Below is a schematic drawing for a three-electrode cell used for cyclic voltammetry:

enter image description here

Image modified from Wikipedia Voltammetry

This is what I understand with the voltammetry working principle from self-study, please correct me if I am wrong.

At WE, a half-reaction, $\ce{Ox + e- <=> Red}$ is modulated by the potential applied on the WE. Electrochemical equilibrium can be assumed at the surface of WE. When the potential at the WE is increased, WE draws electrons out from the analyte, equilibrium position shifted towards the Ox. When potential decreased, WE 'injects' electrons into the analyte and the equilibrium position shifted towards the Red. In voltammetry, the potential varied continuously, and it is this continuous shifting of equilibrium position (that involves giving or accepting electrons) leads to motion of electrons and hence gives rise to Faradic current.

Using the quoted understanding above (which might be wrong), I couldn't answer the following questions:

  1. When you increase the potential, electrons got drawn into the WE then go towards the CE via external circuit. What will happen when the drawn electrons reach the CE? Will they get injected back into the solution? If yes, what species will accept the electrons?

  2. A typical reversible CV will have an onset potential, beyond which the current increases sharply. If the argument above is correct, even a slight increase of potential should already lead to shifting of equilibrium and hence Faradic current, and so there should be no onset but rather a direct increase in current as soon as potential is applied.

  3. Say you keep changing the potential applied, the equilibrium position will be shifted continuously and hence Faradic current observed. What if you stop changing the potential at one point, but hold the overpotential at a non-zero constant? What will be observed? 0 current?

  4. When there is no redox-active species in the electrolyte, current will increase very slightly when potential increased, and that is due to non-Faradic current (capacitive current), due to ions accumulating at WE surface. I don't see how ions accumulating at WE surface will lead to current flowing between WE and CE and hence gives rise to non-zero current.


1 Answer 1


Consider for a moment an electrochemical reaction where we can (for small overvoltages) us the Butler–Volmer equation to predict the current.

I = Io {(exp ka E)-(exp -kc E)}

Where E is the overvoltage and ka and kc are two constants (in real life they are the combination of many constants for the electrode). As long as the current is not limited by mass transport being too slow we can use this equation. This current is the Faradic current which flows as a result of the chemical change.

The surface of the electrode is constantly bombarded with either the reduced or oxidized form of the electroactive "thing", the probaility of the thing reacting when it hits the surface is a function of the potential of the electrode vs the bulk of the solution it is in. This current is a function of the potential of the electrode.

As we change the potential of the electrode relative to the bulk liquid which it is immersed in then we will have a current (Non - faradic) which flows due to the change of the potnetial difference between the two terminals of the capacitor which is present. The electrode is one of the two terminals of the capacitor while the liquid it is immersed in should be regarded as the other one.

Here for this current we have

I = C (dE / dt)

This current is a function of the rate of change of the electrode's electrical potential.

So for a real electrode with a small current flowing we can write the following equation for the total current.

I = (C (dE/dt)) + Io {(exp ka E)-(exp -kc E)}

Now lets deal with sub questions 1 to 4

  1. Imagine we have a solution of [Fe(CN)6]4- (1 mM) in 1 M NaNO3, now imagine that at the working electrode one molecule of [Fe(CN)6]4- is oxidized to [Fe(CN)6]3-. What will happen is that one electron from the counter electrode is needed to perform some electrochemical event there. If by luck some [Fe(CN)6]3- has diffused there then this could be reduced. As an alternative a proton could be converted into a hydrogen atom. Depending on the nature of the counter electrode this could be adsorbed onto the surface (if it is Pt) or it could combine with another hydrogen atom to form a hydrogen molecule. There are other things which could be reduced at the counter electrode such as dissolved oxygen if the chemist has failed to deoxygenate the cell.

  2. As I wrote above the Butler-Volmer equation predicts that the current will change suddenly as we pass through the point where E is zero.

  3. If we maintain the potential at a fixed value then the non faradic current will become zero, however the faradic current will continue to be dictated by a combination of mass transport and the Butler-Volmer equation. When a reaction is occuring at the electrode then the small volume of liquid around it will have different concentrations of the reduced and oxidized forms as the bulk. What will happen is that diffusion will move molecules of both forms between the bulk of the liquid and the special zone around the electrode. The more the cell is stirred the thinner the stagnant zone is around the electrode.

  4. If there is no electroactive species in the cell and as we increase the potential of the workeing electrode then ions are repelled or attracted to / from it then at the counter electrode a similar but opposite number of ions will be repelled / attracted thus allowing charge to flow through the charging effect.


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