I've never really given this much thought since it's never really been something of attention in any reaction I've ever done or read about.
What would happen if by some reaction you have a salt in an aqeuous solution, and the anion disappears? Sodium Formate can give up its hydrogen, and
the carbon dioxide will evaporate, leaving only the sodium cation.
What would be left if all the water were to be evaporated if there was only sodium cation? woelen - 17-6-2014 at 22:51
This is not possible. Where would the charge go? Indeed, the ion HCOO(-) can lose hydrogen, but then oxalate ion, C2O4(2-) remains behind.
In a normal chemical reaction there never will be charge buildup in a confined space. aga - 17-6-2014 at 23:02
Wonder what would happen if you generate a huge electric charge at one side of a reaction vessel, using a van der graaf or similar.
One way to find out ...JefferyH - 18-6-2014 at 00:25
This is not possible. Where would the charge go? Indeed, the ion HCOO(-) can lose hydrogen, but then oxalate ion, C2O4(2-) remains behind.
In a normal chemical reaction there never will be charge buildup in a confined space.
Sorry for being unclear. Formate's are commonly used as a hydride/hydrogen donor in transfer hydrogenations. The Formate anion gives up a hydrogen and
carbon dioxide is produced.
Another hydrogen donor used in transfer hydrogenation reactions is ammonium formate, which in the presence of Pd/C, generates hydrogen, ammonia and
carbon dioxide (eq 3).
When Ammonium Formate is used... Ammonia is formed. What is formed when Sodium Formate is used? Sodium Hydride, which reacts to form Hydroxide?woelen - 18-6-2014 at 01:13
Somewhere the charge has to go as well. The reaction would be
HCOO(-) --> [H] + CO2 + e, where [H] is the donated hydrogen atom, but e also should go somewhere. In the case of ammonium formate, ammonia is
formed and then it is conceivable that two hydrogens are donated:
NH4(+) + HCOO(-) --> 2[H] + NH3 + CO2
There is no net charge to be consumed.
I can imagine that with sodium formate such a hydrogen transfer also can occur, but then there must be something which absorbs the electrons. Maybe in
an electrolytic cell?
Anyway, never can a chemical reaction occur, if charge is not consumed as well. Either the charge must be removed by an electrode (in an electrolytic
cell), or there is a redox reaction in which charge is transferred from one species to another. Macroscopic charge buildup cannot occur.
@aga: Generating a high charge at one side of a vessel will just lead to electrolysis, with production of a lot of heat and possible breakdown of
molecules due to ionisation. I once did an experiment in which I electrolysed a salt solution with 15 kV plasma beams as anode. This leads to
formation of ozone, chlorine and a lof of heat (spattering of liquid, where the plasma beam hits the surface of the liquid).FireLion3 - 18-6-2014 at 01:32
No net charged being consumed makes sense, since the mechanism is a transfer hydrogenation; charges being transferred.
If ammonium formate gives up two hydrogen then so should sodium formate, seeing as both are just the neutralization product of formic acid. Carbon
Dioxide has to be produced in both of them.
There are a number of ways to form sodium formate, but only a one or so that I know of to form ammonium formate......so, what would Sodium Formate
form as reaction products? I think Sodium Hydroxide is out of the question. I have read well over 30 papers on the use of Sodium Formate as a transfer
hydrogenation agent and none of them mention anything about Sodium Hydroxide, Carbonate, or Bicarbonate, being reaction products, but a few did
mention CO2. The lack of this is probably because the aqeuous layer is just separated and tossed out, but it still leaves the question unanswered.
Edit:
On a side note, does Sodium Formate really offer two hydrogen in a transfer reaction? I was under the impression it only gave up one, for many months.
I haven't found any source claiming it can transfer two.
[Edited on 18-6-2014 by FireLion3]HgDinis25 - 18-6-2014 at 04:25
Well, the closest thing I can remember that reseambles what you're saying is the solvation of an electron to form an anion. For instance, Lithium can
dissolve in anhydrous ammonia, donatinhg one electron to form Li+. The electron doesn't have anywhere to go so it remains in solution as an anion
(e-).
Guess it depends on what is considered as 'isolated'.
According to maxwell electric monopoles can and do exist. I recall someone purposing a thought experiment about spinning a tube containing an
electrolyte at high speeds in a high magnetic field to drive the cations and anions apart in a statistically significant way. Could then the tube be
cut in half?
In the example given in the OP, why wouldn't a hydroxide anion be involved in the mechanism to create sodium hydroxide? Or could you be most explicit
on the supposed mechanism? Perhaps with a primitive bond dissociation energy assessment of each step.thesmug - 18-6-2014 at 10:23
In the case of solvated electrons you have to keep your anions and electrons in solution. I don't know what happens if you evaporate the solvent. You
most likely get an anion hydride in the case of water.
[Edited on 6/18/14 by thesmug]DraconicAcid - 18-6-2014 at 10:32
In the case of solvated electrons you have to keep your anions and electrons in solution. I don't know what happens if you evaporate the solvent. You
most likely get an anion hydride in the case of water.
In liquid ammonia, you will get your alkali metal back (unless it reacts to give an alkali metal amide, which I believe requires a catalyst such as
iron). You can't form solvated electrons or hydride ions in aqueous solution- you only get metal hydroxides.
