DrMario
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Strange result from cuprous iodide experiment with EDTA
I synthesized cuprous iodide (CuI) by adding a solution of potassium iodide to Cu(II) sulfate solution. I obtained the tell-tale dark brown solution
of elemental iodine in water, and a precipitate that ended up being (after many decantations) off-white.
When I reacted this precipitate with diluted ammonium hydroxide, the precipitate dissolved and I obtained a deep royal blue solution which I guess is
some sort of complex with the cuprous ion (Cu(I)) I have never heard about before.
But this is not what surprised me the most: when I added a solution of EDTA disodium salt to another sample of CuI precipitate, I obtained no
significant dissolution, apart from a light green tint. The precipitate stayed mostly in the water, unaffected, except as I said, the water became
very lightly green tinted. HOWEVER... what shocked me is that the solution started smelling rather pleasantly of lemon or lime. And it was quite
distinct.
What up with that?? What sort of organic compound did I obtain that it smells like lime? And perhaps contains iodine? Maybe?
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DraconicAcid
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Quote: Originally posted by DrMario ![](images/xpblue/lastpost.gif) | I synthesized cuprous iodide (CuI) by adding a solution of potassium iodide to Cu(II) sulfate solution. I obtained the tell-tale dark brown solution
of elemental iodine in water, and a precipitate that ended up being (after many decantations) off-white.
When I reacted this precipitate with diluted ammonium hydroxide, the precipitate dissolved and I obtained a deep royal blue solution which I guess is
some sort of complex with the cuprous ion (Cu(I)) I have never heard about before.
But this is not what surprised me the most: when I added a solution of EDTA disodium salt to another sample of CuI precipitate, I obtained no
significant dissolution, apart from a light green tint. The precipitate stayed mostly in the water, unaffected, except as I said, the water became
very lightly green tinted. HOWEVER... what shocked me is that the solution started smelling rather pleasantly of lemon or lime. And it was quite
distinct.
What up with that?? What sort of organic compound did I obtain that it smells like lime? And perhaps contains iodine? Maybe? |
The deep royal blue is actually the cupric ammine complex- the cuprous one is colourless, but oxidizes on exposure to air.
The EDTA result is surprising. Then again, Ag(I) and Cu(I) tend to form two-coordinate complex ions, which EDTA doesn't really lend itself to.
[Edited on 23-11-2014 by DraconicAcid]
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DrMario
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Quote: Originally posted by DraconicAcid ![](images/xpblue/lastpost.gif) |
The deep royal blue is actually the cupric ammine complex- the cuprous one is colourless, but oxidizes on exposure to air.
The EDTA result is surprising. Then again, Ag(I) and Cu(I) tend to form two-coordinate complex ions, which EDTA doesn't really lend itself to.
[Edited on 23-11-2014 by DraconicAcid] |
I would very carefully (and very, very respectfully) tend to disagree on the cupric ion formation... at least the mechanism you describe doesn't sound
credible: the precipitate completely dissolved within a fraction of a second, in the NH4OH solution. There was no time for meaningful interaction with
air.
[Edited on 23-11-2014 by DrMario]
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DrMario
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Another thing I noticed: while I didn't measure carefully the amount of precipitate I created, the amount of ammonia necessary to dissolve even a
large amount of this precipitate seems to be tiny. About an order of magnitude less than is the case for cupric hydroxide.
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DraconicAcid
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Quote: Originally posted by DrMario ![](images/xpblue/lastpost.gif) |
I would very carefully (and very, very respectfully) tend to disagree on the cupric ion formation... at least the mechanism you describe doesn't sound
credible: the precipitate completely dissolved within a fraction of a second, in the NH4OH solution. There was no time for meaningful interaction with
air.
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It dissolves quickly because it forms [Cu(NH3)2]+, but this (being a d10 system) is colourless. It turns
blue because of oxidation- either through exposure to air, or oxygen dissolved in solution.
[Edited on 23-11-2014 by DraconicAcid]
[Edited on 23-11-2014 by DraconicAcid]
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DrMario
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According to this ChemWiki:
http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Descriptive_...
the [Cu(NH3)2]+ complex is stable!
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DraconicAcid
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Yes, just not in the presence of air. The website says it will not disproportionate (i.e., react with itself to give Cu metal and Cu(II)), but says
nothing about not reacting with air.
