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On Arrhenius plots in cyclic voltammetry

  1. Oct 13, 2009 #1
    I am doing a catalytic study on my Pt nanoparticles. My experiment set-up is a three-electrode cell with sulfuric acid as electrolyte for methanol electrooxidation reaction. Now, i want to calculate the apparent activation energy and for that I need to get the voltammograms at various temperatures (eg. between 30oC - 60oC). After that I have to plot log i(current density, mA/cm2) vs 1000/T(in K) at a certain potential (eg. 0.6V) and then from the slope i can calculate the activation energy (this is based from a paper i read). I am having some doubts as to how to settle with the values of i because I`m kind of confused with the conversions and units like if I "log" the unit what will happen to "mA/cm2" and the like. Also, should I use the actual current density reading from the CV plot as is or do i need to consider other things? anyone with experience on this area, please kindly share some insights. Thanks.
     
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  3. Oct 13, 2009 #2

    Mapes

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    If you're just looking at the slope, then you'll be calculating [itex]\log I_2-\log I_1=\log(I_2/I_1)[/itex] and the units will cancel out.

    What paper are you referring to (just curious)?
     
  4. Oct 13, 2009 #3
    Thank you for the reply.
    This is the paper I am referring to : Electrooxidation of methanol on platinum doped polyaniline electrodes: deposition potential and temperature effect

    here is the linkk :
    http://www.sciencedirect.com/scienc...d=130270&md5=5914e803ab0f43e6ca00f8fbe6780dda

    Based from your equation, I should be getting two I`s. If I2 is the current density from the CV plot at a certain potential, where can I get the corresponding I1? Is I1 related to the double layer current or something else?
     
  5. Oct 13, 2009 #4

    Mapes

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    [itex]I_1[/itex] comes from the same experiment at a different temperature.
     
  6. Oct 13, 2009 #5
    if that is the case, for example if my temperatures are 30oC, 35oC,40oC,...,60oC;
    I will have I(35oC)/I(30oC), I(40oC)/I(30oC),...,I(60oC)/I(30oC)?
    that being the case, there would have been no negative values on the log i - axis since an increasing temperature normally increases current density. but from the graph on the paper that i was reading, log i - axis has negative values.
     
  7. Oct 14, 2009 #6

    Borek

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    It depends on the reference chosen - if you start with I(30°C) there are no negative values, but if you start with I(40°C) you will get them.

    Not that it matters.

    --
     
  8. Oct 14, 2009 #7
    i see. thanks for the enlightenment.
     
  9. Oct 14, 2009 #8
    by the way, if ill get negative values that way, is it supposed to be between -1 and 0 only? but from the paper, the range in the y-axis (log i) is -2.5 to -4. it doesnt seem to add up...
     
  10. Oct 14, 2009 #9

    Borek

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    Why -1..0? That will mean current ratio between 1 and 0.1 (assuming log10).

    I don't have access to the paper.

    --
    methods
     
  11. Oct 14, 2009 #10
    I was not really sure about that. For now I will continue with the measurements and try to see whether I will arrive at a similar result.
     
  12. Nov 3, 2009 #11
    good day.
    I have settled the issue, thanks for all the inputs from you guys. i used impedance spectroscopy for the activation energy calculation and it did suit for me well.
     
  13. Nov 4, 2009 #12
    This is exactly my point and I1 always taken for different temperatures everytime.

    Thanks!
     
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