Sciencemadness Discussion Board - Doubts about H2O2 reduction potential

Doubts about H2O2 reduction potential

Metallus - 22-2-2015 at 18:35

Looking at standard reduction potentials, the following reaction has a standard reduction potential of 1,78 V

H₂O₂ + 2H⁺ +2e^- <=> 2H₂O

which would make H₂O₂ a formidable oxidant. Yet, whenever I have an acidic solution of H₂O₂, most of the time what happens is that the H₂O₂ gets oxidised to H₂O and O₂, even when reacting with supposedly "weaker" oxidants.

When I put two oxidants together, what I expect is that they either don't react or that the stronger oxidant oxidizes the weaker one, IF they can further be oxidized. In the case of H₂O₂, any mild oxidant will do the job.

Why? According to that table, H₂O₂ should be a monster of oxidant and yet it is used as a reductant/oxidant depending on the situation. Why such misleading behaviour?

Also, I used H₂O₂ back then to oxidize Pb²⁺ to PbO₂ and it actually worked. I had a solution of Pb(NO₃)₂ in diluite HNO₃ to which I added a solution of H₂O₂ and I actually saw the formation of a red solid (PbO₂) which was accompained by the bubbling of what supposedly was oxygen. But this made little sense to me. If indeed the H₂O₂ had oxidized Pb(II) to Pb(IV), then H₂O₂ should have been reduced to water and not oxygen. This is what should have happened:
H₂O₂ + 2H⁺ +2e^- <=> 2H₂O
Pb²⁺ +2H₂O <=> PbO₂ +4H⁺ +2e^-
----------------------------------
H₂O₂ + Pb²⁺ ---> PbO₂ + 2H⁺
but instead the reaction was accompained by the bubbling of oxygen, as if the freshly formed PbO₂ was oxidizing the H₂O₂ to O₂, what the fuck?

Moreover, the only time I saw H₂O₂ acting as a "true" oxidant was when it was in basic solutions and not acid ones.

Can you shed some light on the ignorant one? Thanks

Oscilllator - 22-2-2015 at 21:43

Perhaps what happened with the lead oxidation is that some of the hydrogen peroxide was decomposed by the lead dioxide. This could give the appearance of the reaction forming oxygen, even though it really didn't.

Although I don't know what's happening with regards to H2O2 behaving in as either an oxidant or a reductant depending on the situation I can tell you that H2O2 oxidises Fe(II) to Fe(III) in acidic solutions, so it definitely can act as an oxidant under acidic conditions.

j_sum1 - 22-2-2015 at 22:10

I am with you on that one Metallus.
Just a couple of hours ago I was demonstrating the production of Cl2 and Br2 from NaCl and NaBr. I was using excess peroxide acidified with sulfuric acid in both cases. I checked the reduction potential table just before hand and reasoned that H2O2 should be more than adequate for the job. Great result on the Br2 but nothing much to speak of with the Cl2. I ended up throwing a spatula of KClO3 into the mix to get things going. Even though ClO3- --> Cl2 has a reduction potential of 1.49 compared with peroxide's 1.78. Hardly a fair test since the chlorate produces its own chlorine gas as well. I could have used permanganate (1.51V) except that I didn't want any colouration.
So I echo your question -- why is the actual performance of H2O2 as an oxidant so much less than what would be expected looking at the published values?
(If I was to hazard a guess at the answer I would go for prohibitive kinetics and that the addition of a catalyst would bring the peroxide into line.)

blogfast25 - 23-2-2015 at 07:35

I hope the following will clarify it, by means of an example.

In alkaline conditions, H2O2 oxidises Cr(III) to chromate (CrO42-, Cr(VI)) easily.

And yet that same Cr(VI), in acid conditions (it is then dichromate, Cr2O72-

H2O2 reduces Cr(VI) back to Cr(III).

What H2O2 does simply depends on pH.

H2O2 + 2 H+ + 2 e === > 2 H2O
2 H2O === > 2 H+ + 2 OH-
---------------------------------------- (sum)
H2O2 + 2 e === > 2 OH-, in alkaline conditions, acts as an oxidiser.

