semiconductive
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Electro-sulfur reduction, color hints.
I'm experimenting with sodium sulphide nonahydrate.
I couldn't get pure Na2S crystals so I bought the nona-hydate. The crystals are quite white, with occasional yellowing (very slight) on one side.
The nona-hydrate appears to be well made, as it has almost no odor.
As a first experiment, I mixed the sodium sulphide with dry methanol to see if it would dissolve.
Na2S dissolves in MeOH, and it emits a mild sulphurout / egg-like odor.
I think it's possible the sodium sulfate is reacting with methanol, and producing some sodium alkoxide and H2S.
I'm not sure how much of the material is reacting, rather than simply physically dissolving, because I think methaol dissolves sodium alkoxide ---
meaning that the whole remains liquid regardless of how much reacts with MeOH.
Next:
I thought, I might try electrolysis to see if I could break down the water that the nona-hydrate carried with it.
I put a carbon fiber anode into the solution, and a tin coated wire as cathode.
The solution is a conductor and i'm getting around 20-40mA current flow at 12V applied.
I figure the carbon anode will not go into solution, and so it either has to break down the water and remove oxygen; or perhaps it will break down
H2S. From EPA literature, I think H2S is soluble up to ~.05 mole fraction in methanol at 1ATM.
I found this in a table. It's tiny compared to the breakdown of Oxygen in water ... but it's also not for dissolved H2S in methanol. But it
suggests reducing H2S to sulphur is easily done. I'm expecting reversing this reaction is going to be easier than breaking down water at the anode.
S + 2e + 2H --> H2S(g) at 0.13V.
When I run the experiment, I do not see oxygen bubbles (or CO2) bubbling off the carbon fiber in my experiment.
The result is that a yellow liquid started forming immediately at the anode, and gas bubbles release only at the cathode. I assume those are hydrogen
bubbles from either water breaking down, or from sodium plating out and then attacking methanol.
So ... my initial expectation is confirmed.
Over time, I think sodium will preferentially destroy water over alcohol as I think the O-H bond is weaker than the O-R bond in alcohol. So, am I
correct in assuming electrolysis ought to still remove water from solution?
I am thinking that (eventually) sodium hydroxide will release release oxygen at the anode once the concentration is great enough; I don't know if
sodium hydroxide will eventually be removed or not ... But ... I'm curious about the yellow color forming at the anode as a side reaction.
The yellow substance is not sticking to the carbon, but goes into liquid clear -- eg: no signs of colloidal cloudiness or of solid sulphur forming.
I'm not aware that solid sulfur is soluble in methanol, so I'm wondering what the yellow substance is.
From a bit of research, I found that most polysufides are red when bonded with H or alkali metals ... and that complete sulphur rings are supposedly
colorless;
I'g guessing that yellow color is generally caused by broken ends of a sulphur chain where an unterminated charge is temporarily present.
So I'm wondering -- What exactly causes polysulfide molecules to tend to be red?
eg: like when I heat pure sulfur, it reddens as well.
Would the yellow color I'm seeing suggest that sulphur is in elemental form with open ends (broken chain) in solution? ( or is it just as likely to
be an ion?)
eg: would H-S-S-+ cations or perhaps H-S-S-H (hydrogen di-sulfide?) -- H-S-S-R or H-S-R molelcules be red or yellow ?
(why/why not?)
EDIT: I just looked up H2S2, and it's pale yellow. I looked up H2S3 ... and it's supposedly yellow as well. So, I'm wondering why several pieces of
literature make the generalization that polysulfides are usually red.
As the reaction has continued, the solution is slowly turning an orange-ish color.
EDIT: New IDEA:
I've been studying the nernst equation, and I realize there is nothing in there which is specific to water. It seems to me that the standard
potential voltages may be different in alcohol, or other liquids (DES's, propylene carbonate, etc); but otherwise the principles of the nernst
equation shold still be correct.
Does anyone know if that's the case?
