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For the Hall-Héroult process, the solubility of alumina in flux seems to be important. But why should I care about the persence of alumina in the bath? I can use feeder to keep the concentration constant because electrolysis decomposes $\ce{Al2O3}$. It can be 0.1% or 10% what would be the difference? Concentrations does not affect decomposition voltages of the components, so the voltage would only decompose $\ce{Al2O3}$ and would not touch the flux.enter image description here


I have found the answer! The limitation is in the current you can apply (limiting current) If you exceed it - the anode effect may occur and NaFAlF3 will start to decompose instead of alumina. So your cell will make you rich faster if the solubility is higher!

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    – Buttonwood
    Commented Jan 31, 2022 at 11:06

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Most of the bath is cryolite ($\ce{Na3AlF6}$) and not alumina ($\ce{Al2O3}$) because for electrolytic reduction, we do not use an aqueous solution, we have to use melt alumina. But alumina melts at a very high temperature ($\pu{2072 ^\circ{}C}$). So we use cryolite which serves two purposes:

  • Cryolite acts as a solvent for alumina and dissolves it.
  • It increases the conductivity of the melt and decreases the melting point of alumina to about $\approx \pu{950 ^\circ{}C}$

Thus a suitable composition of the bath is required to make the process possible and feasible.

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    $\begingroup$ I do know that cryolite bath should dissolve alumina, but what would be the advantage of having say 15% solubility instead of 1% solubility of alumina in the bath? Whenever the flux is untouched by the applied voltage we only need to dissolve SOME fraction of Al2O3 and we can run the cell while adding new Al2O3 from the feeder to keep this fraction constant. Why should I want to dissolve more % of alumina? What is the effect of alumina concentration in the bath? $\endgroup$
    – Demidrol
    Commented Jan 31, 2022 at 12:12

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