H2O2 === > O2 + 2 H+ + 2 e, in acid conditions, acts as a reducing agent.

It behaves also like that with Ce(III) to Ce(IV) which happens with H2O2 in alkaline conditions. But Ce(IV) is reduced back to Ce(III) by H2O2 in acid conditions.

[Edited on 23-2-2015 by blogfast25]

deltaH - 23-2-2015 at 08:08

Quote: Originally posted by blogfast25

I hope the following will clarify it, by means of an example.

In alkaline conditions, H2O2 oxidises Cr(III) to chromate (CrO42-, Cr(VI)) easily.

And yet that same Cr(VI), in acid conditions (it is then dichromate, Cr2O72-

Are you sure about this? The reaction

"H2O2 + 2 e === > 2 OH-, in alkaline conditions, acts as an oxidiser."

produces hydroxide, so I would imagine that this is favoured under acidic conditions where the concentration of hydroxide is low.

That same reaction can also be equivalently written as:

H2O2 + 2 e => 2 OH-
2H+ + 2OH- <=> 2H2O (water dissociation)
------------------------------------------(sum)
H2O2 + 2e- + 2H+ => 2H2O

Which makes it clear that this is favoured under acidic conditions.

Similarly, "H2O2 === > O2 + 2 H+ + 2 e, in acid conditions, acts as a reducing agent." I believe is favoured under alkaline conditions.

This is why hydrogen peroxide decomposes rapidly into oxygen and water under alkaline conditions and is reasonably stable under acidic conditions (free of catalysts).

blogfast25 - 23-2-2015 at 08:37

deltaH:

Both assertions are based on direct, multiple observations with Cr(III)/Cr(VI) and Ce(III)/Ce(IV) oxidations/reductions with hydrogen peroxide. Verify them by all means.

The preparation of chromate with peroxide has that drawback that if there's any peroxide left after the oxidation, it's part reversed when you acidify!

How alkaline is alkaline and how acidic is acidic? How long is a piece of string?

Having said all that, H2O2 works also as an oxidiser in acid conditions, see e.g. Fe(II) to Fe(III). It's the overall cell potential that must be > 0 for anything to work.

[Edited on 23-2-2015 by blogfast25]

Metallus - 23-2-2015 at 17:06

Quote: Originally posted by blogfast25

Yes, that is indeed what I observed sperimentally:

Quote:

Moreover, the only time I saw H2O2 acting as a "true" oxidant was when it was in basic solutions and not acid ones.

But then, how do you explain the fact that the reduction potential table lists H2O2 in acidic conditions as one of the best oxidants while the basic version is lower on the table and supposedly not even strong enough to oxidize Cr III to Cr VI? It's like it was reversed!

blogfast25 - 23-2-2015 at 17:31

Quote: Originally posted by Metallus

For starters, see if you can find the half potentials for the reduction (as an oxidiser it gets reduced!) for H2O2 in acid and alkaline conditions and compare.

Metallus - 23-2-2015 at 17:50

Quote: Originally posted by blogfast25

For starters, see if you can find the half potentials for the reduction (as an oxidiser it gets reduced!) for H2O2 in acid and alkaline conditions and compare.

[Edited on 24-2-2015 by Metallus]

blogfast25 - 23-2-2015 at 18:06

Metallus:

Very nice, ta.

The case of Cr(III) may be a bit of an anomaly. I know that despite listed values for reduction potentials, aqueous Cr(III) is very difficult to REDUCE to Cr(0). This may be due to complexation (Cr(III) is a prolific complex former). Perhaps this explains also why Cr(III) is difficult to OXIDISE in ACID conditions? In alkaline conditions it's mainly Cr(OH)4-, maybe that helps?

It's an interesting point you raised.

[Edited on 24-2-2015 by blogfast25]

j_sum1 - 23-2-2015 at 18:18

From Wikipedia http://en.wikipedia.org/wiki/Standard_electrode_potential_(d...

