Sciencemadness Discussion Board

Modification of Castner–Kellner process - bromine

biomechem - 13-1-2017 at 07:50

Is it possible, and relatively easy to obtain bromine in the following process.
In a vertical tube, on the bottom is mercury electrode, above it NaBr solution, and carbon electrode. The mixture is heated above 58.8 °C (boiling point of bromine).
1. Will Br3- formation will be the same as in lower temperature or it will degradate to Br2 and Br-?
2. Does Br3- react (violently) with sodium amalgam?
3. Is it better to use saturated (or conc.) NaBr solutions, or diluted like 0.1M will be better?

[Edited on 13-1-2017 by biomechem]

Melgar - 13-1-2017 at 10:42

Br3- would mostly stay away from the cathode, what with them both having like charges. The trouble happens when Br2 comes into play, and falls to the bottom due to its density. Then you have mercury bromide in solution. Sure, most of it will be reduced again, but not all of it, and you can't depend on the solution temperature to remove bromine immediately as it's formed. Maybe a divided cell with sodium carbonate on the mercury side, and sodium bromide on the other side?

biomechem - 13-1-2017 at 12:51

Quote: Originally posted by Melgar  
Br3- would mostly stay away from the cathode, what with them both having like charges. The trouble happens when Br2 comes into play, and falls to the bottom due to its density. Then you have mercury bromide in solution. Sure, most of it will be reduced again, but not all of it, and you can't depend on the solution temperature to remove bromine immediately as it's formed. Maybe a divided cell with sodium carbonate on the mercury side, and sodium bromide on the other side?


Your solution is definitely better and much more simple, thank you, and please forgive me this poor attempt to reinvent the wheel.