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j_sum1
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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.
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blogfast25
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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?
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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.
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blogfast25
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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).
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Molecular Manipulations
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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 Keq 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]
-The manipulator
We are all here on earth to help others; what on earth the others are here for I don't know. -W. H. Auden
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blogfast25
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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 ΔG<sub>reaction</sub> calculations: the ΔG<sub>formation</sub> 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...
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Molecular Manipulations
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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]
-The manipulator
We are all here on earth to help others; what on earth the others are here for I don't know. -W. H. Auden
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blogfast25
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| 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].
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It's a bit more confusing than that!
Remember that for H<sup>+</sup> + e === > 1/2 H<sub>2</sub>(g) the E<sub>red</sub> = 0. This means consequently
that the Free Energy of Formation of H<sup>+</sup> must be 0 because conventionally we set the Free Energy of Formation
of elements (like H<sub>2</sub> to 0! Note that
E<sub>red</sub> = 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<sup>-</sup>.
Try and use ΔG<sub>formation</sub> for reagents and reaction products where possible, they already account for entropies.
[Edited on 25-2-2015 by blogfast25]
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Metallus
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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
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Sulaiman
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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 ?
CAUTION : Hobby Chemist, not Professional or even Amateur
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Metallus
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https://we.tl/5HcL8WYb1w
Let me know if you know other more permanent upload websites for this lightweight stuff :c
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RawWork
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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.
https://www.scribd.com/
[Edited on 17-4-2018 by RawWork]
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Metallus
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| 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.
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.
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RawWork
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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.
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