H2O2 as a reducing agent
H2O2 --> O2 + 2H+ + 2e- -0.70 volts

H2O2 as an oxidising agent
H2O2 + 2H+ + 2e- --> 2H2O 1.78 volts

Now what?
Why did my bromide to bromine oxidation work (-1.066V) and my chloride to chlorine not work (-1.36V).

Both were highly acidic solutions with a feisty amount of conc H2SO4 added to 6% H2O2.
What am I missing?

@blogfast25
[I follow deltaH's reasoning on the H+ equilibrium -- that is that the first reaction I have written is favoured in alkaline conditions and the second in acidic. I will also concede that you probably have empirical evidence on your side with Cr and Ce. But I don't understand why this should be the case. And I also fail to understand why Br and Cl should behave so differently. Are you saying I should have dumped some sodium hydroxide in there instead of sulfuric acid? I don't get that. The reaction needs the H+. Tell me how confused I am. It wouldn't be the first time with redox (this week).]

Edit
[tried to make] link work. [That's just annoying!]
And I see trhat metallus and blogfast have continued discussion in the cross-post.

[Edited on 24-2-2015 by j_sum1]

[Edited on 24-2-2015 by j_sum1]

blogfast25 - 23-2-2015 at 18:25

J_sum1:

I haven't go time to address your entire post right now but where you write:

Quote:

And I also fail to understand why Br and Cl should behave so differently

That one's easy: bromide is easier to oxidise to bromine than chloride to chlorine, see the relevant oxidation potentials, which I don't have at my fingertips right now (I prefer my CRC to Wiki for that kind of data).

BUT, from memory, H2O2 SHOULD be able to oxidise chloride to chlorine, though. IF memory serves me right... (note caveat, please)

My point about the oxidation/reduction of Cr(III) with H2O2 you'll also find in textbooks. I have one that explains what I observed and how I explain it too.

[Edited on 24-2-2015 by blogfast25]

j_sum1 - 23-2-2015 at 18:31

That's alright. Take your time. I have other things to do too.
I agree: Chlorine should be possible. Both reactions are on the same side of the magic 1.78V figure. That's why I jumped at a kinetic explanation. In practice, no appreciable amount of Cl2 was produced in a time frame of a few minutes. Throwing some chlorate in on top of the H2O2 got the job done quickly enough that the demo wasn't a disaster.

blogfast25 - 23-2-2015 at 18:33

Quote: Originally posted by j_sum1

That's alright. Take your time. I have other things to do too.
I agree: Chlorine should be possible. Both reactions are on the same side of the magic 1.78V figure. That's why I jumped at a kinetic explanation. In practice, no appreciable amount of Cl2 was produced in a time frame of a few minutes.

I suggest to repeat your chloride to chlorine with H2O2 oxidation experiment. It should indeed work. Take close notes of what you do.

[Edited on 24-2-2015 by blogfast25]

DJF90 - 24-2-2015 at 04:44

Chlorine should be generated from the Cl- and acidified peroxide; the Ecell for this is +0.40 V (cf. 0.69 V for the analogous oxidation of Br-). This only applies to standard conditions, and it may be worthwhile to account for differences in concentrations by using the Nernst equation.

As for the comments on the redox chemistry of hydrogen peroxide, please be aware that there are four half equations involved.

Under acidic conditions:
H2O2 + 2H+ + 2e- => 2H2O (oxidant, +1.776 V)
O2+ 2H+ + 2e- => H2O2 (reductant, +0.695 V)

Under basic conditions:
O2 + 2H2O + 2e- => H2O2 + 2OH- (reductant, +0.146 V)
HO2- + H2O + 2e- => 3OH- (oxidant, +0.878 V)

That last equation may be written more familiarly as H2O2 + 2e- => 2OH-

I guess the only way to determine whether hydrogen peroxide acts as an oxidant or reductant under given conditions would be to calculate delta G = -nFE for the two different scenarios (in acid or base, as appropriate) and see which one is more favourable.

