Potassium metal synthesis Alkoxide solubility and NaOH/Mg reaction mechanism

Quote: Originally posted by AJKOER

I did some research, and per a source on the action of H2 on NaOH (which I think may parallel KOH), we have: H2 + NaOH = NaH + H2O (see Eq 1 on page 2192) and at 360 C : NaH + Heat --> Na + 1/2 H2 Source: See 'Revisiting the Hydrogen Storage Behavior of the Na-O-H System' by Jianfeng Mao , et al., in Materials 2015, 8, 2191-2203; doi:10.3390/ma8052191 . Link to download file: https://www.mdpi.com/1996-1944/8/5/2191/pdf . To quote: "The DTA profile for NaH shows one endothermic peak at ca. 360 °C, which can be assigned to the decomposition of NaH to Na metal and hydrogen." -------------------------------------------------- Actually, in my opinion, the actual chemistry is likely a function of the highly anodic metal (like Mg, Zn, Al, Fe, Sn,.., see http://www.zygology.com/cms/upload_area/pdf/Zyg-Anodic-Index... where Zn and Al are the frequent cited sources of so called 'nascent hydrogen' with HCl) reacting with water as follows: H2O = H+ + OH- Al --> Al3+ + 3 e- e- + H+ = .H = 1/2 H2 (as .H + .H = H2) Hence, the cited Equation (1) above could actually be the net of the following reactions: .H + OH- = e- + H2O = e-(aq) (see Eq 37 and Table 2 at https://pdfs.semanticscholar.org/d696/b35956e38351dd2eae6706... ) e- + .H = H- (see http://pubs.rsc.org/-/content/articlelanding/1966/tf/tf96662... ) Net of the two above: 2 .H + OH- = H- + H2O or, seemingly (but not so at very high temperatures as H2 + Heat = .H + .H): H2 + OH- = H- + H2O Further: H- + Heat --> .H + e- = 1/2 H2 + e- (as .H + .H = H2) and: e- + Na+ = Na My point is if one is going to give credit to the hydrogen, and ignore the role of the highly anodic metal, it is really the mono-atomic hydrogen radical (.H formed courtesy of the anodic metal or high temperatures), or as historically sometimes referred to, in my estimation, as 'nascent' hydrogen. Here is a recent account of the surface chemistry of adsorbed .H on aluminum and applications, see https://books.google.com/books?id=1etfSdk55SYC&pg=PA817&... . This last reference per Equation (5) appears to support my reaction presentation in terms of 2 .H (which is adsorbed on an aluminum surface) and not H2, to quote: " PbS + 2 .H = Pb + H2S (5) " [Edited on 3-5-2018 by AJKOER] [Edited on 3-5-2018 by AJKOER]