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Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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DrMario
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Quote: Originally posted by DraconicAcid ![](images/xpblue/lastpost.gif) | Quote: Originally posted by DrMario ![](images/xpblue/lastpost.gif) |
I would very carefully (and very, very respectfully) tend to disagree on the cupric ion formation... at least the mechanism you describe doesn't sound
credible: the precipitate completely dissolved within a fraction of a second, in the NH4OH solution. There was no time for meaningful interaction with
air.
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It dissolves quickly because it forms [Cu(NH3)2]+, but this (being a d10 system) is colourless. It turns
blue because of oxidation- either through exposure to air, or oxygen dissolved in solution.
[Edited on 23-11-2014 by DraconicAcid]
[Edited on 23-11-2014 by DraconicAcid] |
As I said, there is no time for exposure to air - the reaction is virtually instantaneous. Now let's assume that some small portion of the
[Cu(NH3)2]+ does react with (the very little) oxygen in the solution, thus giving it the royal blue colour: if that were the case, then shaking the
bottle in which most of the ions are [Cu(NH3)2]+ should cause oxidation of more of them, thus creating cuprous ions, thus increasing the intensity of
the colour. But this is NOT what happens - the colour of the liquid remains the same in intensity, no matter how much and how long it's shaken.
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DrMario
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Well, I'll be continuing to shake bottles with ever increasing concentrations of [Cu(NH3)2]+ and see if I can spot a change in colour.
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unionised
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Quote: Originally posted by DrMario ![](images/xpblue/lastpost.gif) | Quote: Originally posted by DraconicAcid ![](images/xpblue/lastpost.gif) |
The deep royal blue is actually the cupric ammine complex- the cuprous one is colourless, but oxidizes on exposure to air.
The EDTA result is surprising. Then again, Ag(I) and Cu(I) tend to form two-coordinate complex ions, which EDTA doesn't really lend itself to.
[Edited on 23-11-2014 by DraconicAcid] |
I would very carefully (and very, very respectfully) tend to disagree on the cupric ion formation... at least the mechanism you describe doesn't sound
credible: the precipitate completely dissolved within a fraction of a second, in the NH4OH solution. There was no time for meaningful interaction with
air.
[Edited on 23-11-2014 by DrMario] |
Did you consider the fact that the ammonia solution would contain air in solution before you added it to the CuI?
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bismuthate
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I performed in experiment much like this in which I added tetrasodium EDTA to CuI. I obtained a yellow powder
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DrMario
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Quote: Originally posted by unionised ![](images/xpblue/lastpost.gif) |
Did you consider the fact that the ammonia solution would contain air in solution before you added it to the CuI?
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Yes - so let's assume that this oxygen will oxidize part of Cu(I) to Cu(II) and turn the solution somewhat blue. If I shake the container after this,
more oxygen will dissolve and more Cu(I) should oxidize to Cu(II), creating a darker blue solution. But this latter phenomenon is not observed.
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blogfast25
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Quote: Originally posted by DrMario ![](images/xpblue/lastpost.gif) | Quote: Originally posted by unionised ![](images/xpblue/lastpost.gif) |
Did you consider the fact that the ammonia solution would contain air in solution before you added it to the CuI?
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Yes - so let's assume that this oxygen will oxidize part of Cu(I) to Cu(II) and turn the solution somewhat blue. If I shake the container after this,
more oxygen will dissolve and more Cu(I) should oxidize to Cu(II), creating a darker blue solution. But this latter phenomenon is not observed.
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Trust me, DA et al are right about this: the copper(I) ammine complex (like basically all Cu(I) compounds) does oxidide quickly to the Cu(II)
ammine complex.
Perhaps it helps to understand this as follows:
The Cu(I) ammine complex is in equilibrium:
Cu(NH<sub>3</sub> <sub>2</sub><sup>+</sup> <
=== > Cu<sup>+</sup> + 2 NH<sub>3</sub>, the formation equilibrium.
The free Cu<sup>+</sup> can then be oxidised by air oxygen and the resulting cupric ions complex with the excess ammonia, to form the blue
Cu(II) tetrammine complex ion. According to the Le Chatelier Principle the removal of cuprous ions further pulls the formation equilibrium to the
right.
So the cuprous diammine complex is quite 'stable', yet still prone to air oxidation at the same time.
Also, the cupric tetrammine complex is so intensely coloured that even small concentrations with create the illusion that all the cuprous has
disappeared.
