Sciencemadness Discussion Board

Misconceptions of electrolysis

sakshaug007 - 9-5-2009 at 17:37

Hello everyone,

So apparently I have some misconceptions about the basics of electrolysis that need cleared up. I'm going to start with the example of water using NaOH as the electrolyte. My first misconception is that the sodium ions increase conductivity by providing electron transport from the cathode to the water as follows: Na(+) + e- => Na and then Na + H2O => Na(+) + OH(-) + H2 (observed at the cathode). I was informed that this is actually a "side reaction" that causes a decrease in efficiency (more current consumed). My question is how is this the case if the sodium produced reacts with surrounding water thus producing more hydrogen and therefore contributing to efficiency.

Also how is it neutral molecules (nonionic) can react with electric current at electrode surfaces anyway? I understand the concept of overpotentials but if the sodium ions are in fact not supposed to react with the electrons at the cathode how is it they are providing conductivity at all because the sodium ions would simply migrate to the cathode where they would remain thus creating a capacitive effect and therefore no current would flow.

Everything I've read simply says the addition of an electrolyte will increase conductivity but they don't provide a mechanism as to how!? Yes ions migrate to their respected electrodes but then what?

I'm sure these are stupid questions for those of you knowledgable on the subject I just need qualitative clarification as to what's going on.

Thank you very much

[Edited on 10-5-2009 by sakshaug007]

hissingnoise - 10-5-2009 at 04:41

Quote: Originally posted by sakshaug007  
Yes ions migrate to their respective electrodes but then what?

Let's see!
A positive ion gains an electron at the cathode and a negative ion loses an electron at the anode.
The cathode loses electrons, while the anode gains electrons and this constitutes an electric current flowing in the solution.
Hydrogen ions, having a lower 'discharge potential' than Na ions, gain electrons and leave the solution as H2.

[Edit] Neutralisation of OH ions at the anode causes evolution of O2.




[Edited on 10-5-2009 by hissingnoise]

[Edited on 10-5-2009 by hissingnoise]

sakshaug007 - 10-5-2009 at 10:49

@hissingnoise: Okay so I understand that the OH- ions from the electrolyte will migrate to the anode and oxidize to O2 and H2O but where are the H+ ions coming from? The water right? But in that case does that mean the H2O is getting directly reduced at the cathode even though it is not a charged molecule?

Thanks for the reply

hissingnoise - 10-5-2009 at 11:15

Yes, even pure water ionises to some small extent, but when NaOH dissolves ionisation is effectively complete so that the solution now contains Na + H3O + OH ions.
There are very few, if any, neutral water molecules in such solutions. . .
Na ions at the cathode remain unchanged because they have a higher discharge potential than H ions.

sakshaug007 - 11-5-2009 at 15:51

Moving on to a more complex neutral molecule such as chloroform, does this also have to self-ionize to some extent in order for the electrolysis of it to occur thus producing poly(hydridocarbyne)? (using acetonitrile as a solvent and sodium chloride as electrolyte).

Arrhenius - 11-5-2009 at 20:30

Quote: Originally posted by hissingnoise  
Yes, even
There are very few, if any, neutral water molecules in such solutions. . .


Water is still nearly half in the form of H2O in a saturated solution of NaOH at room temperature. The calculation is complicated slightly by the change in volume of the solution due to the solute.

sakshaug007: No, a molecule does not need to ionize first before it can undergo a reaction at either electrode. However, it's redox potential comes into play relative to what else is in solution; this will determine what is electrolyzed first. Frankly, most electrochemical reactions with organic molecules don't work terribly well in practice; hence they're not widely used.

woelen - 12-5-2009 at 12:54

Quote: Originally posted by hissingnoise  
Yes, even pure water ionises to some small extent, but when NaOH dissolves ionisation is effectively complete so that the solution now contains Na + H3O + OH ions.
There are very few, if any, neutral water molecules in such solutions. . .
Na ions at the cathode remain unchanged because they have a higher discharge potential than H ions.
I totally disagree with this. In a soluiton of NaOH, there are VERY few H3O(+) ions, much less than in plain water.

Arrhenius' explanation is correct. Even although Na(+) has a positive charge, water molecules more easily accept an electron than Na(+) ions.

In general, there are many candidates of compounds, capable of accepting electrons (from the cathode) and dontating electrons (to the anode).

At the cathode the candidates for accepting electrons are:
1) water molecules
2) cationic species, in this example Na(+)
3) anionic species, in this example OH(-)
4) the cathode material

ad 1) Water molecules would accept electrons, going to H2O(-), which at once falls apart to OH(-) and H.
ad 2) Na(+) would accept electrons, going to Na
ad 3) OH(-) would accept electrons going to OH(2-)
ad 4)The cathode material would accept electrons and become M(-), where M is the cathode metal.
Of these 4 possibilities only 1 and 2 seem plausible and from these 2 the most easiest is 1. At very high electrolysis voltage, reaction 2 also will occur to a minor extent.
With some cathode materials, reaction 4 also can occur. E.g. if a tellurium cathode is used, then telluride ions are formed.

