Sciencemadness Discussion Board

Copper(I) Iodide Synthesis

moonfisher - 18-2-2016 at 11:49

Hi y'all!

Just to make sure, and to pick anyone with experience in this, would Copper(II) acetate solution + LiI + NaI yield a precipitate of CuI? Planning to do this today.

DraconicAcid - 18-2-2016 at 11:57

Why use both lithium iodide and sodium iodide? Either one will work just fine.

moonfisher - 18-2-2016 at 12:28

Its a dumb story, but a long while ago I accidentally mixed LiOH and NaOH. I got a new batch, but I couldn't bring myself to throw the contaminated one out.

I used these salts to isolate iodine from Povidone for some fun (its a fun color change, an the plastic gunk produced is fun to play with :P) so now I have a lot of NaI/LiI salt!

Crowfjord - 18-2-2016 at 14:15

Those sound fine for getting the iodide in there, but I don't see a reducer anywhere? How do you plan to reduce Cu(II) to Cu(I)? If you hadn't thought of that, sodium metabisulfite works well ;)

DraconicAcid - 18-2-2016 at 14:20

Quote: Originally posted by Crowfjord  
Those sound fine for getting the iodide in there, but I don't see a reducer anywhere? How do you plan to reduce Cu(II) to Cu(I)? If you hadn't thought of that, sodium metabisulfite works well ;)


Iodide is the reducing agent. 2 Cu(2+) + 4I(-) -> 2 CuI + I2

Metacelsus - 18-2-2016 at 16:58

You also could add a reducing agent to regenerate iodide from iodine.

moonfisher - 20-2-2016 at 12:35

Thanks everyone! Worked like a charm (~95% yield on copp. acetate) about 10g of CuI. (Ill post the pics once I get a decent camera.)

MeshPL - 20-2-2016 at 12:46

Adding a reducing agent may not work as intended, as some reducers e.g. bisulphite will reduce copper to Cu(I) and some maybe even to Cu(0). A careful choice of reagent is necessary.

woelen - 20-2-2016 at 14:13

Sulfite is a perfect reductor for this purpose. It leads to formation of more CuI, otherwise you lose half of your iodide as iodine and you get a brown precipitate instead of a white one. With sulfite you do not need to worry about reduction to the copper(0) state.

moonfisher - 20-2-2016 at 14:31

Yah, forgot to mention, used sodium sulfite + weak sulfamic acid. Thanks for reminding me! lol (the sulfamic acid came from "phosphate-free rust remover")

BTW, woelen, your website is beautiful.

[Edited on 20-2-2016 by moonfisher]

byko3y - 20-2-2016 at 22:11

I have no idea why sulfamic acid was needed, maybe because you've used copper(II) acetate, which is neutral, while copper(II) sulfate or copper(II) chloride is acidic. This trick is not obvious, some people might think they need to add acid even in the case of CuSO4. In fact, I'd say you could use any strong acid to lower the pH (acid should not give insoluble salt with copper).

Leiem - 25-2-2016 at 06:16

It's easy to seperate Li+ from Na+.

MrMario - 25-2-2016 at 06:36

Can't you put some copper metal in solution and boil it a little to go from Cu(II) --> Cu(I)?

I saw that this works when going from CuCl2 -> CuCl

byko3y - 25-2-2016 at 07:35

MrMario, reduction with sulfite is more reliable. Maybe copper powder will also work well, but not just a large piece of copper metal.

Amos - 25-2-2016 at 07:47

Reduction of copper(II) isn't even necessary in the first place. Copper(II) iodide is unstable and rapidly disproportionates into copper(I) iodide and molecular iodine. Any reducing agent present is only there to convert the molecular iodine formed back into iodide.

[Edited on 2-25-2016 by Amos]

woelen - 25-2-2016 at 12:08

Iodine/iodide is quite expensive, so it is better to use a reductor. In that way, all iodide is used for formation of CuI. Without reductor, you lose half of your iodide.