This video is probably the best and most straight forward explanation for 3 electrode systems I've found (after finding a lot of misinformation and looking like an idiot in my lab)
Thanks for this very informative video Gary! I have a question about the slide at 9:42 How can a working electrode that is far from the reference electrode create a potential V1+Vref at the reference electrode? Are they connected in some way?
My question comes down to: how can you bring a solution to a certain potential? Is it because the inert electrolyte will create the potential through the solution?
And if the potential is indeed created through the electrolytes, does the distance to the working electrode matter, or is every point in the water at the same potential?
Jonas, Good questions. Your reasoning is also correct on both counts. It is a bit of a challenge to follow the operation of a three-electrode system, but basically the reference and counter electrodes work together to pump charge into or out of the solution when the power source (potentiostat) applies a voltage to the cell. (It takes very tiny amounts of charge to do this.) This excess charge migrates very quickly to the interface between the solution and the working electrode. The fact that the solution conducts charge (in the form of ions) keeps the potential throughout the solution uniform in a manner similar to the way a metal conductor would if you where charging up a pair of plates in a capacitor. We say that the reference electrode senses the voltage on the solution side. Really what the potentiostat monitors is the voltage between the reference electrode and the working electrode (which is being held at ground or zero volts). The surface of the working electrode is like the second plate in the capacitor and it is in contact with ground (that is, it is held at zero volts). So, the applied voltage is a measure of the voltage between the capacitor plates. It is the electrochemical potential energy drop across this interface between the solution and the working electrode surface and it is that energy that is available to drive a chemical reaction. You pointed out an important idea about the placement of the reference electrode. That can be important if either the electrical resistance of the solution or the current going through the system is big. It takes energy for the ions to move the charge through the solution. That energy is lost as heat. The potentiostat reports the total voltage applied to the cell. If that energy spent moving ions is a significant fraction of the energy applied to the cell, then the actual energy applied across the working electrode/solution interface will be less than the applied cell voltage. That means we will have an error in the applied voltage (the x-axis) on our voltammogram. The error is equal to the product of the current times the solution resistance. So, if we keep the supporting electrolyte concentration high (0.1 M KCl has a resistance on the order of 100 Ω per cm of solution) and the current below 10 µA, then errors are on the order of 1 mV or less. That's usually acceptable. When either the solution resistance is high or the currents are greater than 10 µA, placing the reference electrode close to the surface of the working electrode helps to reduce this type of error (also called "iR-drop" loss).
@@garymabbott4064 A big thank you for your answer Gary! This really helps me clarify this concept. I had another question: We are interested in the potential and current flowing from the working electrode for obvious reasons. The counter electrode however is mostly there to get rid of the electrons our working electrode produces. I was wondering how the potential of this counter electrode is regulated? dependent on which reaction happens there, it seems to me that a different voltage would be needed at the working electrode to perform both the half reactions? If a reaction at the counter electrode happens that needs a high driving force voltage wise, it would seem to me that the working electrode would need to work harder (have a higher potential) to make sure the reaction happens? Or is there some sort of mechanism in the technical part that senses how many volts are needed at the counter electrode for that half reaction to happen to make up for it? I have also been struggling a bit with the concept of an electrolyte. To me, it seems that as long a reduction reactions happens at the cathode, and an oxidation reaction happens at the anode, current is flowing. Does the electrolytes facilitate the transition of the electrons from the electrodes to the species? why does the 'current' wants to go through the solvent instead of taking the electrodes as starting and ending point of a circuit?