Quote: Originally posted by AJKOER

In my defense, what we have here, in my opinion, is not a classic metal/O2 battery, but a related lesser known rendition with (well, at least prior to revision, see Thesis link below) the same anodic half reaction cited in the Al/O2 battery. For those interested, one of my favorite thesis is from 2008, "Alkaline dissolution of aluminum: surface chemistry and subsurface interfacial phenomena", by Saikat Adhikari, link: https://www.google.com/url?sa=t&source=web&rct=j&... I believe a similar electrochemical reaction scheme is occuring with Magnesium. Some extracts of interest for the brave, to quote: "In addition to being a primary corrosion process, dissolution behavior of aluminum and its alloys in alkaline solutions is of considerable interest because it is the anode reaction in aluminum-air batteries.[4] ......The anodic half-reaction at the Al electrode is Al + 4 OH − → Al (OH)4− + 3 e− (1.1) which exhibits an electrode potential of -2.35 V in alkaline solutions(vs. NHE). "2 Al + 6 H2O → 2 Al (OH)3 + 3 H2 (1.2)" ....."dissolution of aluminum in alkaline solutions at open-circuit also leads extremely high rates of H-absorption into the metal, [9-14] "..... "Another study of the dissolution of aluminum in aqueous solutions by Perrault revealed that the open circuit potential of aluminum in strongly alkaline solutions corresponds closely to the Nernst potential for oxidation of aluminum hydride to aluminate ions [25] AlH3 + 7 OH− (aq) → Al (OH)4− + 3 H2O ( aq ) + 6 e− (1.3) This suggests a role of surface aluminum hydride as a reaction intermediate in the dissolution process. Additional evidence for the presence of aluminum hydride was provided by Despic and co-workers.[26, 27] They found that aluminum hydride formation was one of the major processes apart from aluminum dissolution and hydrogen evolution, during the cathodic polarization of aluminum. Titanium corrosion in alkaline solutions is also thought to proceed through a hydride mediated mechanism.[28-30] " "He found that the open-circuit potential in strongly alkaline media was determined by the equilibrium of the reaction AlH3 + 7 OH− (aq) → Al (OH )4- + 3 H2O ( aq ) + 6 e− (3.7) He obtained a standard chemical potential of 25 kcal/mol for AlH3 from his data, which was in reasonable agreement with prior thermochemical calculations done by Sinke et al who obtained a value of 11.1 kcal/mol for the chemical potential.[80] ...." "The anodic reaction 3.7 is accompanied by the cathodic reduction of water to form hydrogen H2O + e- → OH- + H (3.8) and the reaction of hydrogen with aluminum to from hydride Al + 3 H → AlH3 (3.9)" So, a bit technical but related half cell reactions I would guess for the likes of Magnesium. ------------------------- One should accept that if some electric current is generated in situ, it is possible for electrons to be solvated in an appropriate organic medium. This could lead to the decomposition of associated organics, resulting in the release of Potassium. Note, even in aqueous settings, long ago there are reports of unexplained reactions of various salts with Mg metal, including converting nitrate into nitrite and finally ammonia (see, for example, p. 314 at "Chemical News and Journal of Physical Science, Volumes 87-88, 1903, p. 312-316, link: https://books.google.com/books?id=jvjmAAAAMAAJ&pg=RA1-PA... ). ----------------------------------------------------------------------- [Edit] Here is an interesting article citing a temperature and Magnesium alloy presence associated with hydride formation: "absorption of hydrogen by magnesium based alloys", in METAL 2014, link: https://www.google.com/url?sa=t&source=web&rct=j&... This short discussion on preparing the Mg for interaction with H2 may be of value: https://books.google.com/books?id=NR3OxpSiA60C&pg=PA491&... [Edited on 24-5-2017 by AJKOER] [Edited on 24-5-2017 by AJKOER]

Quote: Originally posted by SWIM

In the potassium thread Nicoderm says beta hydride elimination reactions are the problem with alcohols with an alpha hydrogen.

That makes sense now that you say it, my organic is fairly rusty.

As for ionization, I can not comprehend that there is any dissociation. I found that Potassium tert-butoxide forms tetrameric cubane-type clusters.

One of the reasons for sodium's inhibition might be the structure of the alkoxide complexes.




[Edited on 10-5-2018 by walruslover69]

[Edited on 10-5-2018 by walruslover69]

Quote: Originally posted by walruslover69

I might able to do some experiments to test if that is the mechanism, or atleast rule it out if it isn't. Running with the reaction with K2O instead of KOH and using already prepared potassium tert-butoxide instead of tertbutanol. Doing it this way there would be no evolution of hydrogen. Does that seem valid to you?

Quote: Originally posted by AJKOER

Quote: Originally posted by walruslover69   I might able to do some experiments to test if that is the mechanism, or atleast rule it out if it isn't. Running with the reaction with K2O instead of KOH and using already prepared potassium tert-butoxide instead of tertbutanol. Doing it this way there would be no evolution of hydrogen. Does that seem valid to you? Per Eq 1, scaling by 2: 2 H2 + 2 NaOH = 2 NaH + 2 H2O 2 NaOH = Na2O + H2O So: 2 H2 + Na2O + H2O = 2 NaH + 2 H2O Or: 2 H2 + Na2O = 2 NaH + H2O I would rewrite as: 2 .H + 1/2 Na2O = NaH + 1/2 H2O In the presence of the highly anodic metal (Mg or Al) forming e-, replacing the KOH with K2O may reduce available hydrogen atoms, and, as such, I would suspect a reduction (but not elimination) in the level of diffusible hydrogen (see, for example, page 41 at http://www.jwri.osaka-u.ac.jp/publication/trans-jwri/pdf/421... ) and a lower yield of the target metal. [Edited on 19-5-2018 by AJKOER]