woelen
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Theoretical question: reaction mechanism of electrolysis of water
I am studying electrolysis of water with different compounds dissolved in it. As most of us know, a solution of H2SO4 in water gives H2 at the cathode
and O2 at the anode in a 2 : 1 ratio.
What I am wondering about is the mechanism of reaction. At the cathode I imagine that a H(+) ion receives an electron and an atom H. is formed. Two of
these combine to H2.
At the anode, things are more difficult. I have some explantion, but it leaves gaps:
H2O - e --> H2O(+) --> H(+) + OH.
2OH. --> H2O + O.
2O. --> O2
But if this were the mechanism, then I also would expect the following to happen:
2OH. --> H2O2
The solution, however, does not contain any hydrogen peroxide. Even the very sensitive reaction with dilute dichromate does not show any blue color.
Is there someone over here, who has some insight in the mechanism behind the simple net reaction at the anode, where
2H2O - 4e --> 4H(+) + O2
This is just a net equation, and definitely does not reflect the real sequence of events.
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Nick F
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Strange, I'm SURE that I've read that electrolysis of aq. H2SO4 can be used to produce hydrogen peroxide.
Perhaps under most conditions the H2O2 is only a short-lived intermediate, and that conditions must be tuned precisely in order to make useable
amounts of H2O2.
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chemrox
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First, try looking at the correct ion form. H+ doesn't exist. H3O+ is the form you should consider. This is Chem 1 stuff I'm sure you forgot.
now with acid you's have 4H3O+ + 4e -> 4H2 + 4OH-
I think this is right.
at the anode: 4OH- -4e -> 2H2O2 (?) (missing data is energy)
I dunno guys but it looks like it balances but does it happen this way?
"When you let the dumbasses vote you end up with populism followed by autocracy and getting back is a bitch." Plato (sort of)
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microcosmicus
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The reaction
2 OH- -> 2 e- + H2O2
would have to happen at the anode to carry away the electrons. However, at the
anode, there are also plenty of protons being formed by the reaction
2 H20 -> O2 + 4H+ + 4 e-
so the first reaction has to compete with
OH- + H+ -> H20 .
In other words, a hydroxyl ion is only going to be able to form peroxide only
if it is not neutralized before it makes it to the anode.
[Edited on 17-2-2008 by microcosmicus]
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12AX7
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Conversely, is free peroxide actually present in a sulfuric acid solution? If the equilibrium is strongly in favor of peroxysulfate type complexes
(perhaps not the most appropriate word to use), it might prevent formation of the peroxychromate complex.
Tim
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chemrox
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I think all the acid does is increase the ionization of the water and so speeds up the electrolysis as conventionally written. Sorry about the
anode/cathode mixup.
"When you let the dumbasses vote you end up with populism followed by autocracy and getting back is a bitch." Plato (sort of)
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microcosmicus
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| Quote: |
at the anode: 4OH- -4e -> 2H2O2 (?) (missing data is energy)
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We can work this out from other potentials. Looking at the table, I don't find this
reaction, but I do find the reaction
H2O2 + 2H+ + 2e- -> 2 H20
listed at +1.76V. From this, we can deduce the potential by considering
free energy. The reaction you are interested in could be the carried
out as the net result of the following reactions:
2 H20 -> H2O2 + 2 H+ + 2 e- (-1.76 V)
H2 + 2 OH- --> 2 H2O + 2 e- (+0.83 V)
H2 -> 2 e- + 2 H+ (0.00 V)
Remembering that, to get free energy, you multiply electrode potential
by the amount of charge moved and Faraday's constant, adding up the
free energies, then dividing by charge and Faraday constant, we get
the potential of the net reaction. In this case, everything happens to
involve two electrons, so we can take a short-cut and add the potentials
to obtain the grand whopping total of -0.94 V for the reaction
2 OH- -> H2O2 + 2e-
By contrast, the reaction
4 OH- -> 2 H2O + O2 + 4 e-
has a standard potential of only -0.40 V. Thus, it is also
favoured thermodynamically.
[Edited on 18-2-2008 by microcosmicus]
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woelen
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@chemrox: Of course I know about the H3O(+) ion (and even that is not correct, reality is much more involved), but I don't think that it does matter
here.
Do you really believe that the following happens: 4H3O(+) + 4e -> 4H2 + 4OH(-)? I would say a definite no to this. Remember, I am working in a
solution of H2SO4, and there certainly will not be hydroxide ions around near the cathode in significant quantities, let alone near the anode! Simple
reduction of H(+) (or H3O(+) if you wish) is much more likely to occur, so:
H(+) + e --> H. (or H3O(+) + e --> H2O + H.)
2H. --> H2
What happens at the anode is the subject of my question. I know the net reaction, but what I am interested in is the mechanism, which leads to the net
result.
@microcosmicus: You write H20 -> O2 + 4H(+) + 4e, but isn't this just a net equation? I don't think that such a reaction can occur at once,
because it would require at the same time transfer of 4 electrons. I think a process is involved, in which electrons are transferred one by
one, and the reaction products lead to oxygen and H(+).
I agree with you that the hydroxide ion discharge mechanism hardly happens in the strongly acidic medium around the anode.
Bottomline is that this question seems to be quite hard to answer  . I think that
any explanation, given here, must be through a seres of single electron transfer processes. And in strongly acidic medium I do not think that OH(-)
ions play a significant role. OH. radicals may be involved though.