blogfast25 - 24-2-2015 at 05:45

Quote: Originally posted by DJF90

Under acidic conditions:
H2O2 + 2H+ + 2e- => 2H2O (oxidant, +1.776 V)
O2+ 2H+ + 2e- => H2O2 (reductant, +0.695 V)

Under basic conditions:
O2 + 2H2O + 2e- => H2O2 + 2OH- (reductant, +0.146 V)
HO2- + H2O + 2e- => 3OH- (oxidant, +0.878 V)

That last equation may be written more familiarly as H2O2 + 2e- => 2OH-

I guess the only way to determine whether hydrogen peroxide acts as an oxidant or reductant under given conditions would be to calculate delta G = -nFE for the two different scenarios (in acid or base, as appropriate) and see which one is more favourable.

Already done in a sense, no? Assume the Eox of the species to be oxidised then the cell potential for the oxidation with H2O2 becomes:

E = 1.776 + Eox in acid conditions and E = 0.876 + Eox in basic conditions. So acid conditions always trump (no matter the value of Eox, assuming E < 0 of course) and yet Cr(III) is NOT oxidised to Cr(IV) acid conditions but it is in alkaline ones.

Possible causes may be that Cr(III) is complexed strongly and that in acid conditions dichromate (not chromate) would form?

[Edited on 24-2-2015 by blogfast25]

DJF90 - 24-2-2015 at 06:40

Its just an acid/base dependant redox reaction:

In acid:
Cr2O7 2- + 14H+ + 6e- => 2Cr 3+ + 7H2O +1.33 V

Whilst in base:
CrO4 2- + 4H2O + 3e- => Cr(OH)3 + 5OH- -0.13 V

Coupled with the appropriate H2O2 half equations above, I'm sure you'll find everything is kosher.

blogfast25 - 24-2-2015 at 13:10

Yep. That explains it.

j_sum1 - 24-2-2015 at 16:09

Nice to know that my understanding was not waay off. (Unlike my previous gaffe!) I can still read a table of reduction potentials and make sense of it.

I would like to know a bit more about Cl2 production and Br2 production and H2O2 as an oxidant. Let me give details of what I did.

This was done as a demo -- outside on a scale to be viewable by 60 people. Amounts of NaCl and NaBr were calculated sufficient to make just over 1L of gas/vapour at STP. (Gasses bubbled through a bucket of NaOH for safety. Windy day, audience upwind.)
Reaction was done in half litre conical flasks which should have been sufficient to allow visibility of gas to spectators.
6% H2O2 was used -- none stronger available at the moment. 20% excess calculated.
Acid was conc H2SO4, 50% excess added to the peroxide so both could be dumped into the flask together.

NaCl did show some bubbling -- faintly. I could not be certain that this was indeed Cl2. I always suspect O2 when I am dealing with peroxide. In any case, it was insufficient to be visible at a distance. An attempt at the same thing the previous day at test tube scale also produced some minor bubbling. I didn't smell it. There probably wasn't enough acid in that one so not a complete test. I scaled up to flask scale merely trusting the redox potentials.

I have used KClO3 and KMnO4to oxidise Cl- to Cl2 before. Rather effective. My question is why H2O2 is so much less effective. Reduction potentials suggest it should be more effective.
Cl2 --> Cl- (deltaV=1.36)
CLO3- --> Cl2 (deltaV=1.49)
MnO4- --> Mn2+ (deltaV= 1.51)
H2O2 --> H2O (deltaV=1.78)

My thoughts again turn to kinetics. Faint bubbling indicates that there is enough energy to drive the reaction, albeit slowly.
Is there a high activation energy at play here? How much does the dissociation of H2O2 come into play? Is the relatively low concentration of peroxide a factor?

[edit -- get the reaction direction correct you idiot]

[Edited on 25-2-2015 by j_sum1]

blogfast25 - 24-2-2015 at 16:23

j_sum1:

Personally I have no explanation. I assume, without verification, that your calcs are correct.

How did you expect to produce 1 l of bromine at RT though?

j_sum1 - 24-2-2015 at 16:36

Even if incorrect the same conditions were in place for Br and Cl.

