Tomorrow I have my project work presentation, and i have to describe cyclovoltametry their....... I saw various video on topic.. But i can say this is the nicest explanation in compact way , I have ever seen.... Thank you so much sir Huge respect for you - from India🇮🇳
@ 9:17 Should it be the "diffusion of reduced species from the anode faster than the ionic migration" instead of "diffusion of oxidised species from the anode faster than the ionic migration"? Because at this point, we are having a cathodic current, signifying that there is a reduction on the anode.
Very nice video. Thanks for the upload. I just had a concern about the use of the mass transport limitation of the oxidized species as the main reason for the formation of a peak. I think CV peaks are more traditionally taught with the reactants being the mass limited species. While ofc both can be true depending on the their diffusion coefficients and charge, I think the explanation of the CV peak using a reactant mass transport limitation is better. This is because it is easier to transfer conceptually to different electrochemical processes such as CV for electroplating processes. What are your thoughts on this?
This was a long journey, and it was worth it! Much thanks for the concise explanations--everything was pretty on point. Now that I know better, it is time to do better. I will be looking forward to the 11th and 12th videos. Regards Biffo
Hi Andrew, thank you for this series. It has provided me with better understanding in electrochemistry. I have a doubt on a statement you made in the 3:00-3:15 of the 10 session where you said that the concentration of a specie in the system can influence the rate of reaction. I argue that this may not necessarily be the case giving that the concentration has no role to play on the Gibbs free energy, exchange current density and standard potential. Could you please shed more light please.
Wonderful video! Very clear and well presented. Do you have a similar video explaining the selectiveness in a electrolytic refinement cell? How do the electrode potentials influence the species being plated out and how important is this voltage vs the requirement to run at higher currents?
Hello Sir! These electrochemistry videos are extrremly useful and comprehensive, covering all the fundamental aspects and easy to undestand. Can you please upload more on electrochemical reactions such as OER, ORR, HER, HOR in water electrolyzers and Fuel cells? That would be really helpful! Thanks!
When you lower the diffusion rate in general of this Voltamogram. What will it look like? given that the concentration is equal as before and the reductionpotentials are also the same. ???
I don't understand how your reagent is reduced from cation to neutral but then oxidized from anion to cation. Also the part at 9.23 I am not sure if is worded right? That said thank you for your insight, this is excellent
Glad you have found the videos useful. In the video we first look at the cathode process (reduction of cations to uncharged species) in the example of linear voltammetry; but in the second part when exploring cyclic voltammetry in depth we look at the anode process (oxidation of anions to uncharged species). Note that in both cases they are either oxidised or reduced to the uncharged species - I can't see anywhere that I reference oxidation from the anion direct to cation; this would involve a two-electron process and the kinetics would be _very_ different. Remember that we are only interested in processes happening at one electrode (while a counter electrode completes the circuit); at one potential we will be oxidising anions, and when we lower the potential we will start reducing cations. These are not necessarily positive or negative potentials, simply different levels of potential. We could plan to do our CV sweep between 0.2 V and 1.4V; never going into a negative potential region, but we may still get reduction happening. What we see very much depends on our experimental set up and the system we are interested in. RE 09:23 I'm not sure what you are getting from the video, so not sure where the confusion comes from. The background to the processes is that the negatively charged anion migrates across the electric field gradient to the anode, where it is oxidised to an uncharged species. This then diffuses across the concentration gradient away from the anode. So at point E, the concentration gradient causes diffusion to be faster than the electric field causes for migration of the anions. I realise this is just a re-wording of what I say in the video, so if this doesn't help to clear things up please do let me know.
@@aw_mckinley hey, thank you for the reply. It turns out you are completely right, I assumed the oxidation and reduction examples were for the same compound. At the point E as far as I understand we get a reverse current due to a reduction process and the oxidation has already stopped. I think I assumed the oxidized species are nil at bulk solution so I imagined the diffusion backwards.
