I've been involved in some analytical electrochemistry experiments to determine the reduction/redox potential of some compounds I'm working with. Cyclic voltammetry of two different analytes has shown them to have differential midpoint potentials of, let's say, +10mV and +150mV (mV vs NHE). Having a rather limited knowledge of electrochemistry, I understand than thermodinamically speaking the first compound with lower potential would be more redox active, as the difference of potential with hydrogen (~800mV) would be larger and ΔG = -nFΔE. At least, this is how I've understood it. However, a collaborator has mentioned that a third analyte with a reduction peak about 100mV more positive (~250mV) would be easier to reduce. I might have misinterpreted what I've been reading, but as a non-chemist I'm quite baffled by this statement, so any clarification would be much appreciated. Thanks! :)
If you can post a generic picture of the CV, it will be helpful. I am afraid that the question is little unclear. Couple of questions: Is it a mixture of two analytes or three analytes. How did you come up with 800 mV?
Apologies if it seemed unclear. I'm talking about 3 different analytes with their respective CVs. My main doubt was if, in general terms, one could argue which of them would be more redox active by comparison of their midpoint potentials.
I'm not sure what you want to say by "redox reactivity". What the redox potential can normally indicate is how easy it is to reduce or to oxidise something, however "activity" can be measured in many ways. The standard potential tells you about how much energy is required for the process, however, if you do cyclic voltammetry, there might be other influences on you "midpoint potential", such as kinetic factors, for example. In any case, without knowing your particular details, I would tend to agree with your collaborator, the more positive the midpoint potential, the easier it is for the reduction to take place (ie the more likely your compound is in the reduced form).
Hi Daniela, thanks for your reply. I said reactivity meaning that the species may be more easily reduced or oxidised, as you mentioned. I'm aware of the influence of kinetic factors, but I mainly sought to understand what midpoint potential indicated. Thanks again.
The lower the value the easier is the reduction. For a reversible system it is the midpoint but for an irreversible voltammogram, you have just the reduction potential and a lower value means easier reduction.
Thank you for your reply, although unless I'm mistaken it sounds like two opposite arguments (i.e., the more positive the easier the reduction vs. the lower value the easier the potential)?
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