If so, it seems likely I could use the voltages of the cell (open circuit it's at 0.75V) to track how the mole concentration of products and reactants
is chaning over time.
[Edited on 9-3-2021 by semiconductive]
[Edited on 9-3-2021 by semiconductive]
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Bedlasky
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Sulfide is oxidized on anode to polysulfide. Polysulfide is yellow, orange, red - depending on concentration. H2S2, H2S3 surely doesn't form, they
aren't stable at room temperature.
Btw. sodium sulfide hydrolyses in solution in to sodium hydrosulfide, hydrogen sulfide and sodium hydroxide - that's reason why your solution smells
like rotten eggs.
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Maurice-VD-3
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At the anode, you get the reaction : S^{2-} -> S + 2 e-, which is immediately followed by the reactions S + S^{2-} -> S2^{2-} and S+ S2^{2-}
-> S3^{2-} , etc. up to S4^{2-} and S5^{2-} These ions are all yellow, or orange, and they are soluble in water and in methanol. The most stable is
S4^{2-} which can be pure obtained in the pure state as Na2S4.
[Edited on 9-3-2021 by Maurice-VD-3]
[Edited on 9-3-2021 by Maurice-VD-3]
[Edited on 9-3-2021 by Maurice-VD-3]
[Edited on 9-3-2021 by Maurice-VD-3]
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semiconductive
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Quote: Originally posted by Bedlasky  | Sulfide is oxidized on anode to polysulfide. Polysulfide is yellow, orange, red - depending on concentration. H2S2, H2S3 surely doesn't form, they
aren't stable at room temperature.
Btw. sodium sulfide hydrolyses in solution in to sodium hydrosulfide, hydrogen sulfide and sodium hydroxide - that's reason why your solution smells
like rotten eggs. |
Right ... I know that if I let the sodium sulfide nonahydrate get wet it hydrolyses and makes the smell.
Though it's curious that when I mixed dry methanol with the nona-hydrate it also released the smell, because the methanol was quite dry. So, I'm
thinking perhaps a metastatis reaction happened ... and 2 R-OH + Na2S <--> 2 R-O-Na + H2S ?
The yellow to orange color, I think is possibly Na2S2, and maybe Na2S3.
Na2S2 is CAS 2868-13-9
B.P is 70C, and freezing point is < -40C ; so that's stable at room temperature.
Note: During electrolysis, the egg odor stopped.
Overnight, the open circuit voltage went from 0.75V (open circuit, and dropping quickly) to a very stable 1.28Volts. It even is a reversable
reaction, for if I touch a red led to the cell terminals, it will light up for about 1 second. The self discharge rate is high, so it's not a good
battery ... but it's data I can use to figure out likely reactions.
I figure I can rig up an ARM microcontroller to control a volt-amp cycle measurement.
Then I can get quantitative measurements of what voltages cause what oxidations.
I'm very curious if the oxidations will happen at approximately the same voltages in water electrolyte, vs. methanol (which is very water like,
chemically), vs. acetone ; or if the electrolyte itself will shift the energy level of the oxidation reactions.
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semiconductive
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| Quote: Originally posted by Maurice-VD-3 | At the anode, you get the reaction : S^{2-} -> S + 2 e-, which is immediately followed by the reactions S + S^{2-} -> S2^{2-} and S+ S2^{2-}
-> S3^{2-} , etc. up to S4^{2-} and S5^{2-} These ions are all yellow, or orange, and they are soluble in water and in methanol. The most stable is
S4^{2-} which can be pure obtained in the pure state as Na2S4.
|
So, let's start at the top:
The Na2S is spontaneously dis-associating into ions in the methanol solution.
That means S^{-2} approaches a graphite electrode in a doubly reduced state.
I don't know how much energy it takes to remove the second electron from S--,
but I do know the energy for S^{-}.
S + e– → S– – ∆H = Affinity = 200 kJ/mol (gas or ion state of Sulfur).
200000 J/mol = 3.32e-19 J/atom = 2.07eV to oxidize the last electron.
So a single electron has to experience 2V of field in order to be removed from the sulfur.
