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Cryolite $\ce{(Na3AlF6)}$ and fluorspar $\ce{(CaF2)}$ are added to alumina to lower the temperature of the mixture from $\pu{2000 °C}$ to about $\pu{900 °C}$ and also to increase the conductance.

But my question is about the specific effect these compounds are producing. I mean, out of the two mentioned effects, which compound is doing what?

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    $\begingroup$ It is my understanding that Hall originally used cryolyte to dissolve alumina so he could perform electrolysis. The molten cryolite was both solvent and electrolyte and electrolyzing cryolite itself was not a problem. My guess is that the fluorite just helped lower the melting temperature. Fluorite (from 'to flow') has been used for centuries to help 'liquify' oxide slag, in metallurical processing, so the slag could be more readily scraped aside. Maybe someone will provide a definitive answer. $\endgroup$ – Ed V Jul 8 '19 at 18:13
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    $\begingroup$ I don't know enough to post a full answer but as Ed V says $\ce{(CaF2)}$ is used "generally" in slag-metallurgical processes to control viscosity. It also provides a surplus of flourine ions, which I guess would be useful for dissolving and solvating the alumina - since the cryolite natively cannot handle much more aluminium ions, you need a surplus of fluoride. $\endgroup$ – Stian Yttervik Jul 9 '19 at 7:36
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    $\begingroup$ @ChakravarthyKalyan What was the point of your edit? Units should be depicted upright and with proper spacing; representation you've suggested contradicts both rules. If you are having trouble with a mobile app not being able to render \pu{…} macros, please see the corresponding Meta post. The short version is not to use a mobile app at all as it's pretty outdated. $\endgroup$ – andselisk Jul 9 '19 at 11:04
  • $\begingroup$ @andselisk ,in my mobile, these values are shown as "undefined control sequence\pu" twice. I can post screen shot of the same.Hence the edit. $\endgroup$ – Chakravarthy Kalyan Jul 9 '19 at 11:48
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    $\begingroup$ @ChakravarthyKalyan I think I already addressed this in my comment, please read the Meta post and suggestions within carefully. If your device doesn't support formatting features available in every web browser (and mobile web browsers, too), it's not a reason to downgrade the quality of the posts. $\endgroup$ – andselisk Jul 9 '19 at 11:55
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According to the CRC Handbook of Chemistry and Physics, melting point of $\ce{CaF2}$ is $\pu{1418 ^\circ C}$, of $\ce{Al2O3}$ (in its corundum form) is $\pu{2053 ^\circ C}$, and of $\ce{Na3AlF6}$ is $\pu{1009 ^\circ C}$ (lowest of the all three). Thus the cryolite should melt first.

I have read mentions of the problem you asked in the book Nonaqueous Electrochemistry (you can find information about it here. For the best of my summarising ability, I can conclude that the role of the cryolite and other salts added to the reaction mixture is that they decrease the melting point of alumina. As cryolite's melting point is much lower than that of alumina as cited above, when the cryolite melts it acts as a solvent to "dissolve" the $\ce{Al2O3}$, by the process of complexation. Recall how silver chloride dissolves in aqueous ammonia or lead iodide dissolves in excess iodide. The book states that:

Cryolite is used as solvent for alumina because it has high chemical stability and a high solubility for alumina.

In detail:

When pure cryolite ($\ce{Na3AlF6}$) melts, it dissociates into fluoride ions and various lower fluoroaluminates:

$$\ce{[AlF6]^3- -> [AlF4]- + 2 F-}$$

$$\ce{[AlF6]^3- -> [AlF5]^2- + F-}$$

(etc.)

However, in the presence of alumina, the dissociation process is complicated by the fact that the products arise from $\ce{Na3AlF6}$ dissociation can react with $\ce{Al2O3}$ creating oxofluoroaluminates, and it is those complexes that directly get electrolysed (cathodically reduced) rather than the $\ce{Al2O3}$ itself. By various means of calculation and experiments (Raman spectroscopy etc.) they proposed some following reactions (or equilibria):

$$\ce{Al2O3 + 4 [AlF5]^2- -> 3 [Al2OF6]^2- + 2 F-}$$

$$\ce{Al2O3 + [AlF5]^2- + F- -> 3/2 [Al2O2F4]^2-}$$

$$\ce{2 Al2O3 + [Al2O2F4]^2- + 4 F- -> 2 [Al3O4F4]^3-}$$

...

