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.

  • 1
    $\begingroup$ But why doesn’t that logic work in hall-Heroult? : You answered it yourself - the $\ce{BaF2}$ mixture is much more heavier than $\ce{Al}$. $\endgroup$ Jan 1 '20 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. $\endgroup$ Jan 2 '20 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. $\endgroup$ Jan 2 '20 at 16:51
  • $\begingroup$ From the wiki link, molten Al is denser than molten cryolite at the H-H operating temperature, so Al is on the bottom. In the Hoope’s process, as per wiki, there are 3 molten layers: a Cu-Al-Si metal anode, a mix of cryolite and barium fluoride middle layer, and molten Al as the cathode on top. Barium fluoride is much more dense than cryolite. So the layering makes sense if enough barium fluoride is used. What is especially interesting is how the Hoope’s process came to be invented: I never would have guessed it would work! Thanks for asking the question! $\endgroup$
    – Ed V
    Mar 7 '20 at 3:32
  • 1
    $\begingroup$ I just upvoted your question and realized a mistake I made: “Hoope’s” is incorrect, since the process is named for its inventor “Hoopes”. The wiki blurb says he invented the process at ALCOA and it was patented in 1925. Probably not worth looking up the patent, since they can be pathetically lacking in important and proprietary details. Anyway, excellent question and maybe you or someone else will post an answer (other than the current answer)! $\endgroup$
    – Ed V
    Mar 7 '20 at 14:12

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.

  • $\begingroup$ 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? $\endgroup$ 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.