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 current density instead of current only, 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 0.64V 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.

enter image description here


The reason the current starts at 0.8x10^-4 A is probably because you have an equilibration time built in to your procedure (say of a few seconds),or even if you do not,there are parasitic currents always occurring, especially if you are starting at 0.8 V (vs SCE). Given that I don't know much more about your system or how you arrived at the onset potential calculations, I cannot comment. Could you please provide more details?

| improve this answer | |
  • $\begingroup$ Thanks for your response. I dont know what information do you require so i am mentioning all the details: the compound is C/N/S/transition metal, medium: 0.5 M H2SO4, 3mm glassy carbon electrode, scan rate = 0.01V/s, quiet time = 2sec, sensitivity=1e-4A/V. $\endgroup$ – m.s87 Oct 18 '18 at 10:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.