peroxide is mainly due to reaction (6)". The quote reaction (6) being:

IOH + H2O2 --> I(-) + H(+) + H2O + O2

LINK:
http://resources.metapress.com/pdf-preview.axd?code=9v432703...

Comments welcomed.

AndersHoveland - 20-12-2011 at 18:39

Quote: Originally posted by AJKOER

Searching also found the following reaction:

"The second step is the oxidation of of iodine by hydrogen peroxide:

5 H2O2 (aq) + I2 (aq) --> 2 HIO3 (aq) + 4 H2O (l)"

Source: http://chemed.chem.purdue.edu/demos/main_pages/22.15.html

This suggests that excess H2O2 will dissolve and oxidize the I2 (or HI to I2) to HIO3. But, barring excess H2O2 or excess HI, will just mostly HI be created as required? A source that supports this view is: "Iodine(+1) reduction by hydrogen peroxide" by G. Schmitz. To quote: "The mechanism is complicated and the reducing action of hydrogen peroxide is mainly due to reaction (6)". The quote reaction (6) being:

IOH + H2O2 --> I(-) + H(+) + H2O + O2

I found this, which is very revealing about the otherwise obscure reaction dynamics:

Quote:

As another example of catalytic decomposition of hydorgen peroxide by compensating reactions, Bray has studied the system iodine iodate-hydrogen peroxide. In solutions of moderate acidity (about 0.1 N) iodate ion decomposes hydrogen peroxide and is itself unaffected at the end. Bray and Caulkins observed that during the reaction the solution if faintly colored with iodine. subsequent work by Liebhafsky showed that provided the iodine was removed as it forms, the iodate can be almost quantitatively reduced to iodine according to the equation:
2IO3[-] + 5H2O2 + 2H[+] --> I2 + 6H2O + 5O2 (I)

On the other hand, in acid solution iodine is slowly oxidized to iodate by hydrogen peroxide. The oxidation can be speeded up considerably by high acidities and also by addition of iodate. In these circumstances the reaction can be made to go quickly and almost quantitatively according to the equation
I2 + 5H2O2 --> 2IO3[-] + 2H[+] + 4H2O (II)

Reactions I and II occuring together will obviously lead to catalytic decomposition of hydrogen peroxide...

The coloration due to iodine was also periodic, being formed during the slow evolution of oxygen and disappearing during the following rapid evolution.

"Advances in catalysis and related subjects, Volume 5",
Walter G. Frankenburg, p43-44

Iron ions also seem to help catalyze the oxidation of iodide by H2O2, perhaps through the intermediate formation of ferrates.

My experience with HBr reacting with H2O2 is that only bromine seems to form.

The chemistry of chlorine/chloride reacting with H2O2 has some significant differences from bromine/bromide reacting with H2O2. While they will both undergo the same reaction mechanisms, chlorine is more electronegetive than bromine.
The main reactions therefore tend to be, respectively,

Cl2 + H2O2 --> H2O + (2)HCl

(2)HBr + H2O2 --> (2)H2O + Br2

Of course, bromine will still continue to catalyse the decomposition of H2O2, and hydrochloric acid will likewise continue to catalyse the decomposition of H2O2, although more slowly.

But it is very very doubtful that bromide could be oxidized to bromate, like iodine. Iodine is simply much more electropositive than bromine. It is the only halogen that can be oxidized by concentrated nitric acid.

Not having personally experimented with hydroiodic acid, I can only suspect that it would behave similar to HBr when reacted with H2O2. Indeed, sodium iodide is a potent, fast-acting catalyst for the decomposition of H2O2, the reaction sometimes being referred to as "elephant's toothpaste".

I do not know the specifics of how sodium iodate can actually be obtained from the reaction of NaI and H2O2.
I would imagine it is not as straightforward as it may appear.

I believe that specific reference you referred to may be technically incorrect, or at least misleading. Iodide ions certainly are initially oxidized by H2O2, first to hypoiodite IO[-], then possibly to iodite IO2[-]. But both of these would tend to immediately react with more H2O2, being reduced back to iodide I[-], with the liberation of O2.

Some sources about the iodine clock reaction, only mention iodide being oxidized to hypoiodite (before being immediately reduced back to iodide). http://antoine.frostburg.edu/chem/senese/101/kinetics/faq/me...
http://www.kbcc.cuny.edu/academicDepartments/PHYSCI/PL/chm12...

From what I have read, it seems that iodine usually only acts as a catalyst towards H2O2, causing its decomposition.

The reaction of ozone with iodide is completely different, however.

Quote:

Some periodate and peroxide have also been reported to be produced by a reaction between the iodide and ozone.

