Moiety
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Separation of AgI
I have some AgI, and wanted to make some I2 and AgNO3.
My first thought was using an acid:
ex.) HNO3 + AgI --> HI + AgNO3
That won't work because of the pKa of HI is pretty low...
Then I thought about a base:
ex.) NaOH + AgI --> NaI + AgOH
But AgI is insoluble in H2O, so the reaction favors the reactants, so that won't work.
Thermal decomposition won't work either because AgI has a BP of 1506ºC...
Anyone have any other ideas to separate these two elements??
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woelen
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Retrieving both the iodine and silver will not be an easy task and I doubt whether the cost of other chemicals is worth the result.
Retrieving only the silver can be done without too much equipment. I would go for oxidation by concentrated HNO3. Add the AgI to the HNO3 and heat. At
a certain moment copious amounts of red/purple fumes escape from the liquid. This is a mix of I2 and NO2. In solution, impure AgNO3 remains behind.
Maybe you can boil away all iodine and you are left with a fairly pure solution of AgNO3 in excess HNO3. Unfortunately, the solutions almost certainly
will contain some iodic acid as well.
Recovering the iodine will be harder. You might be able to let it condense on a cold glass surface, but keep in mind that it will be very impure due
to the presence of the NO2 and you'll also have iodate in the solution.
Purification of the silver nitrate can be done by precipitatig the silver as AgCl by adding dilute hydrochloric acid to the liquid and boiling in
order to make the precipitate more compact and easier to separate from the liquid. To this AgCl you can add a solution of glucose and NaOH, which
convert the AgCl to black metallic Ag. This metallic Ag must be rinsed very well and then redissolved in HNO3 to get AgNO3 of better purity.
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Moiety
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Thanks for the fast reply. That sounds like a good plan. I Thought that HNO3 and heat might work, but you have confirmed it. 
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AJKOER
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OK, here is another idea to work on, based on the apparent property that AgI is actually more light sensitive than AgCl (but less soluble), and
perhaps could be made to efficiently decompose into Ag and atomic I upon light exposure. To quote "The Columbia Encyclopedia", 6th ed., 2011:
"Silver chloride
silver chloride chemical compound, AgCl, a white cubic crystalline solid. It is nearly insoluble in water but is soluble in a water solution of
ammonia, potassium cyanide, or sodium thiosulfate ( "hypo" ). On exposure to light it becomes a deep grayish blue due to its decomposition into
metallic silver and atomic chlorine. This light-sensitive behavior is the basis of photographic processes (see photography , still ). Since silver
bromide, AgBr, and silver iodide, AgI, react similarly, all three of these silver halide salts are used in making photographic films and plates. Both
the bromide and iodide are less soluble in water and more sensitive to light than the chloride. The bromide forms light yellow cubic crystals; the
iodide forms yellow hexagonal or yellow-orange cubic crystals, depending on the temperature. Besides use in photography, silver chloride is used in
silver plating, and silver iodide is used for seeding clouds. The chloride, bromide, and iodide occur naturally as the minerals cerargyrite,
bromyrite, and iodyrite, respectively. Silver fluoride, AgF, forms colorless cubic crystals; it is much more soluble in water than the other silver
halides."
LINK: http://www.encyclopedia.com/topic/Iodides.aspx
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weiming1998
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Your first formula AgI+HNO3====>AgNO3+HI is not valid. The HNO3 will reduce the HI to I2 according to the following reaction:
2AgI+2HNO3===>I2+H2O+AgNO3+AgNO2. At least, the reduction of HI happens when you react sulfuric acid with an iodide salt.
The silver can then be recovered by the thermal decomposition of the two salts, which will yield metallic silver because Ag2O decomposes at a lower
temperature than AgNO3.
[Edited on 13-3-2012 by weiming1998]
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LanthanumK
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When I reacted ascorbic acid (any other reducing agent may do, but I almost always use ascorbic acid) with silver iodide, a gray precipitate was
obtained of what I suspected was silver metal. The remaining solution should contain all of the iodide.
hibernating...
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woelen
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Quote: Originally posted by weiming1998  | Your first formula AgI+HNO3====>AgNO3+HI is not valid. The HNO3 will reduce the HI to I2 according to the following reaction:
2AgI+2HNO3===>I2+H2O+AgNO3+AgNO2. At least, the reduction of HI happens when you react sulfuric acid with an iodide salt.
The silver can then be recovered by the thermal decomposition of the two salts, which will yield metallic silver because Ag2O decomposes at a lower
temperature than AgNO3.
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There are many errors in this post:
1) HNO3 does not reduce anything, it is an oxidizer.
2) No AgNO2 is formed at all. The way HNO3 works as oxidizer is very different.
3) Recovery of the silver by merely heating AgNO3 is not a good idea. You have to heat very strongly and you hardly can expect any pure silver to be
formed in this way.
When AgI reacts with HNO3, then expect formation of NO and NO2 and I2. The silver goes in solution as Ag(+) ions and on boiling down of the liquid you
get impure AgNO3. The main contaminant will be AgIO3 and remains of HNO3 and HIO3. Part of the iodide and iodine almost certainly will be oxidized to
iodic acid/iodate.