@@jonasmortier9056 Jonas, In our experiment, we really don’t have a direct measure of what is happening at the counter electrode, but we can infer a few things. We know that if the working electrode is oxidizing some species in solution, then it is pulling electrons out of the cell. In order to maintain the balance (hold the potential that the instrument is designed to control at the working electrode) the counter electrode will pump the same number of electrons into the solution. That means that the potential of the counter electrode surface will be negative with respect to the solution (just the opposite polarity that appears at the working electrode). Now, you were right to think that there must be some reaction at the counter electrode that will allow it to transfer electrons into the solution. Often there is no obvious species in solution that can be easily reduced. Then the instrument will drive the potential difference at the counter electrode more negative trying to achieve the balance (or really to keep the potential across the working electrode interface at the target value). Eventually, the potential difference at the counter electrode becomes negative enough to reduce the solvent or some species in the supporting electrolyte. In aqueous solutions we observe the reduction of hydrogen ions to hydrogen gas. (You may have noticed bubbles forming on the counter electrode.) In the opposite situation (when a reduction is taking place at the working electrode) the counter electrode is likely oxidizing water to form molecular oxygen (bubbles again) and hydrogen ions. So, the counter electrode potential is driven in the opposite direction compared to the working electrode. We don’t have a measure of the actual potential that the counter electrode reaches. It will be whatever value it needs to reach in order to pump electrons in the opposite direction compared to whatever is happening at the working electrode. The supporting electrolyte acts to carry charge through the cell in the form of ions. Our experimental cell must be a part of a complete electrical loop that provides charge to move from one side of our instrument, down a cable to an electrode, through the cell, out another electrode, up a second cable to the other side of the instrument (completing the circuit inside the instrument). The conductivity of the solution goes up (and resistance goes down) with the concentration of ions dissolved in solution and that is good for the experiment. (Recall that we want to minimize errors caused by “iR loss”-the energy wasted in moving ions through solution.) In most experiments, we expect the supporting electrolyte to be inert and not participate directly in solution. An important exception is the oxidation or reduction of organic molecules that usually are accompanied by the incorporation or release of hydrogen ions from solution. There are also very interesting reactions that involve oxidation or reduction of the solvent or supporting electrolyte that turn around and exchange electrons with the analyte species (and are re-generated back to their original state). This, of course, is a catalysis process that can be of enormous interest for theoretical and practical purposes. However, we generally expect that the supporting electrolyte is not involved in electron transfer at the working electrode in simple electroanalytical experiments (until we scan to the limit of our useful voltage range where we see increasing background current even in blank solutions containing only supporting electrolyte.) I am happy to continue answering your questions, but I would encourage you to recommend that your library purchase this book: www.amazon.com/Electroanalytical-Chemistry-Principles-Monographs-Applications-ebook/dp/B084GM6C3R/ref=sr_1_6?qid=1673106623&refinements=p_28%3AElectroanalytical+Chemistry&s=books&sr=1-6 (Yes, it is my book. I apologize for the shameless self promotion, but it could be helpful to you and others interested in this field.)
What is the potential at the reference electrode in a 3-electrode setup? is it 0.197V or is it whatever V_in we are driving into the opamp shown at the end of the video? Are we driving the reference electrode potential?
Wherever there is a boundary between two different phases or materials there is the possibility of a potential energy difference. Between the silver chloride coated wire and the solution surrounding it, there is a step up in potential of 0.197 V. A voltage, V1, is applied to one of the inputs of the amplifier. The amplifier is designed to drive the voltage at its output (where the auxiliary electrode is attached) to whatever value is needed so that the other input of the amplifier also reaches a value of V1. So, the wire attached to the reference electrode is at a voltage of V1. Moving from there into the silver wire and then across the interface of the AgCl and into the solution, the voltage steps up to a value of V1+0.197. So, (V1 + 0.197) volts is the potential of the solution. “Looking” from the solution toward the working electrode, an electro-active molecule “sees” the working electrode at zero volts. Consequently, the potential difference across the working electrode/solution boundary is a step of (V1 + 0.197) volts in the negative direction. The reference electrode is not really driven to change the potential across its boundary. That potential difference stays at +0.197 V, but it gets added to the applied potential coming into the input of the amplifier.
@@garymabbott4064 thank you, though I have two more questions: 1. is the 0.197V potential step between the reference electrode and the solution independent of whatever solution the reference electrode is in? 2. If I've understood correctly, we set the solution to a voltage of V1+0.197 with respect to earth ground, is this correct? In such a case, why is there no reaction at the reference electrode, is it because we are blocking mass transport to the reference wire?
