Sciencemadness Discussion Board

Cu from Cu(I)Cl

ONeill - 7-10-2012 at 11:06

Hello folks,

I´m trying to get elemental copper from Cu(I)Cl. I have tried thermal decomposition in inert gas with temperatures between 200 an 600 °C, but the result was just a grey powder that decomposes in air within some hours to a green solid. At higher temperatures the salt just evaporates.

In the next days I will try to decompose the copper chloride in degassed water to get Cu(II) and Cu.

Any new ideas, maybe a reducing agent like NaBH4?

Thanks for your comments.
Cheers ONeill

Mumbles - 7-10-2012 at 11:24

Your best best is probably to convert to a water soluble form, and reduce it with aluminum, iron, or zinc. Oxidizing it to the generally much more soluble copper (II) is one possibility, followed by reduction. Copper (I) Chloride can be made soluble by a variety of things. Dissolving it in HCl or another halide source would probably be my first attempt as far as this goes.

If you want to reduce it with heat, you might try to run the thermal decomposition in a hydrogen stream. This works for Copper (I) oxide, and several other chlorides. I just don't know off hand if it definitely works for copper (I) chloride though.

[Edited on 10-7-2012 by Mumbles]

kristofvagyok - 7-10-2012 at 11:51

Have tried thermite reaction? CuCl could be also converted to Cu with Al and it could be washed with cc HCl to remove impuries.

Or heat it up in a H2 steam... I think that it will also work... It worked for me with CuO and Cu2O.

Hexavalent - 7-10-2012 at 11:59

Yesterday I prepared some copper powder by the reduction of copper (II) sulfate with ascorbic acid (using dilute ammonia solution to control pH and 'coagulate' the fine powder into larger particles). The reduction is more favourable than formation of a complex so the process works rather well.

blogfast25 - 8-10-2012 at 05:20

Copper is so easy to make it's not worth sacrificing an expensive (and unstable) compound like CuCl (I) on it.

'Themiting' chlorides doesn't work. Generally speaking AlCl3 has far too low Heat of Formation, it is also very volatile (sublimes). Generally speaking reducing chlorides pyrometallurgically requires closed reactors to confine the volatile chlorides (see e.g. Kroll Process)

[Edited on 8-10-2012 by blogfast25]

ONeill - 14-10-2012 at 09:05

Hello guys,
thank you for you answers!

Quote: Originally posted by Mumbles  
Copper (I) Chloride can be made soluble by a variety of things. Dissolving it in HCl or another halide source would probably be my first attempt as far as this goes.

I will try to make a solution with typical solvents, any idea what works good? HCl and ammonia should work good (building of complexes), but I woul prefer a neutral solvent.

Quote: Originally posted by Mumbles  
If you want to reduce it with heat, you might try to run the thermal decomposition in a hydrogen stream. This works for Copper (I) oxide, and several other chlorides. I just don't know off hand if it definitely works for copper (I) chloride though. [Edited on 10-7-2012 by Mumbles]

I´m shure that this would work, but I would like to find an alternative for hydrogen as reducing agent.

Quote: Originally posted by kristofvagyok  
Have tried thermite reaction? CuCl could be also converted to Cu with Al and it could be washed with cc HCl to remove impuries.

Shure this would work but I would prefer to use a homogenous solution as reducing agent. Thank you.

Quote: Originally posted by Hexavalent  
Yesterday I prepared some copper powder by the reduction of copper (II) sulfate with ascorbic acid (using dilute ammonia solution to control pH and 'coagulate' the fine powder into larger particles). The reduction is more favourable than formation of a complex so the process works rather well.

This could be a good alternative, thank you.

Quote: Originally posted by blogfast25  
Copper is so easy to make it's not worth sacrificing an expensive (and unstable) compound like CuCl (I) on it.

I´ve got a lot of Cu(I)Cl and If I don´t use it it will be thrown away ;)

Thanks again for your answers.
Cheers ONeill