Hi. I am working on electrocatalytic hydrogen production reaction. I am wondering how I can calculate iR drop for my system. I did LSV and EIS measurements. But I do not know which current I should use to calculate iR and then overpotential.
Dear Yasaman Sheidaei
To correct your polarization curve you should use R determined by EIS (Z' at high frequency) and correct every potential point for the corresponding current, eg. Ecorected = Actual potential + IR. For the hydrogen evolution Eact is usually negative, so corrected potential will be more positive than actual. Check on your potentiostat software does it have automatic mod for IR correction.
As prof. Branimir N. Grgur .
Also, the parameter of the polarization value (VDC) in your diagnostic measurement of the EIS (Zreal=RHF, at High Frequency>100kHz) should be very far (as over-Potential>>0.1V) from the specific Potential (VH) of the hydrogen production reaction. Otherwise, the system will be in a NON-Ohmic regime, and that (bad-)RHF value will be highly misleading for your LSV.
Dear prof. Branimir N. Grgur
Thank you for your answer. As I checked the software, it has an "iR compensation" option. It asks for test potential. Which potential should I use? Actually I am using CH instrument for my measurements.
I am not sure what means test potential. Looking in a CH600 potentiostat description it has Automatic and manual iR compensation. For the automatic in my Gammry potentiostat I only check CI (current interupt) and he automaticaly correct polarization curve. For the manual the resistance obtained from the impedance has to be added in the box. It is best to ask CH instruments what does it means, or to find on the internet better manual.
Dear Bapi Bera
Thank you for your answer. Actually my problem with manual correction is that I do not know which current I should use for calculating iR.
Dear Yasaman Sheidaei
When you do the LSV and export the data you will get two column with one voltage and one current column. Then multiply the current column with R (from EIS) value to get the iR value. Then do below calculation:
Ecorected = Actual potential + iR.
Regards,
Bapi
Dear Bapi Bera
Thank you, I was really confused whether I should apply any changes to the current I got. Thank you again.
Dear Yasaman Sheidaei ,
For your case take positive sigh of your current and multiply with R, will get the iR value. When you add that value in you actual measured voltage, your iR corrected LSV curve should shift little positive direction.
See the the attached image.
Regards,
Bapi
For iR calculation, i is the current you obtained from the LSV/CV and convert it to Amperes (as normally this current is in mA), Solution resistance is R that should be in ohm and then calculate V (V=iR (ampere*ohm)). corrected potential will be E-V(iR).
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