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weiming1998
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| Quote: Originally posted by woelen | | Quote: Originally posted by weiming1998 | Your first formula AgI+HNO3====>AgNO3+HI is not valid. The HNO3 will reduce the HI to I2 according to the following reaction:
2AgI+2HNO3===>I2+H2O+AgNO3+AgNO2. At least, the reduction of HI happens when you react sulfuric acid with an iodide salt.
The silver can then be recovered by the thermal decomposition of the two salts, which will yield metallic silver because Ag2O decomposes at a lower
temperature than AgNO3.
|
There are many errors in this post:
1) HNO3 does not reduce anything, it is an oxidizer.
2) No AgNO2 is formed at all. The way HNO3 works as oxidizer is very different.
3) Recovery of the silver by merely heating AgNO3 is not a good idea. You have to heat very strongly and you hardly can expect any pure silver to be
formed in this way.
When AgI reacts with HNO3, then expect formation of NO and NO2 and I2. The silver goes in solution as Ag(+) ions and on boiling down of the liquid you
get impure AgNO3. The main contaminant will be AgIO3 and remains of HNO3 and HIO3. Part of the iodide and iodine almost certainly will be oxidized to
iodic acid/iodate. |
Sorry, I did realise HNO3 was an oxidizer, probably a typo, or just a brain freeze. What I meant is that H2SO4/HNO3 oxidizes the I- ion to I2, not
reduce.
If the heating of AgNO3 requires very high temperatures, try precipating with sodium carbonate or sodium hydroxide. Ag2O decomposes at 280 celsius,
according to Wikipedia and AgCO3 decomposes at 210 celsius, which should be easily achievable by a stove or a bunsen burner. The advantage of this
reaction is that it will precipate the contaminant AgIO3, making NaIO3. React all the remaining solution with sodium carbonate/hydroxide, dry the
solution and react the iodate with a reducing agent (insoluble like carbon), then dissolve the remaining iodide, concentrate the solution and then
pass chlorine gas through it. That should probably give you the remaining iodine
Note: If iodates can be reacted with chlorine to form chlorates and iodine, simply skip the reducing agent step.
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woelen
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Only a small fraction of the iodine will end up as iodate, most of it will escape as iodine.
Chlorine does not make chlorate from iodate and no free iodine will be released.
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Moiety
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Ok, so it sounds like my best bets are on AJKOER's light sensitive decomposition and Lanthanum's reducing agent idea.
What does the reaction look like with ascorbic acid?
What about an electrochemical reduction, could I use a carbon anode and silver cathode to get I2 and Agº?
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unionised
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Hell's teeth!
Do you lot all have some sort of obsession with nitric acid?
Mix the AgI with excess NaOH (which is cheap) and heat it till it melts.
2 AgI +2 NaOH ---> Ag2O +H2O + 2NaI
2 Ag2O ---> 2Ag + O2
The first (disproportionation) reaction is driven to completion by the removal of water and oxygen
You need to be careful- molten caustic is nasty stuff, but so is hot HNO3
Let it cool, carefully add water, and filter off the silver. I guess you can dissolve that in HNO3 if you are suffering withdrawal symptoms
Some of the NaI will get oxidised to higher oxidation states like iodate.
That won't matter much if you plan to recover the iodine by acidification and oxidation. And yes, HNO3 would do- if you like NOx fumes. But H2O2 is
also good.
(Personally, I use O3)
[Edited on 20-3-12 by unionised]
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woelen
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Ascorbic acid is a perfect reductor to convert silver(I) to metallic silver. In order to make it work, you add a solution of ascorbic acid in dilute
NaOH to a precipitate of a silver salt. Metallic silver will be formed quickly.
@unionised: I like the method with HNO3 better than the method with molten NaOH. In theory this sounds as easy as the HNO3 method, but in practice it
is more involved. The required temperatures are much higher (~300 C instead of ~100 C), you cannot work in glass, but need a suitable (nickel?)
crucible and the risks are higher. Hot HNO3 is nasty, but molten NaOH is much worse. I also can imagine that because of the need of a metal crucible
there also is the issue of contamination with metal from the crucible. Molten NaOH, combined with hot AgI and O2 makes a perfectly strongly oxidizing
mix, which might eat some of the hot metal.
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unionised
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Last time I used an old tin can, heated on the gas cooker.
I'm generally happier with reactions which give off O2 than those which give off NOx and I2 (though I2 is prettier).
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Nicodem
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Am I missing something here? From what I understood, the question is how to prepare Ag and I2 from AgI, so what's wrong with its thermolysis? The Ag
remains, the I2 sublimates to the cold surface. According to Kirk-Othmer's, the temperature required is 552 °C which is easily reachable. A small
scale experiment can be done even in a flame heated test tube.
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Arsole
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I am curious if the Silver Chloride and Bromide salts have similar decomposition temperatures? Wikipedia lists the boiling temp for the Iodide at
1506C. How would that work if it decomposes at 552C? Any info on the thermolysis temperatures for the other salts?
Thoughts, like fleas, jump from man to man, but they don't bite everybody.
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