In a class of electrolysis, my instructor told me that Hg forms Na-Hg in the electrolysis of dilute NaCl aqueous solution. For this reason, sodium cations are reduced in the cathode instead of hydronium ions. However, he did not give me any reason for how creating an amalgam affects the reduction of the metal or, why Hg creates amalgam in the first place. So, I would like to know why Hg creates amalgam whereas other metals like Pt don't form them ? And also how this affects the reduction of Na.

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    $\begingroup$ It has to something with the hydrogen overpotential of Hg electrode. $\endgroup$
    – AChem
    May 17, 2020 at 23:21
  • $\begingroup$ Well,what do you mean by "hydrogen overpotential" ?And what has it got to do with Hg? $\endgroup$
    – Habib
    May 17, 2020 at 23:31
  • $\begingroup$ Umm, proper question would be rather why Na is reduced with Hg electrode and is this even really the case. While Na might perhaps be reduced also with other electrodes, it obviously won't become part of amalgamate, as only solutions in Hg are called like that. $\endgroup$
    – Mithoron
    May 18, 2020 at 0:13
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    $\begingroup$ Please see “Castner–Kellner process” in wikipedia. Definitely sodium can be electrolyzed from brine. The reason it works is the overpotential for hydrogen reduction at a mercury cathode. In other words, there is a kinetic hindrance to hydrogen reduction and sodium forms an alloy with mercury, so back-reaction with the aqueous solution is impeded. Traditionally, mercury alloys with other metals are called amalgams. Liquid gallium, for example, also readily forms alloys with many metals, but gallium has not been known since ancient times. $\endgroup$
    – Ed V
    May 18, 2020 at 0:24
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    $\begingroup$ In the case of sodium ion reduction at the liquid mercury cathode, the sodium atoms get “dissolved” in the liquid mercury. So they move around like in a ordinary solution. But this keeps them from contacting the brine, so they do not get to react with the brine. Those at the interface of brine and mercury can so react, producing hydrogen gas and NaOH. For platinum, for example, the sodium would, at best, form a plating: the atoms could not get sheltered, as it were, by getting into the interior of the platinum. So it is a matter of getting the sodium away from the brine and mercury does it. $\endgroup$
    – Ed V
    May 18, 2020 at 0:56

2 Answers 2


The mental conflict occurs because two processes could theoretically happen (electrodeposition of sodium, or electrodeposition of hydrogen), and the theoretically unfavorable one occurs - and not only occurs, but is so favorable that it was used commercially to produce NaOH as EdV noted.

So why is the theoretically favored process not favored in this instance? Hydrogen evolution is a multi-step process, and if you wave your hands a lot, you can call it overpotential and be off to other projects.

When (or if) sodium is deposited (on mercury), it can dissolve and be removed from contact with water. So this process is a slam-dunk if you push with a high enough voltage.

But the hydrogen evolution should take place at a lower voltage. Why not? If/when hydrogen is deposited, its first appearance is atomic: one H atom sticks somewhere on the mercury surface; the second step is deposition of another H atom somewhere else. The next step, for the two atoms to form a molecule, requires some mobility on the part of the hydrogen and the mercury surface. At this point, conventional wisdom suggests that the molecule could dissolve into the solution, but eventually the aqueous solution becomes saturated. Then the first molecule must meet another molecule, and another, to begin the process of forming a bubble which will eventually rise to the surface of the aqueous solution. Somewhere in there is an impediment to the formation of bubbles, and it may involve mobility of atoms or molecules, or attachment of hydrogen to mercury, but the process is difficult enough to allow the voltage to be raised high enough to allow deposition of sodium to proceed.

And then we say that mercury has a high overpotential for deposition of hydrogen.

Now suppose we do the same, but with gallium. Sodium should deposit and become dissolved as with mercury, if the voltage is high enough. (Sodium alloys with gallium; amalgamation is a chemical term that should be reserved for mercury alloys.) But the hydrogen overpotential on gallium is less than on mercury (Ref 1), and gallium and hydrogen are deposited together from a solution of $Ga^{+3}$ (Ref 2). So, in the case of liquid gallium, it seems that hydrogen would be more easily deposited than sodium (unless, perhaps you used a very high voltage, which would override the thermodynamic preference because diffusion couldn't keep up with the current demands).

Ref 1. https://pubs.rsc.org/-/content/articlelanding/1966/tf/tf9666203524/unauth#!divAbstract

Ref 2. https://link.springer.com/article/10.1007/BF01016217


If you heat metallic sodium with let’s say gold / silver / mercury / platinum etc.. it will form an amalgam.

I have tried this with a tiny bit of gold some years ago.

I did heat Na in an inert argon atmosphere to several hundred degrees in an quartz tube and drop the gold inside.

It did take only fractions of a second till the amalgam was formed.

The result is a green looking amalgam when it is hot - when it cools down it is a brown solid substance.

Given the fact that gold is a pretty inert substance, I guess that sodium will do the same with Pt, Rh, Ir, Ru, Os, Ag, Hg ...

I guess during electrolysis you have metallic Na formed. I don’t mean junks of Na but Na atoms and some of them may live long enough to amalgamate with the Hg.

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    $\begingroup$ Not every metal amalgamates. See chemistry.stackexchange.com/q/117633/79678. $\endgroup$
    – Ed V
    Jun 23, 2020 at 15:02
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    $\begingroup$ The question was not about Na amalgam nor amalgams in general, but specifically why it or metallic Na is formed at all during NaCl electrolysis with Hg as the cathode. And yes, other metal than mercury obviously cannot form amalgams as they are not mercury. $\endgroup$
    – Poutnik
    Jun 23, 2020 at 15:11

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