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My background is in electrical engineering. Recently I saw instructions on how to manufacture electrolytic capacitors at home by mixing sodium bicarbonate (baking soda) to distilled water, submerging two pieces of aluminum foil into the mixture and applying electrical current through the pieces until current dropped to almost zero. The instructions I followed are here:

https://rimstar.org/science_electronics_projects/make_electrolytic_capacitor.htm

The end result was satisfactory.

However, as I have only very basic knowledge in chemistry, I have only a very vague understanding of what happened, and I would like to understand it in more detail. The article did not explain it very well and I haven't been able to obtain good explanation elsewhere either.

I understand that the current somehow oxidizes the aluminium to form an insulating aluminium oxide layer on top of another one of the plates.

So I would like to pose the following questions regarding this experiment:

  1. Why is distilled water required? I understand that distilled water is basically tap water except its impurities such as charged ions have been removed. This, I believe, makes the water much less conductive. But don't we need the water to conduct? I believe the water, as the electrolyte, acts here as the other plate of the capacitor, with the oxide layer insulating it from the other plate in the final product.

  2. What is the role of the sodium bicarbonate? What exactly does it do in the chemical reaction? We are using distilled water, but then we add sodium bicarbonate to it.

  3. What is the reaction formula of the oxidation? Could you show me how the water, sodium bicarbonate and the aluminium react together to form aluminium oxide, and what is the role of the electric current? I understand that this oxidation of metals, such as aluminium, happens in air as well, and this method simply "accelerates" this process to produce a thicker, more insulating layer.

I would greatly appreciate if somebody could help me understand this process!

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  • $\begingroup$ Excuse me for commenting when I can only guess: 1. distilled water is used to exclude hazardous contaminations 2. sodium bicarbonate makes the water an electrolyte, without Cl that could react with Al in an undesired way. 3. depending on the voltage, Al on the outside will be converted to an insulating layer of Al_2O_3 $\endgroup$
    – Gyro Gearloose
    Commented Dec 31, 2017 at 16:27
  • $\begingroup$ @GyroGearloose I am aware of the molecular formula of aluminium oxide - what I'm interested in is the actual reaction between all the chemicals involved. On points 1 and 2, I would like more detail. But thank you! $\endgroup$
    – S. Rotos
    Commented Jan 2, 2018 at 20:34
  • $\begingroup$ Yes, distilled water has its ions removed. It will still conduct cause you have added sodium bicarbonate to it-those are ions. Regular tap water has many other ions which might interfere in electrolysis, might be electrolytically deposited on your electrodes, etc. $\endgroup$
    – getafix
    Commented Jan 3, 2018 at 1:23

1 Answer 1

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You are performing an electrolysis. This means that on the (+) electrode de reaction is Al $\rightarrow$ Al$^{3+}$ and on the (-) electrode the reverse occurs.

  1. Distilled water is required to avoid the presence of chloride ions which will form chlorine at the (+) electrode and attack your aluminum foil.

  2. Sodium bicarbonate is used to raise the pH so that the formed aluminum ions precipitate as the oxide instead of dissolving as Al$^{3+}$ ions.

  3. The oxidation is Al $\rightarrow$ Al$^{3+}$. Then, the basic pH due to the dissolution of bicarbonate:

    NaHCO$_3$ $\rightarrow$ Na$^+$ + HCO$_3^-$

    HCO$_3^-$ + H$_2$O $\rightarrow$ OH$^-$ + H2CO$_3$

    the aluminium oxide precipitates:

    2 Al$^{3+}$ + 6 OH$^-$ $\rightarrow$ Al$_2$O$_3$ + 3 H$_2$O

The electric current forces the oxidation reaction to continue. In air a thin layer of oxide is formed and then the rest of the metal is not in contact with the air anymore. This is called passivation.

Since the aluminium ions are precipitating as oxide, in the (-) electrode the reduction is of sodium ions:

$\hspace{1cm}$Na$^{+}$ $\rightarrow$ Na

which reacts with water to form hydrogen and OH$^-$ so that the cycle is complete:

$\hspace{1cm}$2 Na + 2 H$_2$O $\rightarrow$ 2 Na$^{+}$ + 2 OH$^-$ + H$_2$

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  • $\begingroup$ As I said I'm not an expert in chemistry so your answer will take me a while to understand completely. A couple of questions meanwhile: How does the increase in pH stop Al ions from dissolving? I understand that pH is a measure of positive ions in the solution, so does their presence (precisely their positive electric field) somehow repel the positive ions making them "stick" to the electrode instead? Second, why does the aluminium ionize in the positive electrode in the first place? Thank you! $\endgroup$
    – S. Rotos
    Commented Jan 3, 2018 at 17:54
  • $\begingroup$ @S.Rotos, it is just Chemistry. I understand that you are not in this field so it will be not to easy to explain in a few words. pH is a measure of H$^+$ ions in solution. When it is above 7 it means that there are more OH$^-$ than H$^+$ ions. Then the neutral species Al(OH)$_3$ will be formed, which, in the case of aluminium, will precipitate as Al$_2$O$_3$, the oxide. This oxide is insoluble so it will stick to the electrode. $\endgroup$
    – Raoul Kessels
    Commented Jan 5, 2018 at 15:28
  • $\begingroup$ @S.Rotos, the positive electrode drains electrons. When you take off electrons from metallic aluminium, you get the positive ion: Al$^0$ - 3 e$^-$ $\rightarrow$ Al$^{3+}$. $\endgroup$
    – Raoul Kessels
    Commented Jan 5, 2018 at 15:33
  • $\begingroup$ Alright, I think I have a much better idea now how it works. I will have to familiarise myself better with some of the concepts involved, but overall your explanation makes sense. I will accept your answer. Thanks! $\endgroup$
    – S. Rotos
    Commented Jan 7, 2018 at 17:32
  • $\begingroup$ Question (also not a chemist): Does this mean, once all aluminum is consumed, the resulting solution is sodium hydroxide + water + possibly leftover bicarbonate, with an undissolved alumina precipitate, and some released hydrogen gas? Am I understanding this correctly? $\endgroup$
    – Jason C
    Commented Oct 21, 2021 at 22:20

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