@@lucha6262 1. The +0.197 V is the value that the reference electrode will have for a saturated solution of KCl. The potential energy difference between the AgCl coated Ag wire is, indeed, dependent on the chloride concentration (actually the activity) level. In a 1 M KCl solution, the potential is +0.222 V at 25˚C. 2. Yes, the V1 +0.197 V is the voltage of the solution with respect to ground. There will be a very tiny amount of a net reaction at the AgCl/Ag electrode because a real operational amplifier is not perfect and will allow some current (maybe a few picoamps--less for better amplifiers) to flow into the input of the op amp. That will not shift the potential of the reference electrode significantly. But, you have the right ide., The whole point of using a 3-electrode system was to avoid too much current (say microamps) going through the reference, because that would shift it away from the +0.197 V.
How about the current in ampere(in both oxidation and reduction)? How the system measure the current? is it related to counter electrode and working electrode? Any equation for it just like the potential plot on x axis is the voltage of working electrode related to references electrode. Thank you.
Hi Canny, You are asking a lot of good questions. This video was only part of the story. The current goes through the working electrode and the counter electrode through separate operational amplifiers in the outer circuit that are part of an instrument called a potentiostat. Here are two other videos that may help with some of your questions about what controls the current for oxidation and reduction. In these videos, I describe the main concepts from which the current-voltage equations are derived. To get another description, complete with equations see the article below. Controlled potential analysis. Be patient. The video starts with a review of potentiometric electrode measurements. Then it describes the principles behind electron transfer experiments (called voltammetry). ruclips.net/video/d9ADCv5vNzQ/видео.html Cyclic voltammetry experiment. This describes the set-up of an instrument and running an experiment. ruclips.net/video/BFefYBEL7wM/видео.html A Practical Beginner’s Guide to Cyclic Voltammetry Noémie Elgrishi, Kelley J. Rountree, Brian D. McCarthy, Eric S. Rountree, Thomas T. Eisenhart, and Jillian L. Dempsey J. Chem. Educ., 2018, 95 (2), pp 197-206.
Can someone please explain to me...if the working electrode cannot pass current (transfer of electrons into or from the solution) then how does it reduce or oxidize the species??
Hi Emaan, I am not sure what conditions you are thinking of where the working electrode cannot pass current. You are correct that its main purpose is to reduce or oxidize species in solution. (We prevent the reference electrode from passing current so we don't disturb its potential. Could that be what you were thinking of?) We force the working electrode to different potentials in order to drive an exchange of electrons with something in solution. That potential must be near or more negative than the E˚' for the redox species in order to reduce any of the oxidized form and we monitor the current that passes in the voltammogram. (We see current for the oxidation of any reduced species when the working electrode is near or more positive than the E˚'.)
@@garymabbott4064 thankyou for replying! And im confused because supposedly, a polarizable electrode cannot exchange electrons with the solution. I dont understand, if its not giving or receiving electrons (as a polarizable working electrode) how is it able to reduce or oxidize species? Im really really new to electrochemistry (my whole career is in bio) so pardon me if im asking something really obvious
@@emaanimtiaz4371 Electrochemistry can be confusing, in part because of all of the jargon. The term “polarizable electrode” means that voltage (the energy for an electron) between the surface of the electrode and the solution can be changed by imposing energy (from a battery or electrical power source) from outside the cell. We need at least one other electrode in the cell in order to do that experiment. A reference electrode, such as a silver/silver chloride electrode, can be used for that purpose. A good reference electrode will not move from its “rest potential” (its value that can be calculated from the Nernst Equation for the reference reaction, such as AgCl + e Ag + Cl-). We say that the reference electrode is non-polarizable. By selecting a non-polarizable reference electrode for a voltammetry experiment, virtually all of the energy (voltage) that we apply from the outside circuit ends up polarizing the working electrode. That is, all of the energy applied to the cell from the outside changes the voltage at the interface between the working electrode and the solution. If that voltage is big enough, it can force an electron exchange process there. A platinum wire or carbon electrodes are good working electrodes since they are polarizable and will exchange electrons readily with species in solution. One point of possible confusion with electrochemical measurements is that there are two common types of experiments. Voltammetry is the type I was referring to in the paragraph above. Voltammetry is based on applying an external voltage to the system which forces an electron exchange at the working electrode. The current is measured and its magnitude is proportional to the concentration of the electroactive species in solution. Voltammetry uses a polarizable working electrode. The other common experiment is potentiometry. The rest voltage of the working electrode is determined by the solution conditions and it is this quantity that we measure in a potentiometric experiment. In that case the working electrode or sensor should not be polarizable. We try to prevent significant amount of current going through the cell, since it could distort the voltage that appears at the working electrode. That is, allowing current in the system would push the working electrode away from the voltage that naturally develops at its surface because of the concentration of the species that it senses. I hope that helps.