At the anode the following candidates exist for donating electrons:
1) The water molecules
2) The cationic species, in this example Na(+)
3) The anionic species in solution, in this example OH(-)
4) The anode material
In the case of electrolysis of NaOH-solution with copper anode, reaction 3 occurs most easily and this results in formation of OH, which in turn results in formation of water and oxygen.

If e.g. sodium chloride were used instead of sodium hydroxide, then the easiest reaction is reaction 4, the copper atoms easiest donate electrons, and the anode erodes. If a platinum electrode is used as anode and NaCl as dissolved material, then the easiest reaction is donation of electrons by Cl(-), but a noticeable side reaction occurs, in which H2O donates electrons, especially if high voltage is applied.


So, summarizing, list all candidates of electron acceptors near the cathode and electron donors near the anode and lookup redox potential tables in order to determine which most likely occur.

hissingnoise - 13-5-2009 at 03:00

Ooops! Looks like I have a few misconceptions of my own. . .
Thanks Arrhenius and woelen for the correction---I let runaway assumptions get the better of me.
sakshaug007, IMO, woelen has a better understanding of electrolysis than anyone here.
I'm learning too. . .

Nicodem - 13-5-2009 at 04:05

Most was already explained, so I’ll be short.
Quote:
My first misconception is that the sodium ions increase conductivity by providing electron transport from the cathode to the water as follows: Na(+) + e- => Na and then Na + H2O => Na(+) + OH(-) + H2 (observed at the cathode). I was informed that this is actually a "side reaction" that causes a decrease in efficiency (more current consumed). My question is how is this the case if the sodium produced reacts with surrounding water thus producing more hydrogen and therefore contributing to efficiency.


Two main effects that reduce electrolysis efficiency are overvoltage (polarization of electrodes to a potential higher than the redox potential of the reaction) and heat loss due to (lack of) conductivity of the electrolyte. The power used in the hydrolysis is like in every electric work just the simple P=U*I equation.
Therefore, lets assume for an example that the overvoltage for the 2H2O + 2e(-) => H2 + 2OH(-) reaction for some reason increases up to that of the Na(+) + e(-) => Na(0) reaction and thus some sodium is being produced while at the same time being quenched by H2O. You now have two competing reaction at the new potential of U(2), both leading to the same H2 evolution. Overvoltage can be raised by increasing the current per electrode surface (A/m^2), by the electrode material or by the solvent used (but still the potentials for the two mentioned reactions are so far away that this can only be a hypothetical situation so treat it as such). So the U(2) must be much higher than the U(1) in the reaction where no Na(+) ions get reduced due to the overvoltage of H2O reduction being lower. The overall power consumed would now be P=U(2)*I which is more than P=U(1)*I since I stays the same for the same flow of produced hydrogen (due to Faraday’s constant and such). As you can see a hypothetical water hydrolysis where Na(+) is involved in electrochemical reactions is much more power wasteful.
Quote:
Also how is it neutral molecules (nonionic) can react with electric current at electrode surfaces anyway?


The same way they react in other redox reaction. For example, the opposite reaction of water electrolysis, that is O2 + 2H2 => 2H2O requires no ionic intermediates. It is a purely non-ionic type of reaction. Redox reaction do not require ions, they require electron flow from one species to the other, or in the case of electrochemical reactions, from the electrode to the substrate or vice versa.
Quote:
I understand the concept of overpotentials but if the sodium ions are in fact not supposed to react with the electrons at the cathode how is it they are providing conductivity at all because the sodium ions would simply migrate to the cathode where they would remain thus creating a capacitive effect and therefore no current would flow.

Everything I've read simply says the addition of an electrolyte will increase conductivity but they don't provide a mechanism as to how!? Yes ions migrate to their respected electrodes but then what?


This was already explained so I'll just add something else. Imagine an electrolytic cell without an electrolyte. Just two electrodes in a noncunducting solvent. Since there can be no current (unless the potential is above the breaking point) there can also be no electrochemical reactions at the electrodes. There will only be an electric field between the electrodes (E=U/d where d is the distance between the electrodes), but since there is no electrolyte to compensate for the charge which would form in the 2H2O + 2e(-) => H2 + 2OH(-) or any other reaction, the electrode is polarized and therefore no reaction can occur. Essentially, any increase at the electrode potential is compensated by the exact same polarization counterpotential unless the solution is conductive.

The case of electroreduction of CHCl3 is in its essence the same as that of electroreduction of H2O. You need a solvent that will not get reduced faster than CHCl3 and you need an electrolyte and electrodes that will not get involved in the electrochemical reactions.

But then, electrochemistry is not my field and this is only from memory of what we were thought at school. Since there is quite likely some nonsense in what I said and since all this is surely better explained in review papers and books, I suggest to rather thrust scientific literature than these second hand explanations.

[Edited on 14/5/2009 by Nicodem]

sakshaug007 - 13-5-2009 at 18:25

Thank you all for your responses, they have really cleared up my misunderstandings of what occurs during electrolysis. As electrochemistry is my intended niche I'm glad to be getting a better understanding of it.

Thanks again.