[Edited on 18-2-08 by woelen]
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12AX7
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| Quote: | Originally posted by woelen
@microcosmicus: You write H20 -> O2 + 4H(+) + 4e, but isn't this just a net equation? I don't think that such a reaction can occur at once,
because it would require at the same time transfer of 4 electrons. |
Indeed, and more evidence is that anodic conditions are far more destructive than mere dioxygen would suggest. Oxygen is a powerful oxidizer, but not
so much when bonded to itself. So how does it come to be bonded to itself? There must be a reactive intermediate which causes corrosion of less
robust anodes, or bonds with itself in the absence of anything else.
Species with various numbers of electrons and labile atoms (e.g., hydrogens) added or missing would make sense, but I've never heard discussion of
electrolytic radicals before.
If radicals are responsible, then it would be possible to suggest, for instance, that alkanes dissolved in solution (with a PTC I suppose) could be
halogenated by anodic chlorine, for instance, or hydroxylated by oxo-species. Might this be one way to confirm the mechanism?
Tim
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microcosmicus
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Of course, it is a net equation as is the one for forming H2O2. Since my aim
was to calculate free energy changes, it didn't matter because of energy
conservation. From a thermodynamic view, I see a satisfactory explanation
of why the net reaction
4 OH- -> 2 H2O + O2 + 4 e-
is favored over
2 OH- -> H2O2 + 2e-
in that not only does the former reaction have lower energy, it is driven
forward by the escape of O2 as a gas. Whatever, the exact mechanisms,
if the conditions are present for both these processes to occur, the
invisible hand of thermodynamics will ensure that they come to a
dynamic equilibrium soon enough. Using Nernst's equation, we
obtain the equilibrium condition
4.1E36 [H2O2]^2 = [O2]
or
[H2O2] = 4.95E-19 sqrt [O2]
So unless the production of the H2O2 happens much faster than
the destruction because of large overpotential or some other
cause, the same mechanisms which turn OH- into H2O2 should
have no problem turning it back in the opposite direction. To say
more, I would need some data about the kinetics, which I do not have.
Having said that, I agree that the question of reaction mechanisms is
an interesting one.
One thing to look at is the hydorperoxyl radical. Looking at
electrode potentials, I notice the reactions
O2 + H+ + e - -> HO2 (-0.13 V)
and
HO2 + H+ + e- -> H2O2 (+1.51 V) .
Also, this radical acts as an acid with pKa 4.88:
HO2 -> H+ + O2-
It is known as an oxygen donor, leaving behind
a hydroxyl radical. (I see this mentioned fairly
often in connection with reactions which happen
in the atmosphere and combustion.)
Thinking of this suggests at some possible mechanisms
for the net reactions discussed above. I suppose that
listing the different radicals and ions of oxygen and
hydrogen, their reactions, and numerical data would
provide a good picture exactly what is going on
when one electrolyzes water. Also, this thinking
about the possibility of free radicals playing a role in
the reaction suggests some ideas like whether shining UV
light would have a noticeable effect by increasing
the concentration of radicals.
As mentioned above, seeing effects ascribable to radicals
would show that are involved in intermediate steps.
Another very good thing for sorting out exactly how the molecules
react to do would be to look up or measure the rate equations
of the net reactions one can observe, but that would involve
slowing down the chemistry with science
[Edited on 18-2-2008 by microcosmicus]
Thinking about this a bit more and looking up material, it occurred
to me that a good way of approaching this systematically would be
by organizing oxygen molecules as follows:
O2___O2-_ _O2-- ___2O-___2O--
____HO2__HO2-__2OH__2OH-
_________H2O2________2H2O
Molecules next to each other in a row are related by
removing an electron and molecules in a column by
adding a proton. Knowing the energies and rate constants
for all these molecules would allow one to figure out what is
going on.
[Edited on 19-2-2008 by microcosmicus]
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garage chemist
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When you electrolyze dilute H2SO4, H3O+ ions (aqueous solutione never contain free H+ ions) are discharged at the cathode, giving H2 and 2 H2O.
Since this is an acidic solution, essentially no OH- ions occur in there, and so there must be something else happening at the anode.
At the conditions you have in your electrolysis setup, this is the oxidation of water to O2 and 2 H3O+ (equalizing the loss of H3O+ at the cathode!).
For the production of peroxodisulfate at the anode, very specific reaction conditions have to be met- such as high concentration of the sulfate ion,
low temperature, and high current density at a platinum anode.
Industry uses a solution of 380-440g/L ammonium sulfate and 0-150g H2SO4 as the anolyte (ammonium because the peroxodisulfate can be precipitated by
cooling) and up to 1A/cm2 at the platinum anodes. You see, the synthesis of peroxodisulfate doesn't even need an acidic anolyte (but it always needs a
diaphragm).
Hydrolysis of the peroxodisulfate to peroxomonosulfate or H2O2 does not occur readily- this would require acidifying and heating further down the line
if H2O2 is to be manufactured.
On the other hand, when you electrolyze a NaOH solution with platinum electrodes, tiny amounts of H2O2 can be formed at the anode because OH- ions are
discharged there and may join together, contrary to the H2SO4 electrolysis.
This is the reason why NaOH solutions are not very suitable for use in gas coulometers- they give slightly less oxygen than water electrolysis is
supposed to give, noticeable when doing precise measurements.
[Edited on 19-2-2008 by garage chemist]
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Magpie
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Perhaps at the anode the following reactions occur:
2OH- ----> H2O + O(-2)
O(-2) ----> O + 2e-
Thereby no peroxide is present.
I know there won't be much OH- in an acidic solution but there is not much in water either.
[Edited on by Magpie]
The single most important condition for a successful synthesis is good mixing - Nicodem
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Pulverulescent
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'Just a reminder, "electrolytic peroxide" is produced by hydration of peroxydisulfuric acid. . .
H2S208 + H20--->H2S04 + H2S05
H2S05 + H20--->H2S04 + H202
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