I am working for PEM fuel cell cathode material. I am confused regarding the onset potential determination for my catalyst (CV attached, fig 1 is complete CV while fig 2 is its zoom image for clear understanding of change in reduction curve direction) and also why the reduction starts from 0.8e-4 A instead of 0 A. I performed this CV in calomel electrode under oxygen in 0.5M H2SO4.
i) If I use Argon and overlap both the graphs then probably the curves start to separate at around 0.58V.
ii) I have read in some discussions to consider the onset potential at which current density approaches 0.1mA/cm2 while in other discussion, I have read to consider 20microA/cm2 as onset potential. If I plot LSV curve for current density (instead of current) vs potential , then in first case (0.1mA/cm2), it would be 0.785V while in second case (20microA/cm2), it would be 0.775V
iii) If I only consider the point from where curve starts to change the direction of reduction peak, it would be o.64V or 0.56V
iv) if I draw tangents then this would be around 0.5V.
So please help me in i) analyzing onset potential ii) guide me why there is positive current in the reduction peak iii) i also don't know what does the background correction mean and should I need to do it for my CV? All potentials are measured in SCE.
So please help me to analyse onset potential in the best way and also guide me why there is positive current in the reduction peak. All potentials are measured in SCE.
Dear Muhammad Shoaib
People measured onset potential, the way you mention in four points. If you see the literature, people do measurement of onset potential in different different way. For your case~ 0.56 V or ~0.58 V (from attached figure) will be the onset potential.
Yes, you have to do background CV correction to get rid of positive current in reduction reaction. For that you have to measure your CV in argon saturated electrolyte and then you have to subtract the current from the oxygen saturated CV. Then you will get negative current for reduction process.
Best Regards,
Bapi
Dear Name fellow,
Bapi Bera is right, I will just add that simply the onset potential is that point where reduction curve is started. Further you have to do background correction, because may be your all current values not due to reduction of oxygen, there may be some other small molecular reactions with in this range of potential. for further question feel free to ask.
Dear Muhammad
One of the problems is that "onset potential" is a pretty poorly defined term. We say its when a reaction "starts happening", but as the current has an exponential increase, what we really mean is when we start *noticing* the reaction happening. So you have to define the onset potential somewhat arbitrarily (e.g. by saying "after this current density I will consider my reaction to have started").
To answer your question about the current not being 0 A to being with; you have double layer capacitance contributing a current at all potentials. Your oxygen reduction current happens "on top of" the double layer capacitance.
https://en.wikipedia.org/wiki/Double-layer_capacitance (although I'm sure there are better descriptions)
Sean
Thanks to all for your valuable answers. I have done background corection as suggested by Bapi Bera and Syed Shoaib Ahmad Shah to get rid of positive current in reduction.
Sean Goodwin you gave a good suggestion regarding onset potential. Can only background subtraction eliminates the double layer capacitance effect? Also is there any effect on onset or peak potential due to double layer capacitance. If yes, then how can I avoid it.
Muhammad Shoaib background subtraction will remove the double layer capacitance, so long as the electrodes have not been changed in any way (capacitance is a function of electrode surface area). Your onset and peak potentials should not be affected by double layer capacitance.
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