White Yeti
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Creating free radicals.
Hello all,
I spent a few days thoroughly researching the subject of free radicals and I was too shy to post this thread in the ever popular "chemistry in
general" category. I finally decided to give this a shot.
I've considered the use of free radicals as oxidisers in batteries for some time. The high reactivities (and thus high reduction potentials) of free
radicals, explain their fleeting existence and short half life.
But even though free radicals don't exist for long (with the exception of ClO2, NO2, Fremy's salt and the like) they can still participate in some
interesting redox chemistry before they disappear.
With that being said, I'm curious to know how many chemical reactions that yield radical products the SciMad community can come up with. There are two
conditions though, many radicals can be formed through photodissociation or the direct addition or removal of electrons:
Br2 ---hv---> 2Br·
CO2 + e- -----> CO2·- -1900mV
(just to name a few)
I don't have a grudge against those particular types of radical reactions, but it's a stupid idea to make a battery that needs an energy input to
function properly
I'll start off with two reactions I can write off the top of my head:
2H2O2---Fe(III)---> H2O +OH· +HO2·
S2O8(-2) --Fe(II)---> 2SO4·-
Before you ask, I did my research, I have a little more than three and a half pages of radical reduction potentials. It's just that I can't find many
reactions that yield radicals as products.
For convenience, the · symbol can be typed by pressing (alt shift 9).
Thank you for reading.
"Ja, Kalzium, das ist alles!" -Otto Loewi
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AndersHoveland
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I have often wondered if the fenton reaction (Fe ions + H2O2) actually produces hydroxyl radicals as the main route of oxidation. In older chemical
literature, for example, it was repeatedly claimed that Zn + HCl produced hydrogen radicals, and that this was the reduction mechanism. But this was
later shown to be false, no hydrogen radicals are produced. Could the same also be true for fenton chemistry?
Perhaps the Fe+3 first gets oxidized to a lower form of ferrate (or even to ferrate VI itself), then this ferrate is reduced back to Fe+3 by reaction
with more H2O2, liberating oxygen.
Fe[+3] + H2O2 --> (2)FeO4[-]
FeO4[-] + H2O2 --> Fe[+3] + O2
(equations are unbalanced)
Some of the ferrate may react with H2O2 to produce traces of ozone (this also happens in the reaction of permanganate with H2O2). The ozone will then
react with more H2O2, through the peroxone reaction, to form hydroxyl radicals. So although the fenton chemistry may produce traces of hydroxyl
radicals, it may be possible that the main oxidation mechanism involves ferrate, rather than hydroxyl radicals. Ferrate is still a powerful oxidizer,
although no quite to the degree that hydroxyl radicals are.
Reduction Potentials
Persulfate (acidified) 2.01v
Persulfate (heat or catalyst) 2.6v
(approximate value as the exact redox potential is difficult to determine, for the catalyst Fe+2 may be added immediately prior to the reaction,
which immediately is oxidized to Fe+3 and precipitates out )
Hydroxyl Radical 2.80v
There is some disagreement about the exact value for the reduction potential of the hydroxyl radical because of the complexities of measurement.
Bard (1985) reported 2.38v and Stanbury (1989) 2.72v
[Edited on 26-10-2011 by AndersHoveland]
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White Yeti
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Interesting, but isn't hydrogen peroxide a weaker oxidiser compared to ferrate (in acid)?
H2O2(aq) + 2 H+ + 2 e− ----> 2 H2O +1.78V
FeO4(-2) + 3 e− + 8 H+ ----> Fe3+ + 4 H2O +2.20V
In that case, how would the ferrate intermediate reaction be favoured over the accepted reactions? I see what you are saying, but that reaction just
looks unfavourable.
Here's the pdf where I got tons of reduction potentials. You might have consulted it already, but it doesn't hurt to mention it:
http://www.nist.gov/data/PDFfiles/jpcrd372.pdf
It has everything except the carbonate radical and a dozen pages on organic radical compounds. I'd suggest skipping ahead to page 1711 because that's
where all the inorganic radicals begin, but that's just me.
"Ja, Kalzium, das ist alles!" -Otto Loewi
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woelen
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I understood that the solution of H2O2, together with iron(II), gives a species, containing iron(IV), which then acts as a very reactive oxidizer. I
have described this in more detail on my website:
http://woelen.homescience.net/science/chem/solutions/fe.html
Look at the part of the page about oxidation state +4.
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White Yeti
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Thanks for the link woelen! I like your picture of ferrate. Is it the same one that appears on the wikipedia page?
On the subject of fenton's reagent though, when I mix an Fe(II) salt with hydrogen peroxide, I get the same colour described on your website, halfway
between yellow and red.
But in the past, I thought this was due to a mixture of Fe(II), (pale yellow) and Fe(III) (blood red), not due to iron(IV). I could look into this
further, but all I've gotten so far with fenton's reagent is an oxidising mess of bubbles and foam (maybe because of copper contamination).
I didn't take pictures of these colours, but I can if anyone in interested.
