Lion850
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Copper reaction with iodine in HCl
Since observing that gallium is easily put into solution when reacted with I2 in HCl, and that Sn reacts with I2 in HCl to give a very stable orange
compound (that I suspect is SnI2), I wanted to do more such experiments. The one described below started last August but such is life that I did not
complete it until today.
- 5g Cu powder, 10g concentrated HCl, and 5g I2 added to a beaker and stirred with a glass rod.
- White ppt formed, beaker heated up to the point where it almost could not be touched, and fumes that smelled like HCl.
- After 10 minutes all I2 appeared to have been consumed and beaker cooled. There was still unreacted Cu.
- Add 5g I2 and stir. Contents heated up again.
- Add 5g HCl and leave for a while.
- Filter and rinse remainder with concentrated HCl.
- Filtrate pale yellow. Remainder a very pale grey powder with minute flecks around the edge of the filter paper of what appears to be unreacted
copper. -
- Separated the "copper" flecks and leave rest on bench to dry (up to near 50C in shed in summer).
- Noticed the powder slowly turned green until after a few months it was completely dry and free flowing and sea green in color. Weight of 6.5g.
- 8 month later....still looked the same, still same weight.
- Put 1 gram in a beaker with 10g water. Shook and stir, but no sign of anything dissolving: settles out as ppt with same green color with water above
clear.
- Add 5 gram 70% nitric acid. A blue solution formed, and some black flakes appeared. Add another 5g nitric acid.
- Filter. Scrape off some of the black solid and drip on HNO3 to see if it is copper: no NOx fumes.
- Add a solution of lead nitrate to the blue remainder to see if any lead iodide precipitates....none.
- Heat the still wet filter paper to dry it and then see if the black solids subimates into I2 vapour.....the filter paper caught fire, beautiful
green flame, and after it was burned out there was copper streaks on it (no I2 fumes observed). Dripped HNO3 onto the copper colored streaks and
observed and smelled NOx fumes.
So, it seams none of the Iodine made it into the solid product, thus it remained in the solution after the initial reaction. My best guess of what
happened:
2HCl + I2 + Cu = HI + CuCl.
If this is true it seems a relatively easy way to make both HI and CuCl in one go, although how pure I don't know.
I want to do this again, sticking to the weights of reagents as per the stoichiometry of the above equation, and then see if the filtrate - which
should be HI - reacts with tin to give orange SnI4/SnI2. And whether the remainder, if dried quickly in a steam bath and bottled, stays pale grey.
Maybe over time the initial remainder being light colored CuCl turned into Copper oxychloride while standing on the bench which explained the final
green color and insolubility in water?
Seems some of the product disproportionated into CuCl2 and Cu when dilute HNO3 was added?
Why did the black flakes (after adding HNO3) not seem to react with concentrated HNO3 until after the paper burned?
Why did Tin preferentially react with the I2 in a mix of I2 and HCl but copper with the Cl2?
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Bedlasky
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CuI is less soluble than CuCl, so it should precipitate from the solution.
https://www.engineeringtoolbox.com/solubility-product-equili...
So your observations seems interesting. But I can't think right know (too tired), so I really don't have any suggestions what happened.
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bnull
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White precipitate that turns green as it ages: copper(I) chloride. I made some copper(I) iodide 5, 6 months ago. It is still a pale cream or grey
colored powder, even in contact with air. No signs of green in it.
Triiodide (I3-). Iodine is soluble in iodide salts, and now we may assume it is soluble in hydriodic acid too (I'll file the
information for later. Thanks).
Quote: Originally posted by Lion850  | So, it seams none of the Iodine made it into the solid product, thus it remained in the solution after the initial reaction. My best guess of what
happened:
2HCl + I2 + Cu = HI + CuCl. |
What I suppose that happened is thisNote:
Iodine is slightly soluble in water and reacts directly with copper: $$Cu(s)+I_2(aq) \rightarrow CuI(s);$$
Hydriodic acid is formed: $$2CuI(s)+2HCl(conc)\rightarrow 2CuCl(aq)+HI(aq);$$
Iodine is more soluble in the presence of iodide ions and forms triiodide: $$I_2(s)+I^-(aq) \rightarrow I_3^-(aq)$$
Reaction (1) is may be quite slow in the beginning because of the solubility of iodine, which increases as reactions (2) and (3) proceed (catalytic
process, so it seems). Reaction (3) also explains the yellow color.
| Quote: Originally posted by Lion850 |
If this is true it seems a relatively easy way to make both HI and CuCl in one go, although how pure I don't know. |
The precipitate may be a mixture of copper(I) iodide and copper(i) chloride depending on the proportions of copper to HCl and iodine. In your case,
all the iodine stayed in the solution. The yield of copper(I) chloride may be higher with copper as limiting reagent and an excess of hydrochloric
acid over iodine.
You can recover unreacted iodine from the solution. Triiodide seems to decompose on heating, liberating iodine fumes. If you can distill hydriodic
acid from the solution while condensing the iodine fumes out of the receiving flask, that's a good idea.
If you store it under inert atmosphere (carbon dioxide, argon, nitrogen) in a hermetically sealed flask, it will stay grey.
| Quote: Originally posted by Lion850 | | Maybe over time the initial remainder being light colored CuCl turned into Copper oxychloride while standing on the bench which explained the final
green color and insolubility in water? |
Yes.
If you don't want to recover the iodine, then add more copper and a source of copper(II) ions to the yellow filtrate. Copper becomes copper(I) and
copper(II) is reduced to copper(I) because of reactions (1) and (2). You'll see a small fog (I can't find the right word) of white copper(I) iodide
sink slowly from the metal to the bottom of the flask. The white precipitate becomes purple as the iodide is oxidised to iodine, or so it seems,
according to the reaction below:
$$2CuI(s)+2Cu^{2+}(aq)+4Cl^-(aq) \rightarrow 4CuCl(s)+I_2(s).$$
Use a copper wire or foil suspended in the solution, or copper powder if you don't mind stirring.
The reaction of copper sulfate with sodium iodide in the presence of hydrochloric acid (I couldn't resist and performed it half an hour ago) produces
copper(I) chloride, minute crystals of iodine, and the supernatant liquid is an amber- or honey-colored solution. On heating, a thin purple smoke
(purple haze?) rose above the surface of the liquid. On cooling, iodine condensed in the walls of the tube. I added a piece of clean copper foil and
more hydrochloric acid. That's why I suggested the procedure above.
Thank you for the report.
Note: I may be making an ass of myself again, of course. It happens from time to time.
Nevermind.
Edit: It seems that, when the concentration of chloride ions is low, the copper(I) iodide begins to precipitate. No oxidation to iodine, no purple
color, no amber solution.
[Edited on 5-5-2024 by bnull]
Quod scripsi, scripsi.
B. N. Ull
P.S.: Did you know that we have a Library?
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Lion850
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Thanks bnull. Interesting stuff. Have a few things in mind to try when time allows. And I need to get a new hot plate stirrer.
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