Please any one help me in solutions this problem One gram of zinc metal dissolved in a solution of hydrochloric acid with a concentration of 6 molars, dilute it to 250 ml. Take 25 ml of it into a polarography cell, and the polarogram shows a voltage wave at -0.65 volts (assigned to Cd impurities), a diffusion current equal to 32 milliamps. If we add to the cell containing the zinc solution 5 ml of Cadmium chloride concentration is 0.0005 molar, and when taking a second crystalogram, the diffusion current was 77.5 milliamps. Calculate the weight percentage of cadmium impurities in zinc metal.
Good question! We work with DC voltages, HOWEVER sometimes we switch the DC from positive to negative. True "AC voltages" switch from +V to -V; whereas our CV experiments switch between V1 and V2. The difference is subtle: - The magnitude of V is constant in AC; so voltage going from +110V to -110V. - In CV, V1 and V2 can have any value, and are not necessarily connected. --- V1 could be +0.5 V , while V2 could be +1.2 V : this would not give an AC condition. --- V1 could be +0.9 V, while V2 could be -0.5 C : while the current would switch direction, it would not have the same magnitude at the extremes of the sweep. True "AC voltages" vary sinusoidally. In our CV experiments, we ramp the voltage linearly up and down, so any wave is more of a saw-tooth rather than a smooth wave.
Tomorrow I have my project work presentation, and i have to describe cyclovoltametry their.......
I saw various video on topic..
But i can say this is the nicest explanation in compact way , I have ever seen....
Thank you so much sir
Huge respect for you - from India🇮🇳
Thank you, just saved me.
That was the most didactic Physical Chemistry class I have watched.
Greetings from Brazil!
You've just saved a Japanese University Student. Thank you!
How old r u?
Needs more people like you to explain electrochemistry, really. thanks a lot.
Superb, Excellent videos, please keep up the good work by uploading such videos.
You really help me with these explanation. Electrochemistry becomes very easy with your lectures. Thank you very much. You did an amazing job.
@ 9:17
Should it be the "diffusion of reduced species from the anode faster than the ionic migration" instead of "diffusion of oxidised species from the anode faster than the ionic migration"?
Because at this point, we are having a cathodic current, signifying that there is a reduction on the anode.
Very nice video. Thanks for the upload. I just had a concern about the use of the mass transport limitation of the oxidized species as the main reason for the formation of a peak. I think CV peaks are more traditionally taught with the reactants being the mass limited species. While ofc both can be true depending on the their diffusion coefficients and charge, I think the explanation of the CV peak using a reactant mass transport limitation is better. This is because it is easier to transfer conceptually to different electrochemical processes such as CV for electroplating processes. What are your thoughts on this?
Please complete videos 11 and 12 of the series, great job!!
This was a long journey, and it was worth it!
Much thanks for the concise explanations--everything was pretty on point.
Now that I know better, it is time to do better.
I will be looking forward to the 11th and 12th videos.
Regards
Biffo
Hi Andrew,
thank you for this series. It has provided me with better understanding in electrochemistry.
I have a doubt on a statement you made in the 3:00-3:15 of the 10 session where you said that the concentration of a specie in the system can influence the rate of reaction. I argue that this may not necessarily be the case giving that the concentration has no role to play on the Gibbs free energy, exchange current density and standard potential. Could you please shed more light please.
Wonderful video!
Very clear and well presented.
Do you have a similar video explaining the selectiveness in a electrolytic refinement cell?
How do the electrode potentials influence the species being plated out and how important is this voltage vs the requirement to run at higher currents?
Thank you Dr. McKinley, very nice and clear explanation
so beautifully explained, thankss
This is very good and helpful...Thanks,
Very clear explanation about electrochemistry. Could we get the part 11 and part 12 videos if possible? Thanks.
Hello Sir! These electrochemistry videos are extrremly useful and comprehensive, covering all the fundamental aspects and easy to undestand. Can you please upload more on electrochemical reactions such as OER, ORR, HER, HOR in water electrolyzers and Fuel cells? That would be really helpful! Thanks!
Just excellent, keep good workings.