Hope this answers all your difficult to answer questions!

[Edited on 21-12-2011 by AndersHoveland]

woelen - 21-12-2011 at 03:54

I have done quite a lot of experimenting with iodide, hydrogen peroxide and acid. Iodide is oxidized to iodine very quickly and quantitatively to elemental iodine. No iodide remains in solution. Iodine in this case forms a precipitate, which quickly settles at the bottom and can be separated without too much hassle.

At very high concentrations, iodine reacts to iodate, but this reaction is slow and certainly not complete. From a preparative point of view, the reaction is useless. In practical situations, with solutions of a few percent concentration, iodine is the end-product and there is no formation of iodate. Making iodine from iodide with dilute acid and hydrogen peroxide, however, works very well. You get the iodine in a form, which can be isolated easily.

ScienceSquirrel - 21-12-2011 at 05:15

Addition of hydrogen peroxide to an acidified solution of potassium iodide results in a brown solution initially. As more is added the solution changes to very pale yellow and iodine precipitates.
Iodine does not react with water at room temperature, tincture of iodine which is an aqueous alcoholic solution of iodine and potassium iodide is stable for long periods.
Members of the same periodic group are similar in their chemistry but not identical.

Ephoton - 21-12-2011 at 05:16

yep I can second that

no HI made so easy to test as well why untested.

simba - 21-12-2011 at 06:28

HI is much more easily oxidized than is HCl. So, though this method may work for making HCl, it will not work for making HI.

blogfast25 - 21-12-2011 at 07:51

Quote: Originally posted by AJKOER

This interestingly theoretical parallels one of proposed reaction with HCl/H2O2 etching solution (namely Cl2 + H2O).

Comments welcomed.

Except of course that chlorine isn't iodine. Same group, quite different animals. 'Theoretical parallels', my *rse.

AJKOER - 21-12-2011 at 09:05

Quote: Originally posted by ScienceSquirrel

Addition of hydrogen peroxide to an acidified solution of potassium iodide results in a brown solution initially. As more is added the solution changes to very pale yellow and iodine precipitates.

Actually, the above observation would be consistent with the creation of HI, which is said to readily dissolve iodine turning brown (see reference below), followed by additional H2O2 destroying the HI and forming iodine water (yellow). Nevertheless, I do now have some doubts as to the process using H2O2 in light of the apparent great affinity of HI for O2 forming iodine.

But for those who wish other existing methods for making HI from I2 and water without the use of Red Phosphorous or smelly (and poisonous) H2S, my research has come up with two other paths:

1. Combine Tin and Iodine to form Tin Iodide. Upon adding a large quantity of water, the Tin iodide is said to completely dissolve leaving aqueous HI and forming a tin oxide precipitate (a white flocculi). Using a small quantity of water creates HI and a silky orange salt.

2. Just heat I2 in H2O. The solution's color reputedly disappears with the formation of HI and HIO3. Apparently, any formed HIO disproportionations upon heating:

I2 + H2O <===> HI + HIO
Note: pH is said to be important as to whether this equilibrium moves to the right or left.

3 HIO --> 2HI + HIO3

My speculation is, as iodine is not very soluble in water but is highly soluble in HI (turning brown), I would guess the reaction proceeds slowly at first. Also, following the path of hypochlorites disproportionation, boiling may not be the optimal temperature for yield.

Reference: The London encyclopaedia: or Universal dictionary of science, art ..., Volume 12, page 116

LINK:
http://books.google.com/books?id=ClkKAQAAMAAJ&pg=PA116&a...

[Edited on 21-12-2011 by AJKOER]

[Edited on 21-12-2011 by AJKOER]

ScienceSquirrel - 21-12-2011 at 09:38

The brown colour is due to the triiodide anion.
As you oxidise more of the iodide to iodine the solubility of the iodine falls as it is a lot more soluble in an iodide solution as this anion than in the absence of iodide.
There is not any HI formed. The solution contains hydrogen and iodide ions.
http://en.wikipedia.org/wiki/Triiodide

ScienceSquirrel - 21-12-2011 at 09:41

BTW chemistry has moved on a bit since 1829!

AJKOER - 21-12-2011 at 10:00

Quote: Originally posted by ScienceSquirrel

The brown colour is due to the triiodide anion.
As you oxidise more of the iodide to iodine the solubility of the iodine falls as it is a lot more soluble in an iodide solution as this anion than in the absence of iodide.
There is not any HI formed. The solution contains hydrogen and iodide ions.

