# Need of refining aluminium and Hoope vs Hall–Héroult processes

Recently I was taught how aluminium is commercially extracted. The ore is first concentrated by leaching either by Bayer's process for red bauxite (impurity: $$\ce{Fe2O3}$$) or by Serpeck's process for white bauxite (impurity: $$\ce{SiO2}$$).

Then alumina is reduced electrolytically via Hall–Héroult's process directly to 99% pure aluminium. But for further purification to remove remaining copper or iron impurities, Hoope's process is executed to get 99.99% pure $$\ce{Al}$$.

Since both processes involve electrolysis, and the materials added are similar in both (except $$\ce{BaF2}$$) with same graphite electrodes, shouldn't we get the same purity in both process (thus not requiring a further refinement)?

Also, the arrangement of the electrodes are exactly opposite to each other. Hoope's argues that pure aluminium is lighter and thus floats so cathode is kept at top, but why doesn't that logic work in Hall–Héroult's? The difference in purity varying by 0.99% and the electrolyte used being same.

Any insight into this matter is helpful because I haven't been able to come to a convincing conclusion.

• But why doesn’t that logic work in hall-Heroult? : You answered it yourself - the $\ce{BaF2}$ mixture is much more heavier than $\ce{Al}$. – Aniruddha Deb Jan 1 at 16:17
• Ok so you seem to be right that BaF2 is heavier than Al, but Na3AlF6(cryolite) is not. So it should possibly float over the pure molten Al in Hoope's process? Making the conditions same as for Hall-Heroults. – Sir Arthur7 Jan 2 at 16:37
• Moreover I feel that adding cryolite is lot more necessary than BaF2, and Wikipedia says that "en.wikipedia.org/wiki/Hall%E2%80%93H%C3%A9roult_process" molten Al is heavier than cryolite, so it should sink down both in Hall-Heroult's and Hoope's processes. I don't think there's anything to argue on that. – Sir Arthur7 Jan 2 at 16:51

Edit: Thinking about it some more, the reason for the mixing may be due to where the $$\ce{CO2}$$ gas is formed. ($$\ce{Al2O3 + C_{electrode} -> Al +CO2}$$) Bubbles formed a the bottom of the cell would rise up and cause more stirring than if they formed near the top of the cell.