It is an interesting question. I would like to follow you through your work to understand the aim and the path you follow to reach the goal.

I give an answer which could work, but there is no defense I can give for that answer except to quote the equation which underlies my answer.

Q = C'V, where, Q is the charge on the electrode, C' is the capacity of the interphase ( to hold charge) and V is the electrode potential VS SHE.

If we take 64 as the capacity (C/Volt), and V as 0.158V, we get Q = 10.112C.

Let me reiterate that this is just a guess with not much of defence.

Even if it works, I will not subscribe to it. I have some alternatives to discuss if you be interested. These alternatives are on non conventional lines (new ways of thinking).

Best wishes

I use the formula for a flat capacitor: Q = CV. On the other hand, C =EE0(S/d), where E and E0 - dielectric constants, S - capacitor plate area, d - distance between plates. So, Q depends of voltage, dielectric permiability and geometry of your system.

I use this formula for calculate of concentration of ions in electochemical cell and are resultes quite accurate.

Do you model your cell as a whole (the two electrodes and the ionic conductor) as a capacitor? Or, do you model an electrode electrolyte interface as a capacitor?

I model the whole cell as a capacitor. This can only be done if the resistance of the boundaries is negligible. Therefore, I first check the resistance of the boundaries.

With such a model you can only simulate transient behavior. You will not be able to simulate the working of an electrolysys cell.

I don't understand the aim of your work. If it is to study electrode reactions (electron transfer reactions), I'm afraid if the path is the right one.

Dear Radhakrishnamurty Padyala ,

First of all, thank you for your interest in my work!

I used this approach to simulate the EMF of the cell M / M + / RxTiX2.

I deal with intercalate compounds based on transition metal dichalcogenides. Some of them have high diffusion mobility, which makes it possible to use the methods of solid-state electrochemistry. However, in many intercalate compounds, such as FexTiSe2, CrxTiS2 and others, the diffusion mobility of intercalated metals in the region of moderate temperatures is negligible. Therefore, I undertook the measurement of EMF relatively to metals that are not part of the compound, but possess high diffusion mobility in these materials already at room temperature. These are alkali metals, copper and silver. The EMF of the cell, for example, Cu / Cu + / FexTiSe2 is created by the introduction of copper ions through eelectrolyte Cu +. According to the formulas described, I calculated what should be the concentration of ions to create the observed potential. It turned out that no more than 10 ^ 15 (1 / cm3). This is below the level of uncontrolled impurities.

As you can see, this task is close to the task of Mikko Oksanen.

Cell EMF is a measure of the free energy (delta G) of the reaction. That has no relation to the capacity of the cell. The voltage of a capacitor and the EMF due to cell reaction have no connection.

Make sure that you are measuring what you are seeking.

You are absolutely right about the EMF of the cell. However, the electric field is created by charged particles - electrons and ions. Here I am a strict follower of Faraday. This means that, having an electric field, we can always calculate the number of particles that created it. Another thing is that the number of charged particles is determined by the difference in chemical potentials in the electrodes. However, in no way prevents to count their number.

It is good that you said you are a strict follower of Faraday - as one must be.

In electrochemistry, Faraday's laws concern only with reactions at the electrodes; in other words, with ideally non polarizable electrodes.

Where as the capacity model assumes (or demands) ideally polarizable electrodes.

EMF of cells is concerned with thermodynamics of electrochemical cells.

Charging of the capacitors and dielectric properties of the ionic conductor don't enter EMF of these cells.

Your argument sounds some thing like balancing the electric potential difference between the electrodes by the chemical potential difference between the two faces (sides) of the ionic conductor. Not only one must know the chemical potential of each ion under the conditions of the experiment, but also their number and spacial distribution, in this approach. With such large number of adjustable parameters one can arrive at what ever value of EMF one would like to get!

Under equilibrium conditions, the difference in electric potentials is equal to the difference in chemical potentials for the atoms participating in the reaction. Therefore, the concentration of ions passing through the electrolyte will be determined by the difference in chemical potentials (or the free energy of the reaction). Having thin layers as a cathode and an anode, one can disregard the spatial distribution of ions and electrons in the electrodes. How is this system different from a flat capacitor? The fact that we should neglect the dielectric permittivity of the electrolyte? But it's easy to take into account. I do not see any difference anymore.

