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How to measure the diffusion coefficient using a rotating disk electrode 

Pine Research Instrumentation, Inc.
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2 окт 2024

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Комментарии : 34   
@vasuguvera5912
@vasuguvera5912 4 года назад
Thank you for your very clear explanation about rotating disc diffusion coefficient calculation. It would be great if you could put video on how to prepare a cell and perform the experiment. Thank you once again 😊
@Pineresearch
@Pineresearch 4 года назад
I agree! When we've got some time, this will be another video we try to put out. Thank you for your suggestion
@hameemlone3253
@hameemlone3253 Год назад
Sir how can we directly measure diffusion coefficient from eis data
@Pineresearch
@Pineresearch Год назад
Hey have several webinars that go over the topic, but you'll get a better understanding of calculating the diffusion coefficient using EIS from our infinite vs finite warburg diffusion video. ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-iK6HmAnD7GQ.html
@mraja4295
@mraja4295 3 года назад
Thanks for your presentation 👍
@Pineresearch
@Pineresearch 3 года назад
My pleasure
@Diana-ug2nv
@Diana-ug2nv 4 года назад
Hello How come your slope units have (rad/s)^1/2 if your X axis is still RPM^1/2? Thanks!
@Pineresearch
@Pineresearch 4 года назад
I will need to check in with one of our software engineers about that. I think we choose to automatically make it (rads/s)^1/2 because we knew that this was what people were going to be calculating so we made the jump to give users the number that's easiest to use for diffusion coefficient calculations even though instrumentally you work with RPM.
@PoeticBugBear
@PoeticBugBear 3 года назад
A great informative video!!! When to use chronoamperometry and voltammetry??
@Pineresearch
@Pineresearch 3 года назад
It depends on what you are trying to do. Voltammetry is great when you are trying to determine at what potential you electrochemical events occur. Chronoamperometry is more quantitative when you already know the potential where electrolysis is occurring. Chronoamperometry has the added benefit where it's less dependent on the electrochemical double-layer, while voltammetry can generate a large background current depending on the scan rate.
@nugoraqutlimurotova223
@nugoraqutlimurotova223 2 года назад
Thank you for presentation
@Pineresearch
@Pineresearch 2 года назад
Glad you liked it!
@Pho3niX162
@Pho3niX162 3 года назад
Why do you not include the point 0|0 in the diagram? Then i get a bit higher slope (0,0000202593) and a higher diffusion coefficient D=1,86x10^-5 cm2/s . Thx for the help
@Pineresearch
@Pineresearch 3 года назад
Sorry for not responding to you earlier. Your comment got caught in the "held for review" section. You could use 0|0 when calculating the slope. I would say that you'll always have some level of noise and possibly get a current that's not equal to zero at zero RPM. The calculated diffusion coefficient you got isn't all that bad when compared to literature.
@DaniL-hl3ip
@DaniL-hl3ip 3 года назад
Hii so im thinking the value you have given in your table for your concentration is off, you guys put (1x10^6) while 1 mM would be (1x10^ -6) i think you just missed the negative sign, but in replicating and trying to practice finding diffusion coefficient and doing lots of algebra we realized the mistake haha. Thanks for all of your videos love you guys at pine research!!!
@Pineresearch
@Pineresearch 3 года назад
Yes, great catch! You are correct, in the table I should have put a minus sign on the concentration. It should be 1x10^-6 M. I'm glad you enjoy the videos. More to come! :)
@hanseulkim6855
@hanseulkim6855 3 года назад
OMG... This video is a really helpful and easy explanation of whatever I did ^_^b I just know my research error such as rpm--> rad/s through the electrochemical book, but your explanation is very useful and clear for me. But I am just curious about the viscosity(v) value, is it the measearment value or the paper value of the solution? Anyway, I'd like to say again "Thank U" ^_________________^
@Pineresearch
@Pineresearch 3 года назад
I'm glad you enjoyed the video! Definitely stay tuned for more videos from us. The kinematic viscosity value is a literature value, we don't have the tools to calculate the kinematic viscosity separately.
@edward_dantonio
@edward_dantonio Год назад
Do you have a “how-to” video on how to operate and collect data from your rotating disk electrode systems? Thank you for your time.
@Pineresearch
@Pineresearch Год назад
Unfortunately, I don't have a video like that. But in short, if you've done cyclic voltammetry, an RDE experiment is basically a cyclic voltammetry experiment performed at a slow scan rate (5-20 mV/s) while rotating the working electrode. So the data collection will be the same as a cyclic voltammogram. I'm thinking about making a setup and perform RDE experiment video in the future.
@이수빈-s3w1y
@이수빈-s3w1y 2 года назад
7:44 at this time, when you calculate the diffusion coefficient, slope's unit consists of A, s, rads. there is a rads unit! but your parameters (n, F, A, C, v, D) don't contain the rads unit. So I'm curious about how to remove the rads unit in your calculation???
@Pineresearch
@Pineresearch 2 года назад
Great question. Radians are unitless, so I simply removed it from the equation. a radian is the length of an arc of a unit circle divided by the radius. Both quantities are dimensions of length and hence they cancel each other out. I hope this helps.
@marivasfra
@marivasfra 4 года назад
Hi! Great Video! One question: do you have any Video on how to Measure and calculate the ECSA Loss of a Catalyst with a RDE?
@Pineresearch
@Pineresearch 4 года назад
Unfortunately, we do not. We would need access to some catalysts and perform some experiments in order to make that video. But this is a good idea for future videos, thank you.
@InsomniackidLin
@InsomniackidLin Год назад
Great video. Thx! One thing to point out. At 7:47, the concentration C should be 1 E-6 mol/cm3 instead of E6 cuz the initial concentration is 1 mM if I am not mistaken.
@Pineresearch
@Pineresearch Год назад
Yes, you are right. I forgot the minus sign. It should be 1x10^-6.
@indrajitpatil5290
@indrajitpatil5290 3 года назад
Nice...Thanks!.
@Pineresearch
@Pineresearch 3 года назад
Thank you too!
@yakshpopat3232
@yakshpopat3232 4 года назад
Nice, thank you.
@Pineresearch
@Pineresearch 4 года назад
Your welcome
@lymlin6491
@lymlin6491 2 года назад
great!!!
@Pineresearch
@Pineresearch 2 года назад
Hope this is helpful! Thanks for watching :)
@VaranasiRangmanch
@VaranasiRangmanch 3 года назад
"5 hours later" 😂😂
@Pineresearch
@Pineresearch 3 года назад
Yup! :D
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