You may be right, I was thinking of prior the reference equations (1) and (2):
http://resources.metapress.com/pdf-preview.axd?code=9v432703...

and assuming that I(-) is from H(+) and I(-) formed upon adding acid to KI. Then, per reference equation (2), the brown tri-iodide anion is formed:

I2 + I(-) <==> I(-3)

or, at least, perhaps, but possibly of no importance, as I am no longer an advocate for the I2/H2O2 process.

blogfast25 - 21-12-2011 at 10:03

AJ:

Again I fail to see the purpose of your quest for HI. The established method of distilling it from an iodide with concentrated phosphoric acid is hardly complicated and even allows to obtain concentrated HI if engineered accordingly. And if I2 is your starting point, synthing (Na,K)I is a matter of following a UToob video.

Using Sn as an intermediate is expensive business. The hydrolysing SnI4 forms SnO2, in many circumstances that forms as a substance that’s both insoluble in acids and alkalis (in which case fusion with pure alkali is required). Recovering tin metal from any of its compounds isn’t that easy either…

Expensive HI that could be obtained relatively cheaply from the equivalent of a simple HCl generator, is that what you’re after?

[Edited on 21-12-2011 by blogfast25]

AJKOER - 21-12-2011 at 10:33

Quote: Originally posted by ScienceSquirrel

BTW chemistry has moved on a bit since 1829!

Yes, I would agree with this comment. We went from honest reporting of observations to what exactly?

Old chemist knew they were blind, modern chemists claim vision as they have X-ray glasses.

[Edited on 21-12-2011 by AJKOER]

simba - 21-12-2011 at 11:31

Quote: Originally posted by blogfast25

Another easy way of making HI from I2 would be I2 + bisulfite + water.

[Edited on 21-12-2011 by shivas]

AJKOER - 21-12-2011 at 13:00

Quote: Originally posted by blogfast25

AJ:

Again I fail to see the purpose of your quest for HI. The established method of distilling it from an iodide with concentrated phosphoric acid is hardly complicated and even allows to obtain concentrated HI if engineered accordingly. And if I2 is your starting point, synthing (Na,K)I is a matter of following a UToob video.

Using Sn as an intermediate is expensive business. The hydrolysing SnI4 forms SnO2, in many circumstances that forms as a substance that’s both insoluble in acids and alkalis (in which case fusion with pure alkali is required). Recovering tin metal from any of its compounds isn’t that easy either…

Expensive HI that could be obtained relatively cheaply from the equivalent of a simple HCl generator, is that what you’re after?

OK, tin is expensive, but you may have some if you have any steel cans (actually tin lined), Bronze (Sn/Cu), Pewter (85% or more tin), or Tin Solder (Sn 60%, Pb 40%).

I noticed you didn't focus on the second method of preparation which requires Iodine, water and a heat source.

blogfast25 - 21-12-2011 at 13:11

Quote: Originally posted by AJKOER

I have tin but wouldn't want to waste it on this.

Second method? I doubt if you could get any reasonable quantities with it.

Same with the 'bisulphite method': if it was that easy we'd be using it NOW. Often these ideas work only on paper and for very weak solutions...

Lambda-Eyde - 21-12-2011 at 13:34

A simple method for preparing HI is mixing a solution of p-toluenesulfonic acid with an iodide salt. The former can be easily made from toluene and sulfuric acid, UTFSE.

preparation of SiCl4 and SiI4

AndersHoveland - 21-12-2011 at 13:46

Here is an idea:
Dissolving CaCl2 into molten zinc chloride (which melts at only 292 °C ), then bubbling in SiF4 (made by reacting silica with hydrofluoric acid). Anhydrous aluminum trichloride has an even lower melting point, 192.4 °C (this surprising number is not an error!)

(2)CaCl2 + SiF4 --> (2)CaF2 + SiCl4

The silicon tetrachloride could then be combined with sodium iodide and used just like anhydrous HI for most reaction purposes.

ZnCl2 can be made by passing dry chlorine gas into ethyl ether with a piece of metallic zinc. The ZnCl2 that forms actually dissolves in the ether. It might work for aluminum also. "Solutions of Aluminum Chloride in Ethers", Gordon G. Evans, Thomas R. P. Gibb Jr., J. Kevin Kennedy, Frank P. Del Greco
J. Am. Chem. Soc., 1954, 76 (19), pp 4861–4862

Aluminum triiodide melts at 189.4 °C, so if this was used as the solvent, silicon tetraiodide could be obtained.

(2)CaI2 + SiF4 --> (2)CaF2 + SiI4

Silicon tetraiodide can be reacted with a limited quantity of water to form anhydrous hydrogen iodide.

SiI4 + (2)H2O --> SiO2 + (4)HI

It is interesting to note that SiF4 has much less of a tendancy to hydrolyse in water than the other silicon-halogen compounds.