@@garymabbott4064 thankyou, thankyou, thankyou! Iv been stuck on this for 2 days straight, watching a dozen youtube videos and googling it in a hundred different ways and NO ONE explained that. You've saved me from another sleepless night. Huge thanks
Electrons that the analyte takes from the electrode are replaced by a power supply (often a battery) that is part of the outer circuit. The arrangement of electrons (their "concentration") at the surface of the electrode is set by the voltage that is applied. If an electron "jumps" from the electrode to the analyte in solution, another electron will move up to take its place in order to maintain the arrangement of charges associated with the voltage applied by the power supply.
This video is probably the best and most straight forward explanation for 3 electrode systems I've found (after finding a lot of misinformation and looking like an idiot in my lab)
Very good explanation, directly, clearly and consistent. Thank you so mutch!
Very good video! It explains the concept in details and covers the basics as you go!
One of the best videos describing CV! Thank you!
Most effective vedio I have watched on voltametry... Thank you
That was superbly explained,thank you very much.
Excellent videos, thank you~
Thanks for this very informative video Gary! I have a question about the slide at 9:42
How can a working electrode that is far from the reference electrode create a potential V1+Vref at the reference electrode? Are they connected in some way?
My question comes down to: how can you bring a solution to a certain potential? Is it because the inert electrolyte will create the potential through the solution?
And if the potential is indeed created through the electrolytes, does the distance to the working electrode matter, or is every point in the water at the same potential?
Jonas, Good questions. Your reasoning is also correct on both counts. It is a bit of a challenge to follow the operation of a three-electrode system, but basically the reference and counter electrodes work together to pump charge into or out of the solution when the power source (potentiostat) applies a voltage to the cell. (It takes very tiny amounts of charge to do this.) This excess charge migrates very quickly to the interface between the solution and the working electrode. The fact that the solution conducts charge (in the form of ions) keeps the potential throughout the solution uniform in a manner similar to the way a metal conductor would if you where charging up a pair of plates in a capacitor. We say that the reference electrode senses the voltage on the solution side. Really what the potentiostat monitors is the voltage between the reference electrode and the working electrode (which is being held at ground or zero volts). The surface of the working electrode is like the second plate in the capacitor and it is in contact with ground (that is, it is held at zero volts). So, the applied voltage is a measure of the voltage between the capacitor plates. It is the electrochemical potential energy drop across this interface between the solution and the working electrode surface and it is that energy that is available to drive a chemical reaction.
You pointed out an important idea about the placement of the reference electrode. That can be important if either the electrical resistance of the solution or the current going through the system is big. It takes energy for the ions to move the charge through the solution. That energy is lost as heat. The potentiostat reports the total voltage applied to the cell. If that energy spent moving ions is a significant fraction of the energy applied to the cell, then the actual energy applied across the working electrode/solution interface will be less than the applied cell voltage. That means we will have an error in the applied voltage (the x-axis) on our voltammogram. The error is equal to the product of the current times the solution resistance. So, if we keep the supporting electrolyte concentration high (0.1 M KCl has a resistance on the order of 100 Ω per cm of solution) and the current below 10 µA, then errors are on the order of 1 mV or less. That's usually acceptable. When either the solution resistance is high or the currents are greater than 10 µA, placing the reference electrode close to the surface of the working electrode helps to reduce this type of error (also called "iR-drop" loss).