1L of Br if it converted to vapour (which a lot of it did). IOW, equivalent molar quantities of Br2 and Cl2: 0.05 moles of each.
RT in Australia can be reasonably warm. About 30°C yesterday which is enough to make Br2 quite volatile.

blogfast25 - 24-2-2015 at 16:43

So the Br generation worked?

j_sum1 - 24-2-2015 at 17:37

My students are right here they will tell you what they saw. But yes, it worked.

Quote:

Hey, how you doing (laughs)
It went red -- very scientific terms. What I will be like one day. (I'm like a minus international hazard. I'm going to shut up now.)
Red brown vapours went through the tube and bubbled into the bucket.

Welcome to our class Blogfast

blogfast25 - 24-2-2015 at 17:59

Well, it remains a mystery why the Cl generation failed. I wish I could try it myself but health problems make it impossible, for now.

Try HCl + H2O2 in a test tube?

Molecular Manipulations - 24-2-2015 at 20:34

Strange, based on my calculations, and on the water present and the temperature, only 7% should have dissolved assuming chlorine's solubility in a sodium sulfate solution is the same as water - my guess is its slightly more soluble.
Since the container was full of air, and chlorine's color is very faint, I'm not too surprised it wasn't noticeable.
Even relatively pure chlorine is quite faint, but a 500 mL flask should show it.
Are you sure you didn't add more water?
This is a complete long shot, and by no means the only or best explanation but perhaps it is oxygen according to these reactions:
Cl₂ + H₂O --> HOCl (aq) + HCl (aq); HOCl + H₂O₂ --> H₂O + HCl (aq) + O₂
I doubt these reactions can happen fast enough, and why would they happen with chlorine but not bromine?

[Edited on 25-2-2015 by Molecular Manipulations]

j_sum1 - 24-2-2015 at 21:34

I think it was just slow. I forgot to mention the peroxide was chilled. We keep it in the fridge.
I tried H2O2 and HCl in a test tube and got similar results -- a slow effervescence. No yellow-green gas was visible in the first few minutes unless it was the product of wishful thinking. I did the same with HCl and KMnO4 and there was some visible gas evolution. Nothing measured so difficult to draw a direct comparison but it seemed the KMnO4 was faster. I had to go to a sport lesson so I put bungs on the tubes and left them in the fume cupboard. When I came back the H2O2 tube had blown its bung. The KMnO4 was full of greenish gas.

So, in summary, I guess it works but it is slow. That might be a good thing in certain contexts but wasn't what I needed for the demo.

blogfast25 - 25-2-2015 at 09:57

Quote: Originally posted by Molecular Manipulations

Cl₂ + H₂O --> HOCl (aq) + HCl (aq); HOCl + H₂O₂ --> H₂O + HCl (aq) + O₂
I doubt these reactions can happen fast enough, and why would they happen with chlorine but not bromine?

I doubt if these reactions do occur at any appreciable rate but even if they did, you can't just 'transpose' these on analogous reactions with bromine: the free energies and enthalpies won't be the same.

blogfast25 - 25-2-2015 at 10:06

I remember carrying out a titration of Sn(II) with KMnO4. The Sn(II) was obtained by reducing Sn(IV) with Al chips and HCl, the reduced solution then diluted to in a volumetric flask and 20 ml aliquots titrated with standardised KMnO4. I obtained the anticipated result.

Later it dawned on me that the chloride could potentially have interfered with the KMnO4. But that didn't happen because the oxidation of Sn(II) to Sn(IV) with KMnO4 was so much faster than the oxidation of acidic chloride with KMnO4.

But If I did it again, I would use sulphuric acid during the preliminary reduction of Sn(IV) to Sn(II).

