I am an amateur in this field of science and I am trying to separate a solution of Sodium Iodide and water. How would I separate the solution so I can obtain sodium?

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    $\begingroup$ You can't and shouldn't obtain sodium. $\endgroup$ Dec 20, 2019 at 22:16

2 Answers 2


You could use electrolysis with a mercury cathode, taking advantage of its huge hydrogen over potential that makes sodium reduction kinetically favored. The sodium is amalgamated into the mercury.

Now all you need is to separate the sodium. Unfortunately, just evaporating the mercury would (1) probably get you into the news headlines when the toxic emissions are identified and (2) leave you with an intermetallic compound (see the phase diagram below from this reference).

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Older chlor-alkali cell technologies used this method to make sodium amalgam, but not to refine the sodium. Instead the amalgam is catalytically reacted with water in a separate chamber to retrieve sodium hydroxide without contacting the anodic products from the electrolytic cell.


You cannot separate sodium from a solution of sodium iodide NaI. If you try to do it by electrolysis, the sodium produced at the cathode will immediately react with water with the following reaction : $$\ce{2 Na + 2 H2O -> 2 NaOH + H2O}$$ and you will just obtain some sodium hydroxide NaOH in your sodium iodide solution

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    $\begingroup$ don't criticize me i might be wrong in what I say but, what if you were to evaporate the solution to leave behind pure sodium iodide. after doing that is there a way to separate it? $\endgroup$
    – diderstar
    Dec 21, 2019 at 0:50
  • $\begingroup$ Nobody criticizes. To separate the sodium Na in sodium iodide NaI, or in sodium chloride NaCl, there is only one way : electrolysis. Electrolysis can be done in solution with a cathode of mercury. Or it can be done with liquid NaI, i.e. at a temperature higher than the melting point of NaI, which is 651°C. It may be done, but it is not really easy to do. $\endgroup$
    – Maurice
    Dec 21, 2019 at 9:39
  • $\begingroup$ Some more details. At 651°C, sodium will burn in air. So it must be done in an argon atmosphere. And don't forget that simultaneously, an intense vapor of iodine will be produced on the other electrode. For 1 g Na, you will also produce 5.5 g of iodine vapor, which is corrosive to sodium and other metals. $\endgroup$
    – Maurice
    Dec 21, 2019 at 9:49

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