Sciencemadness Discussion Board - Recovering iodine from copper I iodide

Recovering iodine from copper I iodide

Boffis - 11-5-2019 at 08:34

I Recovered some rather impure copper I iodide from a series of experiments I did on converting dicyandiamide into biguanide. These basically involved heating dicyandiamide with an ammonium halide or other salt. Most salts give a fairly consistent yield of about 15-20% but ammonium bromide and iodide are supposed to give yields of 30 and 50% respectively. However, the biguanide is precipitated as its copper complex but there is a tendency for the iodide salts to complicate this precipitation procedure and I was left with a small amount (about 20g) of copper I iodide as a result.

Does anyone have any suggestion as to how best to recover the iodine either as the free element or as an alkali iodide? I was wondering if I could oxidize the CuI suspended in say 2M sulphuric acid with a layer of DCM or similar to collect the iodine (incase the Cu+ salt doesn't dissolve so the iodine cant be seperated by filtration).

Bedlasky - 11-5-2019 at 09:54

Maybe HNO3 or acidified H2O2?

Boffis - 11-5-2019 at 10:04

Hi Bedlasky, I thought about nitric acid but I think it will require heating and thus risk vapourizing some of the iodine or, if stronger nitric acid is used, oxidize it to iodic acid. While the latter is not necessarily a problem if I could separate it from the copper nitrate without it forming some sparingly soluble copper hydrogen iodate type salt that would require more nitric acid (precious stuff now!).

Another idea I had was to fuse it with sodium hydroxide to give sodium iodide, copper oxide and water.

Ubya - 11-5-2019 at 12:18

why not adding sulphuric acid and evaporate/sublime the formed iodine to purify it all in one step?

Boffis - 11-5-2019 at 12:35

@UBya; Because in strongly acidic environments Cu+ act like a reducing agent so I can most of the iodine coming off as HI unless I use conc acid when I will get SO2 etc as well. Messy

Ubya - 11-5-2019 at 16:31

Fusion with NaOH seems messy as well, easy and clean doesn't seem doable

wg48temp9 - 12-5-2019 at 02:03

Dissolve the CuI by complexing it with an alkali iodide or ammonia. You can then probably reduce the Cu+ to metal with zinc dust. Then extract the iodide as iodine or add alkali hydroxide then filter off the zinc hydroxide and you have a solution of alkali iodide.

The CuI can also be complexed with hexamine to form an interesting strongly fluorescent compound.

PS Can anyone tell me what causes the strong color of some transition metal iodides? The only thing I can find is that some are semiconductors, perhaps its a bandgap effect.

[Edited on 12-5-2019 by wg48temp9]

Bedlasky - 12-5-2019 at 03:40

Quote: Originally posted by wg48temp9

add alkali hydroxide then filter off the zinc hydroxide and you have a solution of alkali iodide.

Better is precipitate zinc with sodium carbonate or bicarbonate - excess of hydroxide dissolves zinc hydroxide.

Quote: Originally posted by wg48temp9

PS Can anyone tell me what causes the strong color of some transition metal iodides? The only thing I can find is that some are semiconductors, perhaps its a bandgap effect.

Which iodides exactly do you mean? Most of the coloured iodides are salts containing soft Lewis acid and soft Lewis base.

[Edited on 12-5-2019 by Bedlasky]

CobaltChloride - 12-5-2019 at 03:44

Quote: Originally posted by wg48temp9

The CuI can also be complexed with hexamine to form an interesting strongly fluorescent compound.
[Edited on 12-5-2019 by wg48temp9]

Huh, I didn't know hexamine made any interesting complexes. Do you have a reference on this complex? I'd be interested in reading more about it.

unionised - 12-5-2019 at 04:57

Quote: Originally posted by wg48temp9

The CuI can also be complexed with hexamine to form an interesting strongly fluorescent compound.

PS Can anyone tell me what causes the strong color of some transition metal iodides? The only thing I can find is that some are semiconductors, perhaps its a bandgap effect.

[Edited on 12-5-2019 by wg48temp9]

https://en.wikipedia.org/wiki/Charge-transfer_complex#Trend_...

