Eddygp - 3-3-2013 at 05:37
I reacted zinc with water and NaOH as stated here:
Zn + 2H2O + 2NaOH = Na2Zn(OH)4
I would like to precipitate zinc from the solution, but I don't know how to get it. Before you say "Well, why did you react the zinc then?", the
pieces of zinc I reacted were that: pieces, and I would like to have a fine powder.
Thanks,
Eddygp
12AX7 - 3-3-2013 at 10:31
Supposedly, zinc can be electroplated from such a solution.
All I've ever gotten was dendrites, since I don't know what, or have any, leveling agent that's suitable for zinc and very high pH.
Sounds like that would be just fine for your purpose -- chew up some "pieces" with electrolysis. Just be sure to collect the dendrites frequently,
lest the cell short itself out or efficiency drop.
Tim
AJKOER - 3-3-2013 at 13:34
More on the electroplating angle per Wikipedia (http://en.wikipedia.org/wiki/Zincate ):
"Zincate - plating processes
In industry it can refer to the alkaline solutions used in a dipping (immersion) process to plate aluminium with zinc prior to electrolytic or
electroless nickel plating. This immersion process is electroless (i.e. not electroplating) and involves the displacement of zinc from zincate by
aluminum:[5]
3 Zn(OH)42− + 2 Al → 3 Zn + 2 Al(OH)4− + 4 OH− "
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However, does this recent citation (source link: http://www.sciencemadness.org/talk/post.php?action=reply&... ) by AndersHoveland give you any ideas?
"Cuprous oxide and soda give a somewhat similar reaction with formaldehyde. This reaction was described by Loew (Ber. 1887, 20, 145) as a catalytic
reaction, but it appears in reality to be a quantitative one, expressed by the equation
Cu2O + 2 NaOH + 2 CH2O = Cu2 + H2 + 2 HCO2Na + H2O
Cupric oxide also gives a similar reaction, two atoms of hydrogen being liberated for each atom of oxygen in the oxide
When caustic soda and then formaldehyde are added to a solution of copper sulphate and the liquid gently warmed, the cupric hydroxide is reduced to
cuprous oxide without evolution of hydrogen, and when the temperature is subsequently raised, the cuprous oxide reacts as described above. When, on
the other hand, caustic soda is added to a boiling solution of copper sulphate, the liquid cooled, and formaldehyde then added, no reduction of
cuprous oxide occurs on warming, but metallic copper is formed, and twice as much hydrogen is evolced as the previous case.
Manganese dioxide does not appear to be reduced by formaldehyde, whilst the oxides of mercury and bismuth are reduced without evolution of hydrogen."
Source:
Journal of the Society of Chemical Industry, Volume 18, p716-717. [July 31, 1839]
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So, one could try the direct reaction of CH2O on Na2Zn(OH)4 (speculation):
Na2Zn(OH)4 + 2 CH2O =?= Zn + H2 + 2 HCO2Na + 2 H2O
which does, however, balance nicely.
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Alternately, one could add an excess of HCl, so one would have:
Na2Zn(OH)4 + 2 HCl --> 2 NaCl + Zn(OH)2 + 2 H2O
Zn(OH)2 + 2 HCl --> ZnCl2 + 2 H2O
and adding more more NaOH with CH2O, it would be interesting if (again, an experiment):
ZnCl2 + 4 NaOH + 2 CH2O =?= Zn + H2 + 2 HCO2Na + 2 NaCl + 2 H2O
Note, we cannot have an excess of NaOH as it would attack the free Zn. Also, any available HCl would also attack the Zinc.
AJKOER - 7-3-2013 at 08:52
Here is another reaction I came across which unfortunately forms Hydrazine, N2H4:
2 NH3 + Na2ZnO2 = N2H4 + Zn + 2 NaOH
which is probably reversible. So, perhaps treating with HCl, not in excess to neutralize the NaOH only, may work, or perhaps using NH4Cl (no
guarantee):
2 NH4Cl + Na2ZnO2 --> N2H4 + Zn (s) + 2 NaCl + 2 H2O
Caution: Per Wikipedia (http://en.wikipedia.org/wiki/N2H4 ): "Hydrazine is highly toxic and dangerously unstable, especially in the anhydrous form. According to the U.S.
Environmental Protection Agency: Symptoms of acute (short-term) exposure to high levels of hydrazine may include irritation of the eyes, nose, and
throat, dizziness, headache, nausea, pulmonary edema, seizures, coma in humans. Acute exposure can also damage the liver, kidneys, and central nervous
system. The liquid is corrosive and may produce dermatitis from skin contact in humans and animals."
[Edited on 7-3-2013 by AJKOER]
AJKOER - 8-3-2013 at 15:23
Today I also can across the reaction:
NaAlO2 + NH4Cl + H2O -----> Al(OH)3 + NH3 + NaCl
which, in the case of Na2ZnO2, implies the formation of Zinc hydroxide and not Zn metal. Again, a tentative reaction:
Na2ZnO2 + 2 NH4Cl --> Zn(OH)2 + 2 NH3 + 2 NaCl
which may be more plausible than the formation of N2H4 and Zn.
[EDIT] Interestingly, it is known that Zinc complexes with ammonia (see, for example, http://www.public.asu.edu/~jpbirk/qual/qualanal/zinc.html ):
Zn(OH)2(s) + 4NH3(aq) <==> [Zn(NH3)4]2+(aq) + 2OH-(aq)
so some of the Zn(OH)2 formed could dissolve and create a complex.
Also, with respect to Zinc complexing with Hydrazine, per this source (https://docs.google.com/viewer?a=v&q=cache:dOqOD7T13QkJ:... ), to quote
"On the other hand, in the bath containing N2H4 as the
only complexing agent, the Zn complex is predom-
inantly in the form [Zn(N2H4]3]2+ with k=105.5 [7,11].
This low k value encourages Zn ion presence..."
so again some of the free Zn (if formed) in the presence of a base, sulfide, CO2,...could form a soluble complex with the N2H4.
[Edited on 9-3-2013 by AJKOER]