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Cyclic voltammetry of GOX

  1. Jun 24, 2014 #1
    Hey, this is not a homework question really but more a research issue my fellow students and I have run into.

    So basically, we have a project where we have cross-linked glucose oxidase to a polypyrrole surface on a gold electrode. The solution additionally contain PBS as well as ferricyanide as a mediator. The electrode with polypyrrole and GOX is the working electrode.

    We run several laps of cyclic voltammetry and at cycle 16, we add glucose. We run several of these experiments at several concentration. three cycles have been handpicked, with 30mM and 40mM glucose have been added for your viewing.

    Now we analyse how the current output at 0.6V(forward scan) depends upon glucose addition when we add it at the beginning of cycle 16 (-0.1V forwards scan), which is the third image. (scan rate is 150mV/s)

    The last image illustrates the peak current when we add glucose compared to right before.

    The problem we are facing is that lower concentrations of glucose added results in a higher output peak relative to higher concentration. This seems pretty counter-intuitive since we would think that a higher conc. of ferricyanide would get reduced at higher conc. of glucose which would lead to a greater oxidation current. We are thinking that there is something kinetics based that we are overlooking.

    Hopefully, you have some ideas.

    Attached Files:

  2. jcsd
  3. Jun 24, 2014 #2


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    Staff: Mentor

    Looks interesting, can't say I understand what is happening in the solution (I mean - what is the chemistry behind the system and the electrode reaction). Care to elaborate?

    I must admit I have not dealt with these things for almost 30 years.
  4. Jun 24, 2014 #3
    Well, hopefully we should have glucose oxidase getting reduced (glucose oxidised). The point of the voltammetry is switching between ferri/ferrocyanide redox couples. Ferricyanide should hopefully work as a mediator (getting reduced to ferrocyanide) transport electrons to electrode and get oxidized to ferricyanide once again. The point was introducing a disturbance into this redox couple when adding glucose and see how this would work out. In our understanding, the dependence should be the exact opposite of what we see. We understand that when we do a forward sweep we are oxidizing any possible species at the working electrode, and when we do the negative sweep we should be reducing species.

    We are also wondering what the nature of the current developing at around 0.350 V (backwards sweep). Something is getting reduced, which in our opinion can only be ferricyanide or maybe H2O2 (byproduct when breaking down glucose)
  5. Jun 24, 2014 #4


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    Staff: Mentor

    So the basic chemistry is

    glucose + ferricyanide -> D-glucono-1,5-lactone + ferrocyanide + H2O2

    and electrode reaction

    ferrocyanide -> ferricyanide

    Or am I still missing something?

    Don't you need oxygen to produce H2O2?
  6. Jun 24, 2014 #5
    yes, you need oxygen but that should not be a problem according to articles. The reactions you listed are correct.
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