When you add calcium fluoride (or magnesium fluoride) into the mixture, the following reactions also occur:

$$\ce{MgF2 + n F- + (2-n) [AlF5]^2- -> [MgF_{(2+n)}(AlF5)_{(2-n)}]^{(4-n)-} }$$

$$\ce{CaF2 + (3-n) [AlF5]^2- -> [CaF_{n}(AlF5)_{(3-n)}]^{(4-n)-} + (2-n)F-}$$

The electrolytes in the electrolysis process is obviously "$\ce{Na3AlF6}$", but in fact they always add an excess of $\ce{AlF3}$ alongside other salts such as $\ce{CaF2}$ ($\pu{3-5 \%}$), $\ce{LiF}$ ($\pu{0-4 \%}$) and $\ce{MgF2}$ (also $\pu{0-4 \%}$). In fact, those salts decrease the solubility of $\ce{Al2O3}$ rather than increase it; that's why they only add a little of those salts.

Then why adding those salts?

There are a couple of reasons for this. Firstly, addition of those salts increase the density of the electrolyte. Because aluminium is already produced in the liquid state (melting point $\pu{660 ^\circ C}$), if the density of the two components are too similar, it is impossible to achieve a good separation-by-weight. Kvande and Rorvik had derived an equation of the melt density as a function of temperature and salt composition:

$\pu{d = 2.64 - 0.0008T + 0.18(wt. ratio NaF/AlF_{3}) - 0.008(wt. \% Al_{2}O_{3}) + 0.005(wt. \% CaF_{2}) + 0.008(wt. \% MgF_{2}) - 0.004(wt. \% LiF)}$

As you can see from the above equation density of the mixture increases with CaF2 content.

The other reason is that those salts increase the interfacial tension at the liquid Al-electrolyte boundary. The higher interfacial tension is, the less does aluminium re-dissolve in the electrolyte (yes it can!), thus better current efficiency. For example: at the boundary this reaction occurs: $\ce{Al + 3 NaF -> AlF3 + 3 Na}$. Adding more $\ce{AlF3}$ (or other fluorides) helps surpressing this reaction, and also may increase the $\ce{Na+}$ concentration in the electrolyte - the researchers found that $\ce{Na+}$ ions are the primary charge carriers.

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    $\begingroup$ You've been a member of the community for some time. There is little excuse for not formatting your (often well-received) posts. I've flagged this as LQ. Also: "AlF63- -> AlF4- + 2 F-"? That's not only wrong, it's indecipherable. $\endgroup$ – Todd Minehardt May 28 at 2:04
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    $\begingroup$ I have formatted your answer. Please make sure to learn formatting by yourself. This meta post should help: chemistry.meta.stackexchange.com/questions/86/… $\endgroup$ – Nilay Ghosh May 28 at 3:44
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    $\begingroup$ @ToddMinehardt I know that it is evil for not formatting the post, but the truth is although I've been a member for some time I rarely spend time on this; and I am also not certainly comfortable of using math formats either - which I can improve. And in addition the thing in your comment is actually not wrong, it is just made a bit difficult in deciphering due to my lack of formating, that's all. $\endgroup$ – ĐỨc Lê Hồng May 28 at 3:46
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    $\begingroup$ @ĐỨcLêHồng Understood. I've up-voted and encourage you to do a little bit of editing at least when you do visit. As I said, your contributions are uniformly good and of value to the community here. $\endgroup$ – Todd Minehardt May 28 at 14:43
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Cryollite is used to decrease the temperature from $2000 °C$ to $900 °C$ . Fluorspar is used to increase mobility by controlling viscosity, hence in turn increasing conductivity.

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    $\begingroup$ Please provide a source of this information. $\endgroup$ – Apurvium Oct 28 '19 at 4:23

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