Molecular Manipulations - 25-2-2015 at 10:14

Yeah, ΔG° of HBr (g) is -53.1 (kJ/mol) and HCl (g) is -94.4 I couldn't find HBr (aq) but whateva.
Since the hydrogen peroxide came out of the fridge (assuming 1.6C°), solubility is slightly greater, about 8.9% of total chlorine should have stayed in solution, so that explains a little more.
But again like both of us said, the reaction seems too slow for that to be a major cause.
Of course the ΔG° for the isolation of bromine by this reaction: H₂O₂ + H₂SO₄ + 2NaCl → Na₂SO₄ + 2 H₂O + Br₂ is slightly smaller (read: more favorable) than the analogous reaction for chlorine.
And since ΔG° is proportional to the K_eq of this (and any) reaction, more bromine will be isolated than will chlorine under the same conditions.
But this is nothing new, and sheds almost no light on the problem I think.

[Edited on 25-2-2015 by Molecular Manipulations]

blogfast25 - 25-2-2015 at 12:55

Quote: Originally posted by Molecular Manipulations

Yeah, ΔG° of HBr (g) is -53.1 (kJ/mol) and HCl (g) is -94.4 I couldn't find HBr (aq) but whateva.

HBr(aq) is almost completely dissociated.

The ΔG of Formation of HBr(aq) is thus that of H3O+(aq) + Br-(aq).

Always be careful with ΔGreaction calculations: the ΔGformation of the reagents and reaction products have to be those of the species in the particular State they occur in the process. Often overlooked, that...

Molecular Manipulations - 25-2-2015 at 12:57

Ah, I see. ΔG° of H₃O⁺ (when balanced with a negative ion) is the same as water, and thus makes no difference correct?

I thought the ΔS° would be lower as a liquid has less enthalpy.

[Edited on 25-2-2015 by Molecular Manipulations]

blogfast25 - 25-2-2015 at 13:21

Quote: Originally posted by Molecular Manipulations

Ah, I see. ΔG° of H₃O⁺ (when balanced with a negative ion) is the same as water, and thus makes no difference correct?

I thought the ΔS° would be lower as a liquid has less enthalpy [edit: did you mean entropy? Blog].

It's a bit more confusing than that!

Remember that for H+ + e === > 1/2 H2(g) the Ered = 0. This means consequently that the Free Energy of Formation of H+ must be 0 because conventionally we set the Free Energy of Formation of elements (like H2

to 0! Note that Ered = 0 is also conventional, for the hydrogen electrode...

From an electrochemical series PoV, the free energy of formation of HBr(aq) is equal to the energy of formation of the solvated bromide ion, Br-.

Try and use ΔGformation for reagents and reaction products where possible, they already account for entropies.

[Edited on 25-2-2015 by blogfast25]

Metallus - 17-4-2018 at 04:32

Sorry for the necro, but I finally found an answer to this question that has been tormenting me for years, so I wanted to share it with you who joined the conversation.

This subject was tackled already in the 1934 by Bancroft and Murphy https://pubs.acs.org/doi/pdf/10.1021/j150363a006 who were puzzled by the multitude of behaviors of H2O2 acting as a reductant or oxidant. They found that the potential of 1.78 V reported in literature is actually bogus and introduced the concept of true electromotive force that for H2O2 in acidic medium is ca 1.15 V.

Moreover they added that H2O2 acts as an oxidizer when its true electromotive potential is higher than the other reactant and acts as a reductant when its true electromotive potential is lower than the other reactant, provided that a redox reaction can occur.

They also addressed the different reactivity in basic solutions where some redox reactions are actually reversed and measured a turn-over pH at which nothing happens.

The article is very savage and shits all over previous findings. It was a very fun and interesting read, beside giving me the answers I was looking for. It bothers me how I didn't find this before.

Enjoy

Sulaiman - 17-4-2018 at 05:49

Thanks for the necro

This subject was tackled already in the 1934 by Bancroft and Murphy https://pubs.acs.org/doi/pdf/10.1021/j150363a006

I can't download the pdf, could you post a copy please ?