"Ja, Kalzium, das ist alles!" -Otto Loewi
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AndersHoveland
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Quote: Originally posted by White Yeti  | Interesting, but isn't hydrogen peroxide a weaker oxidiser compared to ferrate? how would the ferrate intermediate reaction be favoured over the
accepted reactions? I see what you are saying, but that reaction just looks unfavourable.
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Yes, indeed. H2O2 is a weaker oxidizer than ferrate. But that does not in itself mean that H2O2 could not oxidize iron ions to
ferrate.
Consider several of the different catalysts for the decomposition of hydrogen peroxide. Manganese dioxide, iron, nickel, silver, chloride. What these
all have in common is that higher oxidation states of each exist. Manganate, ferrate, NiO2 (nickel 'peroxide'), Ag2O2, and HOCl. All these can react
with H2O2 to liberate oxygen. What I am proposing is that H2O2 first oxidizes the catalyst, and that the oxidized intermediate then in turn reacts
back with more H2O2 in a cycle, the net effect being decomposition of the H2O2.
Reduction Potentials
Ferrate (acidified) 2.20v
Ferrate (alkaline) 0.72v
Perhaps the H2O2 is able to oxidize iron ions to ferrate ions, since the reduction potential of ferrate is only 0.72v, if the ion is left alone. But
ferrate is not stable in neutral aqueous solution, because there are inevitably hydronium ions (in equilibrium) floating around that will trigger the
decomposition of the ferrate. What I am saying is that both reduction potentials, 2.20 and 0.72, are valid in such a reaction where ferrate
is being simultaneously formed and decomposed. One experiment that could be done to test this idea is to see if the oxidation rate or efficiency of
H2O2 with a trace of Fe(OH)3 catalyst is significantly lower when plenty of NaOH is dissolved. If this idea is correct, the redox potential for the
main oxidation reaction should not be greater than 0.72v in the NaOH solution, although there will still be small traces of hydroxyl
radicals. One problem with this type of experiment would be that the outer layer of Fe(OH)3 particles might dissolve significantly more in the highly
alkaline H2O2 solution, which could increase the reaction rate relative to the neutral solution. If an another form of iron ions had been used (such
as Fe(NO3)3), the iron would precipitate out as Fe(OH)3 particles in the alkaline solution, but not the neutral solution, again affecting reaction
rates and making a comparison between the two solutions difficult.
This is, of course, only a theory, but the complexity may explain why it has not been described before. The traces of hydroxyl radicals in both
situations would obscure any obvious conclusion.
[Edited on 27-10-2011 by AndersHoveland]
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White Yeti
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I'm not disagreeing with you Anders, I'm just trying to poke holes into the theory to see if it still stands.
It just occurred to me, the reduction potential of ·OH is higher than both peroxide and ferrate...
It almost seems like you are violating a law of conservation of oxidisingness (if that makes any sense). How can mixing a weak reducing agent, FeII,
with a moderately strong oxidiser, H2O2, yield one of the strongest oxidisers known to man, ·OH? There must be a huge incentive for the peroxide to
decompose.
You mentioned that most catalysts can exist at higher [or lower] oxidation states.
Copper compounds also catalyse the decomposition of hydrogen peroxide. It seems like this decomposition pathway also yields radicals:
H2O2 <----> H+ + HO2-
HO2- + Cu+2 ----> HO2· + Cu+
HO2· <----> H+ + O2-
O2- + Cu+2 ----> Cu+ + O2
The problem with these reactions is that the Cu+ doesn't turn back into Cu+2, and therefore it is not a catalyst and behaves rather like a reagent.
Does anyone have a set of reactions, where CuII acts as a catalyst?
It doesn't yield ·OH radicals though, which are far more useful and interesting than HO2·. This reaction seems to proceed much more quickly than
Fe(III) in peroxide, which indicates that it is much more favourable. Unfortunately, it just seems like most of the peroxide is wasted.
"Ja, Kalzium, das ist alles!" -Otto Loewi
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AndersHoveland
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| Quote: Originally posted by White Yeti |
It just occurred to me, the reduction potential of ·OH is higher than both peroxide and ferrate. It almost seems like you are violating a law of
conservation of oxidisingness (if that makes any sense).
The problem with these reactions is that the Cu+ doesn't turn back into Cu+2, and therefore it is not a catalyst. It doesn't yield ·OH radicals
though. |
Yes, I understand what you mean about "oxidisingness". It might seem to defy common sense, but that is how some reactions work. One way to view this
is that oxidation by acidified H2O2 does not occur all at once, but rather in distinct steps, and this is why the maximum redox potential is lower.
This equation might solve your problem:
(2)Cu[+1] + H2O2 + (2)H[+] --> (2)Cu[+2] + (2)H2O
Yes, the reaction mechanism for copper ions decomposing H2O2 is probably different from the other catalysts. Copper-ammonia complexes with H2O2
produce the transient
[+]Cu-O-OH which can oxidize alkanes.
I'm not saying let's go kill all the stupid people...I'm just saying lets remove all the warning labels and let the problem sort itself out.
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