@@garymabbott4064 A big thank you for your answer Gary! This really helps me clarify this concept.
I had another question: We are interested in the potential and current flowing from the working electrode for obvious reasons. The counter electrode however is mostly there to get rid of the electrons our working electrode produces. I was wondering how the potential of this counter electrode is regulated? dependent on which reaction happens there, it seems to me that a different voltage would be needed at the working electrode to perform both the half reactions? If a reaction at the counter electrode happens that needs a high driving force voltage wise, it would seem to me that the working electrode would need to work harder (have a higher potential) to make sure the reaction happens? Or is there some sort of mechanism in the technical part that senses how many volts are needed at the counter electrode for that half reaction to happen to make up for it?
I have also been struggling a bit with the concept of an electrolyte. To me, it seems that as long a reduction reactions happens at the cathode, and an oxidation reaction happens at the anode, current is flowing. Does the electrolytes facilitate the transition of the electrons from the electrodes to the species? why does the 'current' wants to go through the solvent instead of taking the electrodes as starting and ending point of a circuit?
@@jonasmortier9056 Jonas,
In our experiment, we really don’t have a direct measure of what is happening at the counter electrode, but we can infer a few things. We know that if the working electrode is oxidizing some species in solution, then it is pulling electrons out of the cell. In order to maintain the balance (hold the potential that the instrument is designed to control at the working electrode) the counter electrode will pump the same number of electrons into the solution. That means that the potential of the counter electrode surface will be negative with respect to the solution (just the opposite polarity that appears at the working electrode). Now, you were right to think that there must be some reaction at the counter electrode that will allow it to transfer electrons into the solution. Often there is no obvious species in solution that can be easily reduced. Then the instrument will drive the potential difference at the counter electrode more negative trying to achieve the balance (or really to keep the potential across the working electrode interface at the target value). Eventually, the potential difference at the counter electrode becomes negative enough to reduce the solvent or some species in the supporting electrolyte. In aqueous solutions we observe the reduction of hydrogen ions to hydrogen gas. (You may have noticed bubbles forming on the counter electrode.) In the opposite situation (when a reduction is taking place at the working electrode) the counter electrode is likely oxidizing water to form molecular oxygen (bubbles again) and hydrogen ions. So, the counter electrode potential is driven in the opposite direction compared to the working electrode. We don’t have a measure of the actual potential that the counter electrode reaches. It will be whatever value it needs to reach in order to pump electrons in the opposite direction compared to whatever is happening at the working electrode.
The supporting electrolyte acts to carry charge through the cell in the form of ions. Our experimental cell must be a part of a complete electrical loop that provides charge to move from one side of our instrument, down a cable to an electrode, through the cell, out another electrode, up a second cable to the other side of the instrument (completing the circuit inside the instrument). The conductivity of the solution goes up (and resistance goes down) with the concentration of ions dissolved in solution and that is good for the experiment. (Recall that we want to minimize errors caused by “iR loss”-the energy wasted in moving ions through solution.) In most experiments, we expect the supporting electrolyte to be inert and not participate directly in solution. An important exception is the oxidation or reduction of organic molecules that usually are accompanied by the incorporation or release of hydrogen ions from solution. There are also very interesting reactions that involve oxidation or reduction of the solvent or supporting electrolyte that turn around and exchange electrons with the analyte species (and are re-generated back to their original state). This, of course, is a catalysis process that can be of enormous interest for theoretical and practical purposes. However, we generally expect that the supporting electrolyte is not involved in electron transfer at the working electrode in simple electroanalytical experiments (until we scan to the limit of our useful voltage range where we see increasing background current even in blank solutions containing only supporting electrolyte.)
I am happy to continue answering your questions, but I would encourage you to recommend that your library purchase this book: www.amazon.com/Electroanalytical-Chemistry-Principles-Monographs-Applications-ebook/dp/B084GM6C3R/ref=sr_1_6?qid=1673106623&refinements=p_28%3AElectroanalytical+Chemistry&s=books&sr=1-6
(Yes, it is my book. I apologize for the shameless self promotion, but it could be helpful to you and others interested in this field.)
