A dilemma or conundrum exists here in regard to the drop in battery voltage upon discharging. It can easily be shown that as we raise the circuit resistance, we experience less polarization of the two electrodes towards each other (and this co-polarization of the electrodes is what causes this apparent battery voltage drop) under loading/discharge. In other words if we short circuit the external battery circuit and/or reduce the internal electrolyte resistance in the battery, there is more polarization of the terminals upon loading which reduces the overall voltage more. So now we have a situation where the less actual IR drop that actually occurs in the circuit causes more, not less, battery voltage drop upon discharge which you are calling "IR drop". So therefore the lower the resistance is, the higher what you call "IR drop" is. This does not make sense.
Hi there, I'm not sure if I understand what you mean. When there's less resistance, the electrode will experience less overpotential due to resistance.
You are not considering activation polarization (sometimes called kinetic polarization). When there is no resistance at all, the potentials of the anode and cathode come together to the same value (there is more activation polarization which is proportional to LogI for both the anode and the cathode). The amount of activation polarization is proportional to the log of current flow. IR drop is shown on Evans diagrams as the difference between the anode and cathode potentials after current flow creates polarization.
@@bobstevens7028I'm still trying to understand the context of your writeup. Activation polarization is indeed not considered in the IR drop. It's an additional contribution on top of IR drop, which was indicated in the figure.
Correction: I mislabeled the axes of the CV on the slide that talks about "Ohmic Overpotential". Sorry for the confusion!
Thank you, very helpful :)
A dilemma or conundrum exists here in regard to the drop in battery voltage upon discharging. It can easily be shown that as we raise the circuit resistance, we experience less polarization of the two electrodes towards each other (and this co-polarization of the electrodes is what causes this apparent battery voltage drop) under loading/discharge. In other words if we short circuit the external battery circuit and/or reduce the internal electrolyte resistance in the battery, there is more polarization of the terminals upon loading which reduces the overall voltage more. So now we have a situation where the less actual IR drop that actually occurs in the circuit causes more, not less, battery voltage drop upon discharge which you are calling "IR drop". So therefore the lower the resistance is, the higher what you call "IR drop" is. This does not make sense.
Hi there, I'm not sure if I understand what you mean. When there's less resistance, the electrode will experience less overpotential due to resistance.
You are not considering activation polarization (sometimes called kinetic polarization). When there is no resistance at all, the potentials of the anode and cathode come together to the same value (there is more activation polarization which is proportional to LogI for both the anode and the cathode). The amount of activation polarization is proportional to the log of current flow. IR drop is shown on Evans diagrams as the difference between the anode and cathode potentials after current flow creates polarization.
@@bobstevens7028I'm still trying to understand the context of your writeup. Activation polarization is indeed not considered in the IR drop. It's an additional contribution on top of IR drop, which was indicated in the figure.
doing good man, good info
Can you please explain ORR?
Thank you!
Thank you, helped me a lot :)
can you make a video on tafel plots??
I can give it a shot. May I ask your field of study, please?
@@ruocun2615 about hydrogen evolution activity
why your CV has y axis of potential and x axis of current?
That is not a CV
@@EChem_Channel I have the same question. if that is not CV, then what is it?
@@shide1473 @Fencros7y Oh, sorry. Are you refer to the right figure at 2:50? It is a CV. I mislabeled the axes. Sorry for the confusion!
I can deal with the pace, but will more appreciate if you talk clearly
Sure, I will try to improve. Thanks.
can you help me in analyzing the CV ?
Book name or available this book in internate
Electrochemical Methods
Can you explain more easily!
I suggest you go read "Electrochemical Methods" by Allen J. Bard.