I have a 12V supply ... so that's easily going to cause the reaction to happen.
And I know that it takes 10.36 eV to make S --> S+, which is also possible.
And I know that it takes 22.335 eV to make S --> S++, isn't possible from power supply alone.
So, I should only need to think about S-, S or S+ occurring at the anode.
Do you know where to find the free energy change for S --> S-- ?
There is thermal agitation of the S-- ions, which I should also take into account ...
but shouldn't I be able able to estimate a potential voltage level where the reaction is no longer possible (or excessively slow) at room temperature
?
eg: a voltage where the yellow color would not form appreciably.
And: When you talk about Na2S4 being the most stable ... how much energy difference is there between Na2S and Na2S4?
[Edited on 11-3-2021 by semiconductive]
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semiconductive
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I'm not sure what you mean by unstable .... They do decompose over time, I guess ...
But why would a slow decomposition preclude them being formed in the first place?
eg: how is it possible to distil H2S2 into a pure liquid without it completely breaking down unless it has a life-time that is on the order of minutes
or hours?
Inside a test tube (sealed), I suppose the pressure of my experiment is just a bit above 800 mmHg at 20C and I don't tend to leak H2S into the room.
Wouldn't that (by leChatlier's principle) slow down any decomposition of a reversible reaction?
https://en.wikipedia.org/wiki/Hydrogen_disulfide
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Bedlasky
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| Quote: Originally posted by semiconductive |
Right ... I know that if I let the sodium sulfide nonahydrate get wet it hydrolyses and makes the smell.
Though it's curious that when I mixed dry methanol with the nona-hydrate it also released the smell, because the methanol was quite dry. So, I'm
thinking perhaps a metastatis reaction happened ... and 2 R-OH + Na2S <--> 2 R-O-Na + H2S ?
The yellow to orange color, I think is possibly Na2S2, and maybe Na2S3.
Na2S2 is CAS 2868-13-9
B.P is 70C, and freezing point is < -40C ; so that's stable at room temperature.
|
Nonahydrate - water of crystallization is released in to methanol during dissolving (so methanol isn't anhydrous anymore).
What? Boiling point of Na2S2 70°C and freezing point -40°C? This is nonsense. Na2S2 is solid at RT.
But even if you meant H2S2, this isn't true. Polysulfanes are very sensitive and very easily decompose. They are prepared at subzero temperatures from
sodium polysulfide and HCl (and separated by fractional distilation).
https://books.google.cz/books?id=QvshwPCy1DkC&pg=PA101&a...
[Edited on 11-3-2021 by Bedlasky]
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semiconductive
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| Quote: Originally posted by Bedlasky | | Quote: Originally posted by semiconductive |
Right ... I know that if I let the sodium sulfide nonahydrate get wet it hydrolyses and makes the smell.
Though it's curious that when I mixed dry methanol with the nona-hydrate it also released the smell, because the methanol was quite dry. So, I'm
thinking perhaps a metastatis reaction happened ... and 2 R-OH + Na2S <--> 2 R-O-Na + H2S ?
The yellow to orange color, I think is possibly Na2S2, and maybe Na2S3.
Na2S2 is CAS 2868-13-9
B.P is 70C, and freezing point is < -40C ; so that's stable at room temperature.
|
Nonahydrate - water of crystallization is released in to methanol during dissolving (so methanol isn't anhydrous anymore).
What? Boiling point of Na2S2 70°C and freezing point -40°C? This is nonsense. Na2S2 is solid at RT.
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You misread the posts' context. I was responding to someone talking about the stability of H2S2.
The boiling point of H2S2 is 70.7C and the freezing point is below -40.
Wikipedia reports freezing point at -89.6.
https://en.wikipedia.org/wiki/Hydrogen_disulfide
| Quote: |
But even if you meant H2S2, this isn't true. Polysulfanes are very sensitive and very easily decompose. They are prepared at subzero temperatures from
sodium polysulfide and HCl (and separated by fractional distilation).
|
That's ONE way they are prepared. I've found videos of others, non-aqueous. The stability issues aren't the same.