This is THE BEST video
What would happen if there were no counter electrode?
counter connects to reference?
Can any of these electrode be used as a porous electrode, to enable metals to be anchored on them forming a bicomposite electrode?
Very useful video and explanation is very easy to grasp
Brilliant! Thanks very much.
What is the potential at the reference electrode in a 3-electrode setup? is it 0.197V or is it whatever V_in we are driving into the opamp shown at the end of the video? Are we driving the reference electrode potential?
Wherever there is a boundary between two different phases or materials there is the possibility of a potential energy difference. Between the silver chloride coated wire and the solution surrounding it, there is a step up in potential of 0.197 V. A voltage, V1, is applied to one of the inputs of the amplifier. The amplifier is designed to drive the voltage at its output (where the auxiliary electrode is attached) to whatever value is needed so that the other input of the amplifier also reaches a value of V1. So, the wire attached to the reference electrode is at a voltage of V1. Moving from there into the silver wire and then across the interface of the AgCl and into the solution, the voltage steps up to a value of V1+0.197. So, (V1 + 0.197) volts is the potential of the solution. “Looking” from the solution toward the working electrode, an electro-active molecule “sees” the working electrode at zero volts. Consequently, the potential difference across the working electrode/solution boundary is a step of (V1 + 0.197) volts in the negative direction. The reference electrode is not really driven to change the potential across its boundary. That potential difference stays at +0.197 V, but it gets added to the applied potential coming into the input of the amplifier.
@@garymabbott4064 thank you, though I have two more questions: 1. is the 0.197V potential step between the reference electrode and the solution independent of whatever solution the reference electrode is in? 2. If I've understood correctly, we set the solution to a voltage of V1+0.197 with respect to earth ground, is this correct? In such a case, why is there no reaction at the reference electrode, is it because we are blocking mass transport to the reference wire?
@@lucha6262 1. The +0.197 V is the value that the reference electrode will have for a saturated solution of KCl. The potential energy difference between the AgCl coated Ag wire is, indeed, dependent on the chloride concentration (actually the activity) level. In a 1 M KCl solution, the potential is +0.222 V at 25˚C. 2. Yes, the V1 +0.197 V is the voltage of the solution with respect to ground. There will be a very tiny amount of a net reaction at the AgCl/Ag electrode because a real operational amplifier is not perfect and will allow some current (maybe a few picoamps--less for better amplifiers) to flow into the input of the op amp. That will not shift the potential of the reference electrode significantly. But, you have the right ide., The whole point of using a 3-electrode system was to avoid too much current (say microamps) going through the reference, because that would shift it away from the +0.197 V.
How about the current in ampere(in both oxidation and reduction)? How the system measure the current? is it related to counter electrode and working electrode? Any equation for it just like the potential plot on x axis is the voltage of working electrode related to references electrode. Thank you.
Hi Canny,
You are asking a lot of good questions. This video was only part of the story. The current goes through the working electrode and the counter electrode through separate operational amplifiers in the outer circuit that are part of an instrument called a potentiostat.
Here are two other videos that may help with some of your questions about what controls the current for oxidation and reduction. In these videos, I describe the main concepts from which the current-voltage equations are derived. To get another description, complete with equations see the article below.
Controlled potential analysis. Be patient. The video starts with a review of potentiometric electrode measurements. Then
it describes the principles behind electron transfer experiments (called voltammetry).
ruclips.net/video/d9ADCv5vNzQ/видео.html
Cyclic voltammetry experiment. This describes the set-up of an instrument and running an experiment.
ruclips.net/video/BFefYBEL7wM/видео.html
A Practical Beginner’s Guide to Cyclic Voltammetry
Noémie Elgrishi, Kelley J. Rountree, Brian D. McCarthy, Eric S. Rountree, Thomas T. Eisenhart, and Jillian L. Dempsey
J. Chem. Educ., 2018, 95 (2), pp 197-206.
@@garymabbott4064 Thank you very much for your generous sharing, Mr. I''m really appreciate it. Have a nice day!