Let us try to answer the question by way of thought experiment below.

Suppose we have an ideal capacitor fully charged. We can insert it in one of the arms of a bridge and measure its voltage. Now we replace the capacitor and introduce the electrochemical cell (say, a fully charged battery) in its place and measure the voltage of the cell.

We now ask the question: Does the voltage of the elecirochemical cell, thus measured correspond to its EMF and delta G, or does it correspond to the voltage of the cell as if it is a capacitor?

The difference between an electrochemical cell and a capacitor is that the charge carriers are concentrated in the capacitor strictly on the surface of the plates, whereas in the cell the ions are uniformly distributed over the entire volume of the electrode under equilibrium conditions. Therefore, the dielectric permittivity of the electrode should be added to the formula for a flat capacitor.

However, if the thickness of the electrode is set to zero, then this contribution will also go there.

This concerns the connection between the concentration of charged particles and EMF.

However, there is another important difference. The emf of the capacitor is set by an external source and can be arbitrary up to an electrical breakdown. The EMF of a cell can not have an arbitrary value, but is determined by the difference in the chemical potentials of the atoms participating in the reaction. This difference leads to a flow of ions through the electrolyte to the cathode and the creation of an electric field determined by the condition: neE = - (M1 - M0), where M1 and M0 are the chemical potentials of the atoms participating in the reaction, in the cathode (M1) and in the standard state ( M0), for example, in a compact metal, n is the valency of the M atoms. In my opinion, the electric field strength E in the limit of thin electrodes can be interpreted as the emf of a capacitor. And the value of neE - as work on the movement of a charged particle in an electric field.

Thank you for your answers, mr. Padyala and mr. Titov. I will get back to this within a month or two and report what I did. Currently I'm struggling with other problems that have to be solved before I can proceed to this.

An electrochemical cell consists of two electronic conductors separated by an ionic conductor. It will have two electron conductor - ion conductor interfaces. Each such interface is treated as a capacitor (M/M+). Therefore, we have two capacitors connected in series in an electrochemical cell.

Are we discussing about one of these capacitors (half cell) or are we discussing about the two capacitors combined in series (the cell)?

If we are interested in one capacitor and a single electrode electrolyte interface, the model will be simpler for analysis (at least theoretically).

Let us get this clear before we go further with our discussion.

For simplicity sake we can assume the the electrode to be planar with only one side exposed to the electrolyte. (This assumption will eliminate the complications associated with size and shape of the electrode).

Basically, you, Radhakrishnamurty Padyala, are right. However, one should bear in mind that the capacitances of series-connected capacitors are summed additively. Therefore, the formulas I used remain correct. It is possible that I make the mistake of assuming that the anode is grounded. Probably, this is not correct. Since the anode is not grounded, the charge of ions that enter the cathode must be equal to the charge of electrons remaining at the anode. Consequently, the cathode charge obtained by me from the formulas above should be divided in half.

Thank you for this remark. Hardly I would have noticed this without a discussion. In my case, this does not play a role, since I determine the concentration of ions to within an order of magnitude, however, for Mikko Oksanen it will be significant.

When you have two capacitors in series, the capacitances get added, but the values of C1 and C2 need not be equal. If you have one capacitor then charges on the two plates are equal and opposite, but when you have two capacitors in series, the end plates need not have equal and opposite charges. Therefore, we cannot assume two identical capacitors in series as you appear to argue. This point needs further consideration.

When you imagine (or model) an electrochemical cell as a capacitor, how do you account for or explain the passage of direct current (DC) through the cell in view of the fact that an ideal capacitor blocks DC?

Mr. Padyala,

I am interested only on the half-cell reaction as on the anode. Theoretically, a 2-electron transfer oxidation reaction from SO2 to HSO4- would happen on the anode at 0.158 V vs. SHE. In practice, this reaction is observed in CV scan at values close to 0.6 V vs. SHE. FYI, the reaction happening at cathode is hydrogen evolution reaction, which is naturally 0 V vs SHE.