I don't know anything about double layer theory in electrolytes. However, if there were no electric field in the electrolyte, how could you account for the difference in electric potential between the electrodes ? You would have to assume that the resistance at the interphase between the electrodes and the electrolyte accounts for all the change of potential and that this resistance is infinite. Indeed, if you fail to connect the electrodes of the electrophoresis tank to the power supply, no matter how high the voltage you set on the power supply, the field in the tank will be zero and your samples will not migrate.

Electrolyte contains equal amount of concentration of positive ions and negative ions dissolved in electrolyte. So, they are electrically neutral. The positive ions and negative ions move randomly across the electrolyte because of the thermal energy. So, these charges are mobile like the electrons and holes in a semiconductor.

On applying electric field on the electrolyte these mobile charges drift under the electric field causing what is called drift current which can be expressed by the the Ohm law:

J= sigma E,

So, to support current in the electrolyte, an electric field must exist.

The most important point here is that this electric field is constant so long J and sigma are constant through out the electrolyte.

Therefore E= consntat, According to the electrostatics principles

dE/dx= 0 = roh/epsilon, where roh is the volume charge density, which means that the bulk of the electrolyte is neutral. That is so long as the electric field is constant the material will be free of net electric charge. which means that it will be neutral since it already contains positive and negative charges with equal amount.

On the other side let us assume that we have an electrolyte we inject in it an excess of positive ions only once , these positive ions will repel the native positive ions and attracts the native negative ions and finally it will be neutralized in the bulk and an equal amount of positive charges will appear distributed at the surface of the electrolyte. This neutralization process takes on the average a time equal the dielectric relaxation time.

The last phenomenon is an electrostatic behavior of the electrolyte

where the first phenonenon is the dynamic behavior of the electrolyte.

Current can flow while the electric field is constant which means the material is having no net charge.

net charge free does not mean zero electric field but constant electric field.

Best wishes

Pierre, Abdelhalim, thanks for your instructive answers. But, they are not solving my conflict.

I do agree with Abdelhalim that if there is an electric current through an electrical neutral electrolyte, there must be an electric field. I agree with Pierre, that a lot of our experiments like electrophoresis, conductivity measurements are only possible when there is an electric field.

But for me, this view is in conflict with the double layer theory (https://en.m.wikipedia.org/wiki/Double_layer_(surface_science)). If a charged electrode is placed inside the liquid, it is screened by the free ions (of the opposite charge) and the electric field some Debye lengths from the electrode becomes zero. I must agree that these models are mostly derived in an electrostatic framework. But the double layer is also assumed to be present in many (quasi) stationary approximations. For instance, during conductivity measurements it is assumed that there is a double layer explaining the high capacitive effect of the electrodes (which is the reason that we can produce electrolytic capacitors (https://en.m.wikipedia.org/wiki/Electrolytic_capacitor).

So my conflict can be restated for the case that there is a DC current flowing through an electrolyte. If there are double layers present, how can we still assume that there is an electric field in the bulk? One of the assumptions must be wrong. Or there are no double layers, or there is no electric field in the bulk.

Dear John,

At no bias, the charge picture in metal electrolyte interfaces is that an electric dipole layer is formed. Associated with this dipole layer is a built in electric field and a potential barrier accounting for the work function difference between the metal and the electrolyte.

Now if if a voltage is applied between the electrode and the electrolyte, this voltage will be partly dropped on the metal electrolyte junction and on the bulk of electrolyte. The voltage distribution between the junction and the bulk depends on the current passing in the electrode.

In case of forward bias, the potential barrier height at the interface will be smaller meaning the decrease of the charges in the dipole layer leading to higher charge transfer rate at the interface and a flow of an appreciable current across the interface layer. This current must continue flowing in the neutral bulk which causes an ohmic voltage drop as i described previously.

This ohmic voltage drop must be supplied also by external voltage source applied on the electrode.

So, the total charge on the electrode is that the charge neutralizing its counter charge in its dipole layer and also sourcing the electric field in the neutral region.

In more clear words, the charge in the electrode must neutralize its counter charge in the dipole layer in the electrolyte in addition it must source the electric field lines in the neutral bulk region to affect the ohmic current.

This picture applies also for may types of junctions including the p-n junction.

Best wishes

Thanks Abdelhalim for your interesting respons. I like very much the analogy with diodes. But there is a problem with that. We have two electrodes (so two diodes) that are oppositely polarised. So if one electrode gets a forward bias, the other one gets an opposite bias.

The net result will be no current (like there is no current running through oppositely biased diodes). The voltage is standing completely over the diodes and the field in the bulk is zero.

So, I fear that something is still missing.