Decomposition is mentioned in the wikipedia article.
But "easily" is not a measurement of time, or rate, or anything else ... it's just a human relative opinion for a water based reaction and no
explanation about 'why' the decomposition happens.
The book search does me no good. When I follow it ... it goes to a Russian page that shows no text. Do you have a downloadable pdf, or other that I
can actually read?
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semiconductive
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Let me re-iterate:
How can something be fractionally distilled, if it has no stability at all?
eg: how can it be re-condensed, from a fractional distillation?
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Bedlasky
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| Quote: Originally posted by semiconductive | | The book search does me no good. When I follow it ... it goes to a Russian page that shows no text. Do you have a downloadable pdf, or other that I
can actually read? |
Link is correct, it is weird that you cannot open it.
Here are some screenshots from pages which I can view.
Attachment: Polysulfany.rar (5MB)
This file has been downloaded 153 times
Here is article about polysulfane preparation methods:
https://cdnsciencepub.com/doi/pdf/10.1139/v68-384
I don't misreaded the post context, I just copy what you wrote. You wrote about Na2S2 and not H2S2.
Boiling point of H2S2 is just estimated from vapour pressure, not actually measured.
| Quote: Originally posted by semiconductive | That's ONE way they are prepared. I've found videos of others, non-aqueous. The stability issues aren't the same.
Decomposition is mentioned in the wikipedia article.
But "easily" is not a measurement of time, or rate, or anything else ... it's just a human relative opinion for a water based reaction and no
explanation about 'why' the decomposition happens. |
Your solution contained significant amount of water. Btw. alcohols also cause decomposition of polysulfanes. And I doubt, that you worked in dust free
environment.
| Quote: Originally posted by semiconductive | Let me re-iterate:
How can something be fractionally distilled, if it has no stability at all?
eg: how can it be re-condensed, from a fractional distillation? |
Vacuum distillation.
Practically everything I said you can find in book above. Believe me - you cannot prepare polysulfanes from aqueous/alcoholic solution at RT, it's not
that simple.
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semiconductive
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| Quote: Originally posted by Bedlasky | | Quote: Originally posted by semiconductive | | The book search does me no good. When I follow it ... it goes to a Russian page that shows no text. Do you have a downloadable pdf, or other that I
can actually read? |
Link is correct, it is weird that you cannot open it.
Here are some screenshots from pages which I can view.
|
Thank you.
Read it ...
I don't see where the new article discusses instability of polysulfides.
I really DO appreciate the explicit listing of possible chemical pathways for formation of polysulfides. That's helpful.
| Quote: |
I don't misreaded the post context, I just copy what you wrote. You wrote about Na2S2 and not H2S2.
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You've just proven that you misread the context again.
Context isn't what "I" wrote by itself.
Context means things that are in the same thread.
You seem to have known that, for you said "even if you meant H2S"; but you didn't check the earlier posts.
The discussion about stability, and the discussion about Na2S2 are in separate paragraphs separated by a space. It was sloppy of me ... but you can
not assume different paragraphs are on the exact same subject according to standard English grammar rules just because they are next to each other in
a post.
I DID write about H2S and S-O-R, as the main topic of this thread (at the top.)
Do you want me to explicitly quote both myself and the other poster whom I responded to? Or have I clarified the thread of thought yet?
| Quote: |
Boiling point of H2S2 is just estimated from vapour pressure, not actually measured.
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Interesting ... I wouldn't know that. The wikipedia article doesn't say so.
It says, the substance is fractionally distilled from a water based solution.
Last time I checked ... water boils at 100C +- a little depending on vacuum.
| Quote: |
Your solution contained significant amount of water. Btw. alcohols also cause decomposition of polysulfanes. And I doubt, that you worked in dust free
environment. |
I came to the same conclusion.
But: The methanol was molecularly dried.
The sodium sulfide had no odor as delivered by mailman.