(3)SiF4 + (4)H2O <==> (2)H2SiF6 + Si(OH)4

Indeed, although it is a very well known reaction, it is otherwise quite exceptional that hydrofluoric acid can react with silica. Silicon tetrafluoride is a gas.

SiO2 + (4)HF --> SiF4 + (2)H2O

This is just a stab in the dark...

What about reacting elemental iodine with anhydrous hydrazine. Of course the main reaction would be the formation of NH4I, but small quantities of HI might also be obtained if more than the stoichometric quantity of N2 is liberated, which is often the case with hydrazine oxidations.

[Edited on 22-12-2011 by AndersHoveland]

AJKOER - 21-12-2011 at 15:40

Quote: Originally posted by shivas

Another easy way of making HI from I2 would be I2 + bisulfite + water.

[Edited on 21-12-2011 by shivas]

Actually, I was thinking one could add a little Aluminum Sulfide, Al2S3, to Iodine suspended in water.

Al2S3 + 6 H2O --> 2 Al(OH)3 + 3 H2S

H2S + I2 + H2O --> 2 HI + S + H2O

To quote from Wikipedia on Al2S3: "The material is sensitive to moisture, hydrolyzing to hydrated aluminium oxides/hydroxides.[1] This can begin when the sulfide is exposed to the atmosphere. The hydrolysis reaction generates gaseous hydrogen sulfide (H2S)."

Actually, it is a very efficient generator of H2S, so be careful you may unintentionally fumigate the entire neighborhood to your embarrassment. Of course, as the H2S is a highly dangerous and toxic gas, embarrassment may be the least of your worries.

[Edited on 21-12-2011 by AJKOER]

AndersHoveland - 21-12-2011 at 15:45

Remember, the chemistry of aqueous hydroiodic acid is completely different from the chemistry of anhydrous hydrogen iodide. It is nearly impossible to isolate anhydrous HI from a hydroiodic acid solution in water.

Elemental iodine cannot oxidize hydrogen sulfide unless water is present.
See the thread "Sulfur is a stronger oxidizer than Iodine!",
http://www.sciencemadness.org/talk/viewthread.php?tid=17604

[Edited on 22-12-2011 by AndersHoveland]

Panache - 21-12-2011 at 16:39

Quote: Originally posted by AndersHoveland

Here is an idea:
Dissolving CaCl2 into molten zinc chloride (which melts at only 292 °C ), then bubbling in SiF4 (made by reacting silica with hydrofluoric acid). Anhydrous aluminum trichloride has an even lower melting point, 192.4 °C (this surprising number is not an error!)

(2)CaCl2 + SiF4 --> (2)CaF2 + SiCl4

The silicon tetrachloride could then be combined with sodium iodide and used just like anhydrous HI for most reaction purposes.

ZnCl2 can be made by passing dry chlorine gas into ethyl ether with a piece of metallic zinc. The ZnCl2 that forms actually dissolves in the ether. It might work for aluminum also.

Aluminum triiodide melts at 189.4 °C, so if this was used as the solvent, silicon tetraiodide could be obtained.

(2)CaI2 + SiF4 --> (2)CaF2 + SiI4

Silicon tetraiodide can be reacted with a limited quantity of water to form anhydrous hydrogen iodide.

SiI4 + (2)H2O --> SiO2 + (4)HI

It is interesting to note that SiF4 has much less of a tendancy to hydrolyse in water than the other silicon-halogen compounds.

(3)SiF4 + (4)H2O <==> (2)H2SiF6 + Si(OH)4

Indeed, although it is a very well known reaction, it is otherwise quite exceptional that hydrofluoric acid can react with silica. Silicon tetrafluoride is a gas.

SiO2 + (4)HF --> SiF4 + (2)H2O

This is just a stab in the dark...

What about reacting elemental iodine with anhydrous hydrazine. Of course the main reaction would be the formation of NH4I, but small quantities of HI might also be obtained if more than the stoichometric quantity of N2 is liberated, which is often the case with hydrazine oxidations.

[Edited on 21-12-2011 by AndersHoveland]

OMG i love it! Write it up for prepublication for shit and giggles,

'Easy HI in your kitchen'
Some quotes from it i could imagine, '....Simply add just a cup of water to the molten SiI4 (not need to isolate from reaction mix)...'

I know you disclaimed that it was a stab in the dark, and it was an interesting read, but really you were swinging at a neutrino in a black hole with a machete made of teflon coated with synovial fluid.

AJKOER - 21-12-2011 at 18:31

AndersHoveland:

Good point.