[Edited on 23-11-2014 by blogfast25]
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DrMario
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Quote: Originally posted by blogfast25 ![](images/xpblue/lastpost.gif) | Quote: Originally posted by DrMario ![](images/xpblue/lastpost.gif) | Quote: Originally posted by unionised ![](images/xpblue/lastpost.gif) |
Did you consider the fact that the ammonia solution would contain air in solution before you added it to the CuI?
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Yes - so let's assume that this oxygen will oxidize part of Cu(I) to Cu(II) and turn the solution somewhat blue. If I shake the container after this,
more oxygen will dissolve and more Cu(I) should oxidize to Cu(II), creating a darker blue solution. But this latter phenomenon is not observed.
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Trust me, DA et al are right about this: the copper(I) ammine complex (like basically all Cu(I) compounds) does oxidide quickly to the Cu(II)
ammine complex.
Perhaps it helps to understand this as follows:
The Cu(I) ammine complex is in equilibrium:
Cu(NH<sub>3</sub> <sub>2</sub><sup>+</sup> <
=== > Cu<sup>+</sup> + 2 NH<sub>3</sub>, the formation equilibrium.
The free Cu<sup>+</sup> can then be oxidised by air oxygen and the resulting cupric ions complex with the excess ammonia, to form the blue
Cu(II) tetrammine complex ion. According to the Le Chatelier Principle the removal of cuprous ions further pulls the formation equilibrium to the
right.
So the cuprous diammine complex is quite 'stable', yet still prone to air oxidation at the same time.
Also, the cupric tetrammine complex is so intensely coloured that even small concentrations with create the illusion that all the cuprous has
disappeared.
[Edited on 23-11-2014 by blogfast25] |
I know the tetraamminecopper(II) ion is intensively blue. I don't dispute this. I don't _directly_ dispute any of the things you wrote, but am posing
the question, and I repeat: why do I not see an increase in intensity of colour, while shaking the bottle with the solution?
I have another question: If Cu+ ---> Cu2+, what is the cation we're talking about?
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DraconicAcid
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Quote: Originally posted by DrMario ![](images/xpblue/lastpost.gif) |
I know the tetraamminecopper(II) ion is intensively blue. I don't dispute this. I don't _directly_ dispute any of the things you wrote, but am posing
the question, and I repeat: why do I not see an increase in intensity of colour, while shaking the bottle with the solution?
I have another question: If Cu+ ---> Cu2+, what is the cation we're talking about? |
Because going from intense royal blue to slightly more intense royal blue is much more subtle than near-colourless to intense royal blue.
The cations we're discussing are Cu(NH3)2 (+) and Cu(NH3)4 (++). The first is colourless or nearly so; the second blue.
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DrMario
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Okay.
I have actually noticed a very slight increase in colour intensity with one of the samples I later prepared.
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Amos
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The same page DrMario mentioned also said that in the presence of water, cuprous iodide slowly turns solutions blue as it is converted to the +2
oxidation state. Case and point, if you ask me.
Where did your EDTA and iodine come from?
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DrMario
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The iodine comes from the disproportionation of CuI2 into CuI + iodine:
2CuI2 ----> 2CuI + I2
And the EDTA:Na2 comes from Sigma Aldrich.
Or maybe I don't understand your question.
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Amos
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Quote: Originally posted by DrMario ![](images/xpblue/lastpost.gif) |
The iodine comes from the disproportionation of CuI2 into CuI + iodine:
2CuI2 ----> 2CuI + I2
And the EDTA:Na2 comes from Sigma Aldrich.
Or maybe I don't understand your question. |
I'm asking what source you obtained the starting reagents from, you answered on the EDTA already; I too am extremely puzzled by the smell you
mentioned.
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DrMario
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Quote: Originally posted by No Tears Only Dreams Now ![](images/xpblue/lastpost.gif) | Quote: Originally posted by DrMario ![](images/xpblue/lastpost.gif) |
The iodine comes from the disproportionation of CuI2 into CuI + iodine:
2CuI2 ----> 2CuI + I2
And the EDTA:Na2 comes from Sigma Aldrich.
Or maybe I don't understand your question. |
I'm asking what source you obtained the starting reagents from, you answered on the EDTA already; I too am extremely puzzled by the smell you
mentioned. |
I didn't buy iodine, I think that's clear by now. If you read the first post you'll see that I started from CuSO4 (a small UK company) and KI (this is
also from Sigma Aldrich).
The smell is very strange. I hope it's not some toxic compound as I sniffed it plenty of times to make sure it's coming from the beaker. And also
because it was surprisingly pleasant.
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