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Author: Subject: Molten Sodium Sulfide Electrolysis
Filippo
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[*] posted on 29-4-2017 at 03:56
Molten Sodium Sulfide Electrolysis


Could it be possible to obtain metallic sodium via Molten Sodium Sulfide Electrolysis?
Is it less dangerous than Molten NaOH electrolysis?
What is the procedure?
I chose Sodium Sulfide because its melting point is very low (50 Celsius degrees).
I'm trying to get every important element to build up a collection but i'd prefer to sythesize Na since it is very expensive while its salts are very cheap.
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Metacelsus
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[*] posted on 29-4-2017 at 04:03


https://en.wikipedia.org/wiki/Sodium_sulfide

Look in the "Properties" table. The anhydrous melting point is the relevant one.




As below, so above.
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Filippo
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[*] posted on 29-4-2017 at 04:17


Whoops,in the italian wikipedia page it is reported it wa the anhydrous melting point.
https://it.wikipedia.org/wiki/Solfuro_di_sodio

Sooo,is there another sodium salt which electrolysis isn't dangerous and with a low melting point?
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j_sum1
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[*] posted on 29-4-2017 at 05:01


This arrived on YT ten minutes ago.
I think it might be what you want.
https://www.youtube.com/watch?v=jCrFFVVcPUI
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Filippo
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[*] posted on 29-4-2017 at 05:43


Thank you,i'll do that as soon as i'll get a distillation apparatus for my birthday
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Jstuyfzand
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[*] posted on 29-4-2017 at 07:29


That video by Nurdrage is phenomenal, although a proper castner cell is still cheaper in the long run I think.
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Harristotle
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[*] posted on 30-4-2017 at 04:31


It occurs to me that the standard enthalpy of formation for aluminium oxide is even higher at -1675 kj/mol. (compared to sodium oxides -416 kj/mol). This suggests that aluminium turnings would be even better*. I guess there is the old stable aluminium oxide problem, but the same is true for magnesium.

Any reason it wouldn't work with Al powder/ turnings?

* yeah ok, kinetics isn't thermodynamics.

[Edited on 30-4-2017 by Harristotle]
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[*] posted on 30-4-2017 at 05:19


Standard enthalpy of formation in kJ/mol:
MgO: −601.6
Al2O3: −1669.8
NaOH: −425.9


So,
2Mg + 2NaOH = 2Na + 2MgO + H2, ΔH = -175.7 per mole of Na
2Al + 3NaOH = 3Na + Al2O3 + 1½H2 ΔH = -130.7 per mole of Na

So as a first approximation (excluding temperature calculations), Mg is more energetically favourable than Al.

I thought of the same thing myself. And since Al is so much easier to obtain that Mg, I think it is worth a try. Quite what effect this will have on the work up, I am not sure. If aluminium works then it brings sodium well into the realm of an OTC synthesis. And that is a game changer.
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