I am not a chemist but I have a question regarding electrolytes in electrolysis. I have done a lot of reading but cannot find the answer I'm looking for regarding Sodium Sulfate (or Sulfide)

I intend to clean a film off a Stainless Steel plate, so that will be my cathode.

For my Anode I will be using a Platinum coated Titanium plate.

At a small scale I have been using Sodium Carbonate (Na2CO3) and it has worked fine, however I intend to scale it up to ~1m^2 plates for the Anode and Cathode, as well as 150A @ 15v DC for ~3min in Deionised water with many repeats, so temperature rise of the water should also be accounted for.

I would like to make this as safe/clean as possible and am considering switching out the electrolyte at this stage to something more stable. I have read that Sodium Carbonate can precipitate directly onto the electrodes at high currents. I believe this is not an issue with Sodium Hydroxide but want to avoid that for obvious reasons. I have stumbled upon Sodium Sulfate as an alternative but very little information on its breakdown and potential pollutants. Is there any other that I am missing/suggestions?

I will be using a fume hood, however Sodium Chloride is definitely not an option.

Thanks for your time,



So I have done some reading and have mostly worked out the half equations of Sodium Sulfate (Na2SO4):

Cathode: (2)H+ + (2)e- -> H2(g)

Anode: (2)O2- -> O2(g) + (4)e-

Remaining in aqueous solution: NaOH (aq) + S (?)

As an update then, can I ask what, if anything, happens to the NaOH and Sulfur? Also, what form will the Sulfur remain in? I assume it will try to re-bond with oxygen, which will be abundant in the system? Will either be outgassed?

Assuming the NaOH is constantly building up, at what point does it become overwhelming in the system and need to be removed?

Thanks so much for reading this far.



If you electrolyze a solution of sodium carbonate, it will be fine at the beginning. Later on, some problem may happen around the anode where $\ce{O2}$ is released. Indeed some acid ($\ce{H+}$) is produced in solution according to the half-equation : $$\ce{2H2O -> 4 H+ + O2 + 4 e-}$$ and these acidic ions $\ce{H+}$ react with the carbonate ions $\ce{CO3^{2-}}$according to $$\ce{CO3^{2-} + H+ -> HCO3^- + H2O}$$ In presence of the ion sodium $\ce{Na+}$, the ion hydrogenocarbonate $\ce{HCO3^-}$ reacts to produce sodium hydrogenocarbonate $\ce{NaHCO3}$ which is not very soluble in water, and may make a white deposit around the anode $$\ce{Na+ + HCO3^- -> NaHCO3(s)}$$ Furthermore if the electrolysis is continued a long time, a supplement of acid ions will react with these $\ce{HCO3^-}$ ions producing some $\ce{CO2}$ by the reaction $$\ce{H+ + HCO3- -> CO2 + H2O}$$

$\ce{CO2}$ is a gas, and it is relatively soluble in water. In the beginning the $\ce{CO2}$ produced will stay dissolved in the liquid. But on the long run, $\ce{CO2}$ will get out of the solution and get mixed with the oxygen produced at this electrode. So instead of getting pure $\ce{O2}$ at the anode, you will obtain a mixture $\ce{O2 + CO2}$.

So you better use a solution of sodium sulfate $\ce{Na2SO4}$ instead of sodium carbonate. The above mentioned drawbacks do not occur with sodium sulfate.

In a $\ce{Na2SO4}$ solution, the ions $\ce{Na+}$ are moving towards the negative electrode (cathode), and the ions $\ce{SO4^{2-}}$ are moving toward the positive electrode (anode). They move but they don't touch their electrodes and they are not discharged. Instead that is water that either give or absorb electrons.

At the negative electrode (cathode) where $\ce{Na+}$ ions are approaching, the following reaction occurs : $$\ce{2 H2O + 2 e- -> H2 + 2 OH-}$$ and the $\ce{OH-}$ ions neutralize the approaching $\ce{Na+}$ ions. Of course some $\ce{H2}$ bubbles are appearing on the cathode.

At the positive electrode (anode), the following reaction happens $$\ce{2 H2O -> O2 + 4 H+ + 4 e-}$$ and the $\ce{H+}$ ions neutralize the approaching $\ce{SO4^{2-}}$ ions. Of course some $\ce{O2}$ bubbles are produced on the anode. Half as much as $\ce{H2}$, because $4$ e- produce $1$ $\ce{O2}$ and $2 \ce{H2}$

To summarize, electrolyzing a $\ce{Na2SO4}$ aqueous solution will produce $\ce{H2}$ and a solution of $\ce{NaOH}$ at the cathode. And it will produce $\ce{O2}$ plus a solution of $\ce{H2SO4}$ at the cathode.

Now if the anodic solution is stirred and mixed with the cathodic solution, the following reaction happens $$\ce{H2SO4 + 2 NaOH -> Na2SO4 + 2 H2O}$$ As a consequence, $\ce{Na2SO4}$ is regenerated, and the only chemical that is consumed in the whole process is water. The only product getting out of the whole process is the formation of $\ce{H2}$ at the cathode and $\ce{O2}$ at the anode.

  • $\begingroup$ Thank you Maurice, for the detailed breakdown. If you have time and for curiosities sake, could you break down what happens with Sodium Sulfate under these conditions? Appreciate your time, tea Edit: Also, would you suggest a ratio of electrolyte to De-Ionised water? I have been running Sodium Carbonate at aprox 5%. $\endgroup$ – Tea1023 Oct 28 '20 at 8:33
  • $\begingroup$ I have updated my question with an edit after doing some more homework. I want to thank you again for your time, and if you have any more please could you give the edit a quick read/comment. $\endgroup$ – Tea1023 Oct 28 '20 at 10:20
  • $\begingroup$ @Tea1023. The electrolysis of a $\ce(Na2SO4}$ solution does not produce any sulfur. And this solution contains neither $\ce{O^{2-}}$ nor $\ce{H^+}$ ions. So all your chemical concepts are pure fantasy. The electolyzing a sodium sulfate solution, the ions $\ce{Na+ + SO4^{2-}}$ are not modified. They jus approaches their respective electrodes without being chemically modified. I continue this comment as an edit to my previous answer. $\endgroup$ – Maurice Oct 28 '20 at 17:34

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