5
$\begingroup$

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.

|improve this question
$\endgroup$
  • $\begingroup$ But why doesn’t that logic work in hall-Heroult? : You answered it yourself - the $\ce{BaF2}$ mixture is much more heavier than $\ce{Al}$. $\endgroup$ – Aniruddha Deb Jan 1 at 16:17
  • $\begingroup$ 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. $\endgroup$ – Sir Arthur7 Jan 2 at 16:37
  • $\begingroup$ 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. $\endgroup$ – Sir Arthur7 Jan 2 at 16:51
2
$\begingroup$

I think the difference is due to the way eddy currents form Magnetohydrodynamics.

As the current flows through the conductive liquid, a magnetic field is produced. This in turn causes the liquid to be pushed (at right angles to the current and field).

In one version of the cell, this current causes stirring of the aluminum, alumina, flux mix. This is more efficient from a time and electrical point of view. The trade-off however is that some of the impurities which would otherwise get captured in the slag (waste), end up mixed into the final product.

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.

|improve this answer
$\endgroup$
  • $\begingroup$ Thanks for providing a new perspective, I studied eddy currents recently but little did I know that its effects are such profound. But according to Ellingham diagram for oxidation reactions of metals and non-metals, its pretty clear that the thermodynamical feasibility of the above mentioned reaction is less. Also if we are going to really account eddy currents, then don't you feel it'd be likewise applicable in Hoope's process? $\endgroup$ – Sir Arthur7 Nov 21 '19 at 17:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.