I found this
https://www.sciencedirect.com/science/article/pii/0022190270...
which confirms that hexamine (one "m") complexes exist.
But I wonder if there may be confusion with hexammine (two "m"s) complexes.

wg48temp9 - 15-5-2019 at 16:20

Quote: Originally posted by unionised

Quote: Originally posted by wg48temp9

Below is a copy of the note describing the fluorescent copper(I) iodide hexamine (Hexamethylenetetramine) complex.
Attachment: hexamine-complexes-copper-iodide-hardt1975.pdf (412kB)
This file has been downloaded 376 times

Boffis - 16-5-2019 at 05:12

I have found that is the cuprous iodide is finely ground and suspended in fairly strong (battery acid c 30%) sulphuric acid and hydrogen peroxide is added (30%) iodine is slowly liberated as a solid. I am waiting to see if all of the iodide dissolves, if so I will filter of the ppt of iodine. If not then....

AJKOER - 23-6-2019 at 12:01

A definite simple path (see first page citing free Iodine and H2O2 at https://pubs.acs.org/doi/10.1021/j100266a038), apply sonolysis (SN) to the acidified aqueous iodide in dilute H2O2 (or pumped in air, see also 'Free radical formation from sonolysis of water in the presence of different gases' by Masahiro Kohno, et al, https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3171683/). In the case of CuI, add NH3 first to complex, then add O2/H2O2 with acid, say HI. This results in an aquaammine cupric salt: [Cu(NH3)k(H2O)(6-k)]I2, where k depends on NH3 concentration employed.

The underlying sonolysis reaction is:

H2O + SN --> H• + OH•

And, per chemistry below both of these radicals react with H2O2 to form, in steps at pH < 4.88, the hydroperoxyl radical (HO2•). The latter radical eventually, is reported with sonolysis to result in the liberation of free iodine (with the likely not so simple chemistry noted below) and even some H2O2 creation per cited reference.
-----------------------------------------------------------------

No sonolysis available, try using a microwave acting on graphite rods (like used in pencils) on H2O2 with an acidified aqueous iodide.

Based on this related work with free bromine formation (and not complexed), see 'Hydroperoxyl radical (HO2•) oxidizes dibromide radical anion (•Br2−) to bromine (Br2) in aqueous solution: Implications for the formation of Br2 in the marine boundary layer', see the work by Brendan M. Matthew, et al, fully available (for free) at: https://agupubs.onlinelibrary.wiley.com/doi/full/10.1029/200... , one probably needs something like HO2• to remove the likely persistent I2•- radical.

Assuming the iodine chemistry is in line bromine chemistry (see https://pubs.acs.org/doi/10.1021/j100398a045), we want hydroxyl radicals (OH•) and therefrom perhydroxyl radical (HO2• ) in acidic conditions, producing the following reaction system interacting with bromide (or, in the current case, iodide) and H2O2 per the cited reference reaction system:

Br- + OH• --> BrOH•-

BrOH•- --> Br- + OH• (the reversible reaction of above)

BrOH•- + H+ --> Br• + H2O

OH• + H2O2 --> H2O + HO2•

Br• + Br• --> Br2

Br• + Br- --> Br2•-

Br2•- + Br2•- --> 2 Br- + Br2

Br2•- + HO2• --> Br2 + HO2-

And apparently concurrently with the above:

Br2•- + HO2• --> 2 Br- + O2 + H+

H+ + HO2- = H2O2

Now, as to best path to HO2•, based on my recent experiments, I recommend the action of the microwave (MV) on carbon (as graphite rods, and not activated charcoal which absorbs iodine), which may form surface zones containing a deficiency and excess of electrons:

C(surface) + MV → C(+) + e-

which together with an acid source of H+ and H2O2, may produce a one electron reduction of the hydrogen peroxide, as follows:

In the cathodic Zone:

H2O2 ⇌ H+ + HO2-

e- + H+ = H•

followed by an attack on the H2O2 leading to radical formations (see http://www.ipb.pt/~htgomes/FCT110088-%5B2%5D.pdf equations (3) and (4), in particular).

My take is as follows:

H• + HO2- = OH- + OH•

OH• + H2O2 ---> HO2• + H2O (see https://pubs.acs.org/doi/abs/10.1021/jp100204z )

resulting in the required HO2•, which in the presence of an acidified aqueous iodide salt, say KI, may result in the liberation of free I2 and not some iodine complex.

I do not recommend the use of H2SO4, as it effectively scavenges the OH• to create sulfate radicals:

OH• + HSO4- = •SO4- + H2O

[Edited on 24-6-2019 by AJKOER]