Just disappointed. Being a chemistry fossil in his 112th semester
who forgot most of what he has been taught I really don’t think that I look & write like a spammer.
(1) Please imagine a polyethylene film of – say – 30µm thickness. Among the physico-chemical parameters describing the properties of this film also is oxygen transfer rate (OTR) and water vapor transfer rate (WVTR). Both describe transport phenomena of small molecules in a polymer. PE has high OTR & WVTR, indicating that there is an easy diffusion of small molecules in PE.
(2) Imagine our PE film is unilaterally covered (by standard methods) with elemental Cu, be it powder, be it by a sputtering technique.
(3) Further imagine that this elemental Cu is covered by a 2nd polymer film, which is virtually sealed against diffusion, e. g. EVOH or something similar, possessing very small values for OTR and WVTR.
(4) Imagine a scenario where on the PE side this sandwich becomes equilibrated with an atmosphere containing H2S. It is important to add here that we are considering a process over many months or even some years. H2S will follow the concentration gradient and diffuse through the PE film. Subsequently H2S will interact with Cu by chemisorption and/or formation of Cu2S / CuS. This process will come to its end as soon as one of the reactants becomes “eaten up”. Do you agree until this point?
(5) Now let’s change the above scenario by replacing H2S by SO2. Again we will see diffusion of SO2 through the PE film. In parallel we will see diffusion of H2O molecules through the PE film.
(6) The 1st molecule SO2, meeting a molecule H2O will form H2SO3. The 2nd molecule H2O passing this will lead to HSO3- + H3O+. Do you agree until this? In any case it is obvious that in the conditions described no pH measurements are possible.
(7) From the literature (electrochemistry, corrosion science) it is known that elemental Cu in the presence of H2O interacts with SO2 in a complex way, ultimately according to the following molecular formula: 6Cu + SO2 --> 2 Cu2O + Cu2S. This is more than surprising (for me at least), because all my life I thought that SO2 is a potent reducing agent.
This said let’s return to the polymer sandwich scenario. In any case it is more than likely that we will see an interaction of SO2 and/or H2SO3 with our elemental Cu. According to the reaction formula given in (7) we also will get Cu2S.
(9) Cu2S as well as CuS form (and are stable) in slightly acidic medium. On the other hand, however, at a lower pH both sulfides become cleaved, forming free H2S. This free H2S – once it is formed in the above scenario – would follow the concentration gradient and diffuse out of the PE film. I discussed the problem with a famous US geoscientist. He told me that in his opinion and derived from his experience with copper minerals the stability border of Cu2S / CuS should be around pH 1,5.
10. So obviously the crucial question is: Can our H2SO3, putatively formed “in situ” from SO2 and H2O, reach an acidity capable of cleaving Cu2S / CuS or – for that matter – prevent formation of Cu sulfides, leading to formation of free H2S?
In summary let’s return to my question, asked in my 1st post: Is it conceivable that aqueous H2SO3 has a pH which can cleave Cu2S / CuS? Or is it possible that the pH of said H2SO3 is such that formation of Cu2S / CuS is impaired, in either way leading to the formation of freeH2S?
Borek, if you still think this thread is of public interest I have no issues in publicly discussing it. However, I still think we should switch to a private channel. Being sort of an admin it should be easy for you to enable this.
Thank you and curiously looking forward to your comment.