Thanks for your clear explanations, it really helped me to understand these basics, hope you keep doing more videos
Thank you so much it's very complete and useful
Glad you found it useful :-)
Really good series. Are there more videos? I see this was video 10 in a 12 part series??
If so is there anyway we can access them? Cheers
Very well explained....👍❤️
bro you literally saved my life🤯.
Great video, thanks
Can I get the pdf of your slides, Please I need them
Very helpful, thanks!
Very useful presentation.
When you lower the diffusion rate in general of this Voltamogram. What will it look like?
given that the concentration is equal as before and the reductionpotentials are also the same.
???
Obrigada!!!! Thank you!
I don't understand how your reagent is reduced from cation to neutral but then oxidized from anion to cation. Also the part at 9.23 I am not sure if is worded right?
That said thank you for your insight, this is excellent
Glad you have found the videos useful.
In the video we first look at the cathode process (reduction of cations to uncharged species) in the example of linear voltammetry; but in the second part when exploring cyclic voltammetry in depth we look at the anode process (oxidation of anions to uncharged species). Note that in both cases they are either oxidised or reduced to the uncharged species - I can't see anywhere that I reference oxidation from the anion direct to cation; this would involve a two-electron process and the kinetics would be _very_ different.
Remember that we are only interested in processes happening at one electrode (while a counter electrode completes the circuit); at one potential we will be oxidising anions, and when we lower the potential we will start reducing cations. These are not necessarily positive or negative potentials, simply different levels of potential. We could plan to do our CV sweep between 0.2 V and 1.4V; never going into a negative potential region, but we may still get reduction happening. What we see very much depends on our experimental set up and the system we are interested in.
RE 09:23 I'm not sure what you are getting from the video, so not sure where the confusion comes from. The background to the processes is that the negatively charged anion migrates across the electric field gradient to the anode, where it is oxidised to an uncharged species. This then diffuses across the concentration gradient away from the anode. So at point E, the concentration gradient causes diffusion to be faster than the electric field causes for migration of the anions. I realise this is just a re-wording of what I say in the video, so if this doesn't help to clear things up please do let me know.
@@aw_mckinley hey, thank you for the reply. It turns out you are completely right, I assumed the oxidation and reduction examples were for the same compound.
At the point E as far as I understand we get a reverse current due to a reduction process and the oxidation has already stopped. I think I assumed the oxidized species are nil at bulk solution so I imagined the diffusion backwards.
excellent explanation
thank you sir
Thanks a Zillion !!!!!
Please do more videos on chemistry
Please any one help me in solutions this problem
One gram of zinc metal dissolved in a solution of hydrochloric acid with a concentration of 6 molars, dilute it to 250 ml. Take 25 ml of it into a polarography cell, and the polarogram shows a voltage wave at -0.65 volts (assigned to Cd impurities), a diffusion current equal to 32 milliamps. If we add to the cell containing the zinc solution 5 ml of Cadmium chloride concentration is 0.0005 molar, and when taking a second crystalogram, the diffusion current was 77.5 milliamps. Calculate the weight percentage of cadmium impurities in zinc metal.
is the applied voltage in case of cyclic voltammetry, DC voltage or AC?
Good question! We work with DC voltages, HOWEVER sometimes we switch the DC from positive to negative.
True "AC voltages" switch from +V to -V; whereas our CV experiments switch between V1 and V2. The difference is subtle:
- The magnitude of V is constant in AC; so voltage going from +110V to -110V.
- In CV, V1 and V2 can have any value, and are not necessarily connected.
--- V1 could be +0.5 V , while V2 could be +1.2 V : this would not give an AC condition.
--- V1 could be +0.9 V, while V2 could be -0.5 C : while the current would switch direction, it would not have the same magnitude at the extremes of the sweep.
True "AC voltages" vary sinusoidally.
In our CV experiments, we ramp the voltage linearly up and down, so any wave is more of a saw-tooth rather than a smooth wave.
Hello Dr... Can you write your e mail please... Thhank you
THank you sir