Dear John,

I am happy for your positive response on my post.

My answer is conceptual . So, an electrolyte cell is composed of two electrodes, one can speak of a half cell which is an electrode. So a cell can be considered as two electrodes connected in series. To pass certain current in the cell one has to drop a specific voltage on each electrode and also on the electrolyte to cover its ohmic voltage drop irrespective of its value. When the current is very small the ohmic voltage drop will be also small when it zero, the ohmic drop will be zero what is foul in that?

Best wishes.

Dear Abdelhalim, there is nothing foolish on the fact that the electric field is zero. The only problem is that that this seems to be always the case. It is not possible to get a reasonable DC Electric field in the bulk water without creating extra ions (like during electrolysis). Indeed, you are describing two half cells, but they are always oppositely polarised, prohibiting a field in the bulk. To my view, there is no such thing as DC electric conduction in electrolytes. Electric conduction is always measured in AC or as a transient phenomenon.

Dear John, isn't diffusion the answer to this? The electrode reactions are creating a concentration gradient which drives the current also when the field is zero. But I think the field is not screened as well as in a metal for instance, so the field is perhaps never zero?

In the meantime I have had the time to think about this question more deeply. I try to formulate my best answer here:

You should consider two cases. 

• Case one :  when blocking electrodes are used (so no charges of the electrodes can come into the water and form H+ and OH- ions). This case is well described in the paper of Morrow (Morrow, R., & McKenzie, D. R. (2012). Proceedings of the Royal Society a-Mathematical Physical and Engineering Sciences, 468(2137), 18-34. doi:10.1098/rspa.2011.0323). Notice that even for this simple case, the double layer is formed by a complex set of equations (2.1 and 2.2) describing charge conservation and consisting of the ionic drift and diffusion terms, the dissociation rate of the water molecules in ions and the recombination rate of H+ and OH- ions. You can see that for this case after a decay time, the electric field is screened by the double layers (see Fig 1-3). This is a rather slow process (time constants of seconds), meaning that in AC (kHz) the double layers can’t be formed fast enough and the whole sample is exposed to the electric field. This is the way the conductivity of bulk water and electrolytes is mostly measured (in AC), so the double layer isn’t important.
• Case 2: when you use normal non-blocking  electrodes and voltages higher than 2V electrolysis will occur near the electrodes (forming H+ and OH- ions and the gasses H2 and  O2). So, there will be extra terms in Equations 2.1 end 2.2 describing the production of H+ ions near the positive anode and OH - ions near the negative cathode. However, due to principle of electroneutrality, in steady state the excess of charges has to be compensated by the migration of the dominant charge carriers (Corti, H. R., & Colussi, A. J. (2009). Langmuir, 25(11), 6587-6589. doi:10.1021/la900723t). In pure water there is only a small amount of compensating free charges present (H+ and OH-). A significant electrolysis current is therefore only possible in electrolytes when other charges are present (eg Na+Cl- or Na+OH-). The question is now: what does this process do to the double layer? Take the anode (positive) and assume that there is a double layer of negative ions present. There is a continuous production of H+ ions near the anode. These charges need to be compensated, which of course will preferably happen with the negative charges of the existing double layer. That is why the existing double layer is destroyed and the electric field will fill the whole specimen. This field will help to bring extra negative ions to the anode allowing a steady state production. So due to the steady state electrolysis, the double layers disappear and there is an electric field all over the liquid. This field will also push the ions produced by the electrolysis inside the liquid. You can see this effect easily if you put some universal indicator in your liquid (see for instance the experiments from Ovchinnikova (Ovchinnikova, K., & Pollack, G. H. (2009). Langmuir, 25(1), 542-547. doi:10.1021/la802430k) (PS: the reasoning  of Ovchinnikova is not correct (see paper of Corti) but the experiment is and is easily reproducible). 

Although this is an old thread, I still will post another answer that has a more ELI5 response, namely a response written by COMSOL that discusses the Double Layer effect on both ends of the two electrodes, and how current flows from one side to the next. It would be interesting to see how this relates to the Debye Length you speak of, as this is not addressed in the article.

https://www.comsol.com/blogs/does-the-current-flow-backwards-inside-a-battery

Thanks Patrick, this is a nice link, explaining very visually what I meant with words.When there is no current, there is a zero electric field in the electrolyte (see Fig 4). When there is a load, there is a current and the extra charge of the electrochemical reactions near the electrodes will changed the zero filed to an opposite electric field (see Fig.5). This will 'pump up' the positive ions from the negative to the positive electrode and vice versa.

The Debye length determines the thickness of the double layers. Whenever the current stops, it is the length near the electrodes where the electric field is nonzero.