Ok, this is one gnawing question that i have in my mind but lack the ingredients neccesary to perform the experiment(i lack ammonia).
Consider, that i have [Cu(NH3)4]SO4. This I pour to 2 separate cups, connected by salt bridge. If i apply an electric current through it, what would i
get? Could someone help me perform this experiment and see? I mean does Copper plate itself to cathode, liberating ammonia? Or does someother reaction
occur?
Does the same thing happen with copper ions with ligands consisting of chloride ions?Marvin - 12-11-2004 at 06:57
I would expect the reducing side to produce colourless Cu(I) and not to produce any metal until its used up, or to produce it and see it dissolve.
Chloride will also stabilise the Cu(I) state I think, but less well.
Oxygen will attack the solution and a basic salt could ppt.
No idea what would happen at the oxidising end if you used an inert anode.guy - 12-11-2004 at 12:44
Quote:
i lack ammonia
That's like the easiest thing to get at like a grocery store to clean things with. Or you can get it from window cleaners.darkflame89 - 13-11-2004 at 00:17
Wait, i didn't know that copper(II) was so easily reduced to copper(I). I would have thought the reducing all the way to copper metal is easier
than stopping halfway at copper(I). Then where does the ammonia go? Does it get reduced?KemiRockarFett - 19-11-2004 at 07:13
Quote:
Originally posted by darkflame89
Wait, i didn't know that copper(II) was so easily reduced to copper(I). I would have thought the reducing all the way to copper metal is easier
than stopping halfway at copper(I). Then where does the ammonia go? Does it get reduced?
Cu(II) will go to Cu(O) if not a solid product is formed or you got an compound with pi-acceptor character in your solution.
Cu(I) is very unstable to dispropotaion due to the high hydration entalphy of the divalent ion.
2Cu(I) -><- Cu(II) + Cu (s)
If you have got the right circumstances so use a low voltage:
E0(Cu+/Cu) = +0.5072 V
E0 (Cu2+/Cu+) = +0.1682 V
At the anod you want the sulphate ion oxidiated to persulphate? An way of making an primary
[Edited on 19-11-2004 by KemiRockarFett]guy - 19-11-2004 at 18:32
Electrolysis of [Cu(NH3)4]SO4. That could probably make copper at the cathode and NH3 will be liberated and H2SO4 will form at the anode which will
combine with the ammonia forming (NH4)2(SO4). I will try it once I finish making my CuSO4.KemiRockarFett - 22-11-2004 at 06:11
Quote:
Originally posted by guy
Electrolysis of [Cu(NH3)4]SO4. That could probably make copper at the cathode and NH3 will be liberated and H2SO4 will form at the anode which will
combine with the ammonia forming (NH4)2(SO4). I will try it once I finish making my CuSO4.
I dont think you have to try because this is whats going to happend.
Why not skip the ammonia complex and elektrolyse CuSO4(aq) instead, using copper anode and try to oxidise the sulphate ion. Afterwards you can add the
ammonia to make the ammonia complex, the ammonia complex will bond stronger to Cu(II) due to that ammonia becomes an stronger induced dipole than
water.
After that you can always get it out with ethanol and filter it.darkflame89 - 23-11-2004 at 00:51
The problem is not that i want to oxidise the sulphate ion, anyway, it can't be done in aqueous solutions; the problem is that what happens to
ligands when you conduct electrolysis.Marvin - 23-11-2004 at 04:27
For water the Cu(I) state is unstable, when you add ammonia it becomes much more stable and the solution will dissolve copper metal to produce Cu(I).
Remeber in water the Cu(II) ion is still complexed. Its present as Cu(H2O)6 2+. When you add ammonia the water is kicked out and the complex turns
into Cu(NH3)4 2+. This is the reason for the colour change to very dark blue/purple when you add ammonia. (The light blue precipitate in between
these is just copper hydroxide).
If copper goes out of solution as a metal ammonia or water becomes free again as part of the solvent.
[Edited on 23-11-2004 by Marvin]KemiRockarFett - 24-11-2004 at 06:53
Quote:
Originally posted by darkflame89
The problem is not that i want to oxidise the sulphate ion, anyway, it can't be done in aqueous solutions; the problem is that what happens to
ligands when you conduct electrolysis.
Caro´s acid is formed as a byproduct during the preparation of H2S2O8 by electrolysis of aqueous H2SO4. (N. Caro, 1898)KemiRockarFett - 24-11-2004 at 07:04
Quote:
Originally posted by Marvin
For water the Cu(I) state is unstable, when you add ammonia it becomes much more stable and the solution will dissolve copper metal to produce Cu(I).
Remeber in water the Cu(II) ion is still complexed. Its present as Cu(H2O)6 2+. When you add ammonia the water is kicked out and the complex turns
into Cu(NH3)4 2+. This is the reason for the colour change to very dark blue/purple when you add ammonia. (The light blue precipitate in between
these is just copper hydroxide).
If copper goes out of solution as a metal ammonia or water becomes free again as part of the solvent.
[Edited on 23-11-2004 by Marvin]
Have you read my post? I just explained these things.
Ammonia have a stronger induced dipolemoment than water, to copper, and therefore bonds stronger to copper than water.
Perhaps it work out with ammonia but I think you must have a stronger electron acceptor than NH3, for example CN-.
