clearly_not_atara - 6-11-2015 at 12:44
The electrolytic production of borohydrides from borates is commonly believed to be impossible due to the extremely low stability of solvated H- ion,
which is believed to be necessary for the formation of BH3. I propose a workaround.
The cell comprises:
* boric acid, B(OH)3
* a Lewis base, which is reactive with boric acid but not with borane
* Mg(OTf)2; triflate is a spectator
* a heterocyclic compound of the type described in Catalytic Synthesis of Magnesium Hydride Under Mild Conditions, possibly with Li(OTf) as a Lewis acid as transition metal salts are incompatible.
* a solvent, tbd
The anode reaction is simple (Lb denotes a Lewis base):
4 (OH)4B- + 8 Lb >> 4 (OH)3B * Lb + 4 LbH+ + 4 e- + O2
The cathode is interesting:
Mg2+ + 2 e- >> Mg
Mg + 2LbH+ >> Mg2+ + H2 + 2 Lb
but also:
Mg + C14H10 (anthracene) >> Mg * C14H10
Mg * C14H10 + H2 >> MgH2 + C14H10
(the critical step)
3MgH2 + 2(OH)3B >> 2BH3 (gas) + 3 Mg2+ + 6 OH-
(borane must evaporate)
(I hope that the disproportionation of ions like H2(OH)2B- will favor borane and borate over heterogeneous
hydridoborates)
with side reactions:
(OH)3B + Lb <--->> (OH)3B * Lb
(OH)3B * Lb + OH- >> Lb + (OH)4B-
Any chance this could work? Obviously you need a system where Mg2+ is reduced more rapidly than LbH+ or B(OH)3, though the concentration of both
should be low. There's also the concern of hydrogen escaping, some magnesium-borate complex precipitate (hopefully not), water being formed from OH-
and LbH+, etc.
The idea is that magnesium is reduced at the anode and dissolved by anthracene, where it reacts to produce hydrogen and reacts again to
produce MgH2, which ultimately reduces borate to borane.
But the production of borohydrides from MgH2 and borates is known, so...