Interestingly, one of the other preparations suggested at my previously cited reference (repeated below) gives me pause as whether the reaction product is possibly anhydrous. The reaction of interest is between Iodine and certain oxides (including MgO and CaO) and H2O. The reaction reputedly decomposes the water with the water's Hydrogen reacting with Iodine to form HI, and the oxygen moving to HIO3 (call this alternate method three). No equation given, so my best speculation is:

3 I2 + H2O + 2 CaO ---> HI + HIO3 + 2 CaI2

Per Wikipedia on CaI2: "Calcium iodide can be formed by treating calcium carbonate, calcium oxide, or calcium hydroxide with hydroiodic acid:[3]

CaCO3 + 2 HI → CaI2 + H2O + CO2"

Thus, any HI created could indeed react with CaO to form CaI2.

Also same source: "Calcium iodide slowly reacts with oxygen and carbon dioxide in the air, liberating iodine, which is responsible for the faint yellow color of impure samples.[4]

2 CaI2 + 2 CO2 + O2 → 2 CaCO3 + 2 I2 "

So even if all the iodine was consumed, the possible decomposition of CaI2 in air could impart a brown coloration to the HI as it would absorb the iodine (which could possibly be anhydrous given the decomposition of the water).

Source: "The London encyclopaedia: or Universal dictionary of science, art ...", Volume 12, Page 116.

LINK:
http://books.google.com/books?id=ClkKAQAAMAAJ&pg=PA116&a...

[Edited on 22-12-2011 by AJKOER]

simba - 21-12-2011 at 18:54

Quote: Originally posted by blogfast25

Same with the 'bisulphite method': if it was that easy we'd be using it NOW. Often these ideas work only on paper and for very weak solutions...

Indeed, it sounds too good to be true. I don't know if it is a good method for producing HI because I haven't tried it, but it does generate HI afterall.

I2 + NaHSO3 + H2O → 2 HI + NaHSO4

AJKOER - 21-12-2011 at 19:01

Quote: Originally posted by AndersHoveland

This is just a stab in the dark...

What about reacting elemental iodine with anhydrous hydrazine. Of course the main reaction would be the formation of NH4I, but small quantities of HI might also be obtained if more than the stoichometric quantity of N2 is liberated, which is often the case with hydrazine oxidations.

[Edited on 21-12-2011 by AndersHoveland]

AndersHoveland:

Some may not appreciate your insights but I do!

Your hydrazine idea is actually mentioned in Wikipedia as a route to the commercial preparation of Hydrogen Iodide:

"Preparation

The industrial preparation of HI involves the reaction of I2 with hydrazine, which also yields nitrogen gas.[5]

2 I2 + N2H4 → 4 HI + N2

When performed in water, the HI must be distilled."

Go AndersHoveland!

AndersHoveland - 21-12-2011 at 19:11

I strongly think the dominant equation in such a reaction is less ideal unfortunately:

(2)N2H4 + (2)I2 --> (2)NH4I + N2

So the yield of free HI relative to the consumption of hydrazine is probably not very good.

Also, sodium iodide does not react to any extent with oxygen, unless an acid is present. Apparently even the CO2 in the air is sufficient enough to act as the acid.

Dilute hydroiodic acid solutions are actually quite easy to prepare, since elemental iodine can be reduced by a wide range of regents in water.

Quote: Originally posted by shivas

I2 + NaHSO3 + H2O → 2 HI + NaHSO4

Appears correct to me, and it is substantiated by
Journal of the American Chemical Society, Volume 109, Issues 14-19, p4871

The oxidation of thiosulfate by iodine in aqueous solution produces tetrathionate.

(2)Na2S2O3 + I2 --> Na2S4O6 + (2)NaI

Quote: Originally posted by AJKOER

No equation given, so my best speculation is:

3 I2 + H2O + 2 CaO ---> HI + HIO3 + 2 CaI2

Actually, the equation is:

(6)CaO + (6)I2 --> Ca(IO3)2 + (5)CaI2
(in water obviously, first hypoiodite is formed but this is unstable and rapidly disproportionates)

Trying to guess what the reaction products are from an unknown reaction, then balancing the equation, can be very challenging, and ones best intuitions are often unreliable. Especially when the reactions involve obscure interelemental combination compounds between the non-metals.

[Edited on 22-12-2011 by AndersHoveland]

AJKOER - 22-12-2011 at 08:07

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)

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.

AndersHoveland - 22-12-2011 at 10:52

Quote:

Your oxygen doesn't balance.

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]

AJKOER - 23-12-2011 at 11:32

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]

AJKOER - 26-12-2011 at 23:57

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]