1st conclusion: The water of crystallization was insufficient to cause decomposition of Na2S and the crystals were stable.
So -- how do you come to the conclusion that my meOH has significant amounts of water? What is significant?
I have a Honeywell air exchanger to remove dust. I don't think dust affected the experiment. Humidity of the air is low. eg: below 11% as it
comes from the air exchange and hepa filter. For short periods of time, eg: 1-3 seconds ... exposure to air doesn't transfer what I think is
siginicant amounts of moisture.
So, when I add Na2S to a freshly opened bottle of dry MeOH ... the MeOH had to act as a dessicant during dissolution. I think it must dilute
concentration of water of crystallization which is already insufficient water to cause decomposition to H2S gas. So, I think MeOH must act like a
(weak) acid, and partially decompose the Na2S.
After electrolysis for two days, solubility of Na2S in MeOH decreased and most of the dissolved salt fell to the bottom of the test tube. Likewise,
all odor stopped.
There is no camphor odor in the electrolyzed test tube. Nothing to indicate presence of H2S2.
The conductivity of the solution is about the same, so Ions are still present in solution. The open circuit voltage has risen from 0.75V at start of
experiment to 2.03Volts. I think this means I have successfully gotten rid of significant amounts of water and the electrochemistry suggests that
whatever is on the tin cathode likes to chage sulfur's oxidation state by a single electron.
I need to re-run the same experiment with no electrolysis as a scientific control; but tentatively I think Na2S solubility is increased by the
presence of water of crystallization. So, as water is removed ... Na2S falls to the bottom. The salt is noticably more orange, so the polysulfides
are affected similarly in this experiment.
| Quote: |
| Quote: Originally posted by semiconductive | Let me re-iterate:
How can something be fractionally distilled, if it has no stability at all?
eg: how can it be re-condensed, from a fractional distillation? |
Vacuum distillation.
Practically everything I said you can find in book above. Believe me - you cannot prepare polysulfanes from aqueous/alcoholic solution at RT, it's not
that simple. |
I'll download and look at the rar archive. I didn't see any of that information in the other article you linked.
I don't expect anything to be simple in chemistry.
I do want to understand why something is the way it is.
BTW:
Google doesn't allow everyone to look at all the books that they have. This is especially true when links cross country borders due to copyright
laws. Often, links to a google host are restricted depending on who is trying to look at it. Google tracks your IP numbers and sees what cookies are
set in a web browser when deciding what parts of books they allow a web browser to access. Whether that's interfering with the first link you posted,
I don't know ... but it's not uncommon.
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Bedlasky
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| Quote: Originally posted by semiconductive |
Read it ...
I don't see where the new article discusses instability of polysulfides.
I really DO appreciate the explicit listing of possible chemical pathways for formation of polysulfides. That's helpful. |
I post it just because there are good descriptions of preparation methods of polysulfanes.
Because you dissolved in it sodium sulfide NONAHYDRATE. It doesn't matter how much well dried is your methanol, you introduce water with sodium
sulfide. Na2S.9H2O contain 67,45% H2O. I don't know how much Na2S.9H2O you dissolved in MeOH, so I can't tell you how much water your solution
contain, but it contain it.
But even if you dissolve in MeOH anhydrous Na2S, MeOH still cause decomposition of polysulfanes at RT.
 
| Quote: Originally posted by semiconductive | | So, when I add Na2S to a freshly opened bottle of dry MeOH ... the MeOH had to act as a dessicant during dissolution. I think it must dilute
concentration of water of crystallization which is already insufficient water to cause decomposition to H2S gas. So, I think MeOH must act like a
(weak) acid, and partially decompose the Na2S. |
Methoxide is stronger base than hydroxide. So even, when sulfide methanolyse in this way:
S2- + MeOH <--> HS- + MeO-
Freshly formed methoxide will react with present water:
MeO- + H2O <--> MeOH + OH-
And this equilibrium is strongly shifted in to the right.
Solid is far more stable than solution.
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