Hello. I'm a bit confused at the electrolysis process.
Here's the story. I added the slightest amount of NaCl to about 0.5l of water and I run 12V DC into it and I use Cu anode and Cu cathode.
Now if I'm correct, there are different types of iones in that electrolyth:
H+
OH-
Cu2+
Na+
Cl-
Can all of these ions combine with each other? So I will have HCl, NaOH, Cu(OH)2 and CuCl2?
In that case HCl and NaOH will neutralize into water and salt immediately (making them impossible to form), leaving me only Cu(OH)2 and CuCl2. The
experiment itself shows, that the solvent turns green (?) after about 10 minutes. Google gives me, that both Cu(OH)2 and CuCl2 are blue. Why green?
So I'm misunderstanding something. Could someone help?
(sorry for the bad english that might have occurred)woelen - 28-5-2006 at 06:54
Electrolysis of NaCl-solution with a copper anode gives very complicated results. You get both copper (I) and copper (II) species. The copper (I)
species are yellow/orange (hydrous Cu2O) and the copper (II) species can be anywhere from cyan to blue. The mix of these can be green, olivegreen, or
even brown, depending on the amount of copper (I) and copper (II) in the mix.
Another thing is that complexes are formed. Both copper (I) and copper (II) are very good at complexing with chloride ions. This also gives rise to
all kinds of colors, ranging from dark brown to cyan.wahur - 28-5-2006 at 07:42
I see. Would it make any difference if some other eletrolyth is used instead of NaCl, NaHCO3 for example?12AX7 - 28-5-2006 at 07:52
Yeah, CuCl2.2H2O is light blue, but a strong solution is green from CuCl4(2-) ("tetrachlorocuprate") ions.
Because "no" Cl2 or HCl or Na or NaOH is *removed* from solution, the average pH remains neutral. Although the cathode collects a high concentration
of NaOH, the anode collects a high concentration of Cl2, which being a strong oxidizer, attacks the copper as it's produced, forming Cu(II) in weak
solution, or Cu(I) in a chloride-rich solution (because Cu(I) is soluble in a strong chloride solution, as Woelen noted). But where the solutions
combine, the pH is neutralized, and Cu drops out as an insoluble hydroxide. Cu(OH)2 of course is blue, Cu(OH,Cl)2 (copper oxychloride) is green, and
for Cu(I), CuOH is unstable at room temperature and dehydrates to orange Cu2O.
The green you are seeing is probably a suspension of Cu(OH,Cl)2. Let it run for a while longer and see if anything settles out, or if the particles
get larger. Colors get skewed by particle size, for instance Cu2O starts out yellow when colloidial, but is actually brick red in "larger" crystals.
If the green remains in solution, then it may be CuCl2 dissolved.
Tim
[Edited on 5-28-2006 by 12AX7]12AX7 - 28-5-2006 at 08:05
Doh, simulpost
Quote:
Originally posted by wahur
I see. Would it make any difference if some other eletrolyth is used instead of NaCl, NaHCO3 for example?
It would.
Cathode is the same of course:
2e- + 2H2O = 2OH- + H2
(Electrons enter solution by flowing out from the cathode, in the process splitting water. The freed hydroxyls make the cathode solution basic.)
At the anode, you would expect carbonate to be released, which either combines with the copper, essentially making verdegris, or else releasing CO2
gas. But what actually happens, in my experience, is the solution becomes striking blue, most likely a cuprate(II) ion or complex (think CuO2(2-)),
despite the solution not being particularly basic.
Timwahur - 28-5-2006 at 09:08
Thanks Tim and Woelen.
This one is much more harder than I thought it would be.garage chemist - 28-5-2006 at 11:11
NaCl is a very bad choice for an electrolyte in homemade electrolysis experiments.
It gives a multitude of reactions, depending on electrode materials (the anode nearly always gets eaten, with consequent reactions of the metal ions
with the alkali from the cathode), and when an inert anode, such as graphite, is used, gives chlorine only for very short time, and then first some
hypochlorite and then slowly chlorate.
Better choices for electrolyte are HCl (pure chlorine and hydrogen with graphite anode, and metal ions without interference of alkali from the
cathode), dilute H2SO4 (gives oxygen and hydrogen, when inert anode is used) and other acids.
Interesting are also metal salts, like copper sulfate (copper plating at the cathode) or zinc salts (zinc plating) etc.
If your target is to get CuCl2, electrolyse some HCl with copper anode, filter from CuCl precipitate (if any is formed) and boil down (outside!).
Cu(OH)2 is made by adding NaOH to a CuCl2 solution.
Cu(OH)2, when heated, becomes black as it turns into CuO (this already happens when an aqueous suspension of Cu(OH)2 is boiled).12AX7 - 28-5-2006 at 12:50
Funny thing about copper, it doesn't dissolve off the cathode (especially with the voltage present), so the equilibrium between copper dissolving into
solution at anode, plating out at cathode, and dissolving from metal into solution (bare metal or the anode or cathode), is rather low. So before
long, you get a stinking, green solution that's plating copper straight across and not increasing in concentration. Other metals though will dissolve
easily, a zinc-copper alloy for instance could be dissolved in HCl by plating the Cu across, while the reactive metal znic stays put more or less.
GC: a green CuCl2 solution will NOT produce blue Cu(OH)2, but rather Cu(OH,Cl)2, which is green, and doesn't decompose to black CuO on boiling. A
dilute CuCl2 solution, which is light blue, may work; I haven't tried.
