Thank you for the informative video! I have a question regarding the definition of formal potential. I am new to electrochemistry and am designing a DIY potentiostat. I did CVA of potassium ferricyanide on my setup, as it is a good and reversible system for testing. I found that the half-wave potential is not equal to the formal potential. Moreover, my CVA is shifted relative to the CVA obtained on a commercial potentiostat. The electrodes are the same. I also looked at the CVA of my compound in the literature and they are all shifted relative to the standard half-reaction potential for ferricyanide (it is 0.36 V). What is the reason? I understand that this point can be used as a reference for other electrochemical systems, but how do you determine the "true" formal potentials then? Thanks in advance for your answer and sorry for my bad English!
A 1.000 g sample of zinc metal is dissolved in 50 mL of 6 M HCl solution and diluted to the mark in a 250 mL volumetric flask. A 25.00 mL portion is transferred to a polarographic cell and oxygen is flushed out. A polarogram in the range 0 - 1 V (vs. the Hg-pool electrode) shows a wave at E1/2 = -0.65 V, id = 32.0 units of galvanometer deflection. A 5.00 mL portion of 5.00 x 10-4 M CdCi2 solution is added directly to the polarography cell which already contains the Zn solution, oxygen is again flushed out, and a second polarogram is taken. The wave shows the same E1/2 but the id is 77.5 units. Calculate the percent by weight of Cd impurity in the Zn metal. Note: Atomic weight of Cd is 112.4 g/mole.
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.
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.
Hi! It depends on the chemical properties of that species. For example, some species after being oxidized are rapidly react with water or undergo further chemical reaction. Those reacted species, most of the time, cannot convert back to the original form.
![[Ch 3.3f] Square Wave Voltammetry](http://i.ytimg.com/vi/v1vAr3pfKSw/mqdefault.jpg)
![[Ch 3.3f] Square Wave Voltammetry](/img/tr.png)
What a brilliant video. Very well presented, explained, and visualized.
Thank you for the informative video! I have a question regarding the definition of formal potential. I am new to electrochemistry and am designing a DIY potentiostat. I did CVA of potassium ferricyanide on my setup, as it is a good and reversible system for testing. I found that the half-wave potential is not equal to the formal potential. Moreover, my CVA is shifted relative to the CVA obtained on a commercial potentiostat. The electrodes are the same. I also looked at the CVA of my compound in the literature and they are all shifted relative to the standard half-reaction potential for ferricyanide (it is 0.36 V). What is the reason? I understand that this point can be used as a reference for other electrochemical systems, but how do you determine the "true" formal potentials then? Thanks in advance for your answer and sorry for my bad English!
Excellent demonstration
So far i watch lectures this is the best
A 1.000 g sample of zinc metal is dissolved in 50 mL of 6 M HCl solution and diluted to the mark in a 250 mL volumetric flask. A 25.00 mL portion is transferred to a polarographic cell and oxygen is flushed out. A polarogram in the range 0 - 1 V (vs. the Hg-pool electrode) shows a wave at E1/2 = -0.65 V, id = 32.0 units of galvanometer deflection. A 5.00 mL portion of 5.00 x 10-4 M CdCi2 solution is added directly to the polarography cell which already contains the Zn solution, oxygen is again flushed out, and a second polarogram is taken. The wave shows the same E1/2 but the id is 77.5 units. Calculate the percent by weight of Cd impurity in the Zn metal. Note: Atomic weight of Cd is 112.4 g/mole.
Thanks great
Great teaching way Ajarn. Thanks
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.
Very well explained!
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.
Would the anodic and cathodic peaks be close enough together to consider the multistep example to also be reversible?
yes
Thany for such a nice lecture
Sir, thanks for this useful video, but what is the reason that species don't undergo in the case of irreversible
Hi! It depends on the chemical properties of that species. For example, some species after being oxidized are rapidly react with water or undergo further chemical reaction. Those reacted species, most of the time, cannot convert back to the original form.
4:41 what is n here.
Best lecture of CV on internet 🤗
really best cv in simple way