Metallus - 17-4-2018 at 05:59

Quote: Originally posted by Sulaiman

Thanks for the necro

This subject was tackled already in the 1934 by Bancroft and Murphy https://pubs.acs.org/doi/pdf/10.1021/j150363a006

I can't download the pdf, could you post a copy please ?

https://we.tl/5HcL8WYb1w
Let me know if you know other more permanent upload websites for this lightweight stuff :c

RawWork - 17-4-2018 at 06:45

Honestly, I am not worried. I don't see anything unusual here. I doubt that any of you used 1 M H2O2, and 1 M other reactant. Also be aware that multiple or different reactions can happen than what you expect. Only for ammonium or sulfate or peroxide there are many possible reduction reactions for each of that ion. Also I found somewhere successful production of hydrogen peroxide using alkaline and acidic water...everything went as planned.

Quote: Originally posted by Metallus

Let me know if you know other more permanent upload websites for this lightweight stuff :c

https://www.scribd.com/

[Edited on 17-4-2018 by RawWork]

Metallus - 17-4-2018 at 07:08

Quote: Originally posted by RawWork

Honestly, I am not worried. I don't see anything unusual here. I doubt that any of you used 1 M H2O2, and 1 M other reactant. Also be aware that multiple or different reactions can happen than what you expect. Only for ammonium or sulfate or peroxide there are many possible reduction reactions for each of that ion. Also I found somewhere successful production of hydrogen peroxide using alkaline and acidic water...everything went as planned.

Quote: Originally posted by Metallus

Let me know if you know other more permanent upload websites for this lightweight stuff :c

https://www.scribd.com/

[Edited on 17-4-2018 by RawWork]

The unusual thing is that H2O2 with such a high reported reduction potential (1,78 V) should never be oxidized by reportedly weaker oxidizers and at the same time it should instead oxidize a lot of things that in reality do not get oxidized.

The introduction of that article thoroughly goes through all the reactions where H2O2 behaves in both ways, citing that some reactions may even be reversed when changing the concentration or the pH. It cites a lot of previous works and actively contradicts many results found previously by other researchers.

The thing that cleared my confusion is their reported red. potential of H2O2 of 1.15 V, which is lower than other common oxidizers and explains why H2O2 gets oxidized in those conditions (whereas the same reactions didn't make sense by taking into account the 1,78 V reported in literature).

It also analyzes the case of lead(II) oxidation to lead(IV) and why oxygen is released anyways and how the reaction rate changes with pH.

You say that only for ammonium, sulphate or peroxide there are many reactions, but you didn't say why. What I wanted to know was the reason as to why they happen despite theory stating the opposite, and that paper gives reasonable answers.

RawWork - 17-4-2018 at 07:52

I am not deeply into that stuff now, but I am trying to say that generally there are lot of optical illusions. Not is everything as it seems. Not only in electrochemistry or chemistry. But in other sciences, topics, life. Consider Fata Morgana and other optical phenomenas which are proven by science. Not to talk about hallucinations which can't be proven by science.

In electrochemistry there are at least 4 explanations which may answer common problems. Presence of other chemicals (like oxygen in water or hydrogen peroxide), intermediate products of reduction, nonstandard situation (concentration, temperature, pressure), overpotential.

In water they may be acid or base, which includes additional element that forms cation or anion such as sodium or sulfur (sulfate), then water itself, then gases from atmosphere. You have to consider all this. I doubt that values are wrong. These were taken from CRC Handbook of Chemistry and Physics 2016-2017.

Possible reactions for peroxide:
HO₂ + H⁺ + e ⇌ H₂O₂ 1.495 V
H₂O₂ + 2 H⁺ + 2 e ⇌ 2 H₂O 1.776 V
O₂ + 2 H⁺ + 2 e ⇌ H₂O₂ 0.695 V
O₂ + H₂O + 2 e ⇌ HO₂^- + OH^- -0.076 V
O₂ + 2 H₂O + 2 e ⇌ H₂O₂ + 2 OH^- -0.146 V
HO₂^- + H₂O + 2 e ⇌ 3 OH^- 0.878 V
OH + e ⇌ OH^- 2.02 V

There are many possible reactions for oxygen, hydrogen and hydronium, hydroxide...which would take long time to write. All possible has to be considered.