Can someone please explain to me...if the working electrode cannot pass current (transfer of electrons into or from the solution) then how does it reduce or oxidize the species??
Hi Emaan,
I am not sure what conditions you are thinking of where the working electrode cannot pass current. You are correct that its main purpose is to reduce or oxidize species in solution. (We prevent the reference electrode from passing current so we don't disturb its potential. Could that be what you were thinking of?) We force the working electrode to different potentials in order to drive an exchange of electrons with something in solution. That potential must be near or more negative than the E˚' for the redox species in order to reduce any of the oxidized form and we monitor the current that passes in the voltammogram. (We see current for the oxidation of any reduced species when the working electrode is near or more positive than the E˚'.)
@@garymabbott4064 thankyou for replying! And im confused because supposedly, a polarizable electrode cannot exchange electrons with the solution. I dont understand, if its not giving or receiving electrons (as a polarizable working electrode) how is it able to reduce or oxidize species?
Im really really new to electrochemistry (my whole career is in bio) so pardon me if im asking something really obvious
@@emaanimtiaz4371 Electrochemistry can be confusing, in part because of all of the jargon. The term “polarizable electrode” means that voltage (the energy for an electron) between the surface of the electrode and the solution can be changed by imposing energy (from a battery or electrical power source) from outside the cell. We need at least one other electrode in the cell in order to do that experiment. A reference electrode, such as a silver/silver chloride electrode, can be used for that purpose. A good reference electrode will not move from its “rest potential” (its value that can be calculated from the Nernst Equation for the reference reaction, such as AgCl + e Ag + Cl-). We say that the reference electrode is non-polarizable. By selecting a non-polarizable reference electrode for a voltammetry experiment, virtually all of the energy (voltage) that we apply from the outside circuit ends up polarizing the working electrode. That is, all of the energy applied to the cell from the outside changes the voltage at the interface between the working electrode and the solution. If that voltage is big enough, it can force an electron exchange process there. A platinum wire or carbon electrodes are good working electrodes since they are polarizable and will exchange electrons readily with species in solution.
One point of possible confusion with electrochemical measurements is that there are two common types of experiments. Voltammetry is the type I was referring to in the paragraph above. Voltammetry is based on applying an external voltage to the system which forces an electron exchange at the working electrode. The current is measured and its magnitude is proportional to the concentration of the electroactive species in solution. Voltammetry uses a polarizable working electrode. The other common experiment is potentiometry. The rest voltage of the working electrode is determined by the solution conditions and it is this quantity that we measure in a potentiometric experiment. In that case the working electrode or sensor should not be polarizable. We try to prevent significant amount of current going through the cell, since it could distort the voltage that appears at the working electrode. That is, allowing current in the system would push the working electrode away from the voltage that naturally develops at its surface because of the concentration of the species that it senses.
I hope that helps.
So, the bottom line is that a polarizable electrode CAN exchange electrons with something in solution, if we push it to the right voltage.
@@garymabbott4064 thankyou, thankyou, thankyou! Iv been stuck on this for 2 days straight, watching a dozen youtube videos and googling it in a hundred different ways and NO ONE explained that. You've saved me from another sleepless night. Huge thanks
Reduction of the analyte on the working electrode requires a source of electrons, where the electron is obtained?
Electrons that the analyte takes from the electrode are replaced by a power supply (often a battery) that is part of the outer circuit. The arrangement of electrons (their "concentration") at the surface of the electrode is set by the voltage that is applied. If an electron "jumps" from the electrode to the analyte in solution, another electron will move up to take its place in order to maintain the arrangement of charges associated with the voltage applied by the power supply.
This was great thank you!!!
Hey guys i need help. topic polymer electrolyte
I run linear sweep voltammetry. Why the decomposition voltage drop as the temperature increase?
how to use glassy carbon?
Very clear, well explained lecture, very helpful, thank you very much...
Thank you !
EXCELLENT
Make it in hindi