Electrochemistry: rate constant dependence on CV peak

[MatthiasS]

Hi there,

I am currently studying charge transfer reactions between metallic electrodes and redox couples in aqueous electrolytes (in a diffusion limited system), and I was wondering about the influence of the rate constant on the reduction/oxidation peak maximum. Typically, it is assumed that a decreasing rate constant results in a greater reduction/oxidation peak separation and also in a reduction of the peak current values (see attached picture). I get why the peak separation increases for slower rate constants, but I do not understand what causes the decrease of the peak maximum (since the diffusion of the electroactive species in the electrolyte should limit the maximum current value). Does anybody have an idea what causes the decreasing peak currents?

Thanks a lot for your help!

Cheers,
Matthias

 

[HAYAO]

I'm not an electrochemist, but I don't think the electron transfer rate constant would say anything about peak separation between red/ox in the microscopic level. The redox potential of a molecule and an electrode is a fixed parameter in absence of external potential. If I were to say anything about rate constant and redox potential, I would probably talk about it the other way around. It is the applied potential that causes change in rate constant.

When you apply voltage to an electrochemical cell, the redox potential of the electrode changes depending on the applied voltage. This allows electron transfer TO the molecule when electrode potential is low enough relative to the molecule's reduction potential, and electron transfer FROM the molecule when the electrode potential is high enough relative to the molecule's oxidation potential. The rate constant itself corresponds to such "potential matching" (I don't know the technical term for this) and electron correlation between the molecule and the electrode. Typically, electron correlation between a molecule and electrode remains the same unless the molecule as a specie changes its property (e.g. after reduction or oxidation). In the particular case of the OP, I would assume that for all measurements, same molecule/electrode combination is used. In another words, it is not the rate constant that changes the CV, but the applied voltage. It is quite awkward when you say "changing the rate constant".Under this in mind, if you are talking about changing the molecule/electrode combination under hypothetical situation where these species have same red/ox potential but have different electron correlation, then I think we will be discussing whether the experimental conditions of CV are appropriate or not. In the case where the CV scan speed is too fast, if:
1) electron correlation is large, then the electron transfer rate between molecule and electrode is fast enough that the CV curve shows the true red/ox potential of the molecule.
2) electron correlation is small, then the electron transfer rate between molecule and electrode is too slow that it cannot catch up to the speed of applied voltage. I think in such case, the CV peaks will be separated than it is supposed to, with much broader peak.

Is this what you are talking about?

 

[Borek]

It is the applied potential that causes change in rate constant.

Observed - yes. Absolute - no. Matter of a point of view.

 

[HAYAO]

Observed - yes. Absolute - no. Matter of a point of view.

Well you are right, but from the OP I had no other way to interpret the question.

 

[MatthiasS]

Thanks aa lot for your answer!

Yes, I was thinking about a hypothetical situation in which we have the same red/ox potential, but different rate constants.
The reason for my question is that I am interested on how a thin surface layer (sufficiently thin to allow for electron tunneling) affects peak separation and peak current of the CV curves. Typically, depending on the thickness of the surface layer, this layer is expected to decrease the rate constant (as far as I know this effect is even used to investigate the charge transfer kinetics of very fast reactions which would otherwise be difficult to measure due to diffusion limitations).

When I checked the literature on CV measurements in presence of thin surface layers, it seems that typically (i) the peak separation increases and (ii) the peak current density decreases (compared to the unmodified electrode). I agree that for slow scan rates the electron transfer rate between molecule and electrode might be too slow to catch up to the speed of the applied voltage (which results in an increased peak separation), but I do not understand why this also affects the peak current density (since the peak current density should depend only on scan rate as well as on concentration and diffusion coefficient of the redox couple).

 

[Borek]

I can be wrong, but judging from what is written in Theoretical Principles of Electrochemical Analysis (by Z. Galus, unfortunately I have it only in Polish), lower peaks are already predicted by the theoretical solutions given by Matsuda and Ayabe in Z. Electrochem, 59, 494 (1955).

http://onlinelibrary.wiley.com/doi/10.1002/bbpc.19550590605/abstract

 

[HAYAO]

Thanks aa lot for your answer!

Yes, I was thinking about a hypothetical situation in which we have the same red/ox potential, but different rate constants.
The reason for my question is that I am interested on how a thin surface layer (sufficiently thin to allow for electron tunneling) affects peak separation and peak current of the CV curves. Typically, depending on the thickness of the surface layer, this layer is expected to decrease the rate constant (as far as I know this effect is even used to investigate the charge transfer kinetics of very fast reactions which would otherwise be difficult to measure due to diffusion limitations).

When I checked the literature on CV measurements in presence of thin surface layers, it seems that typically (i) the peak separation increases and (ii) the peak current density decreases (compared to the unmodified electrode). I agree that for fast scan rates the electron transfer rate between molecule and electrode might be too slow to catch up to the speed of the applied voltage (which results in an increased peak separation), but I do not understand why this also affects the peak current density (since the peak current density should depend only on scan rate as well as on concentration and diffusion coefficient of the redox couple).

I don't understand what you are saying. You've just answered your own question.