Видео

de nada

Hi Kumodgurav, I am going to make some assumptions about the information that you gave me. You are reporting a retention time that suggests that your data is from an LC/MS run. Is it possible that you used an electrospray ionization (ESI) detector in the negative ion mode? ESI frequently produces adduct ions that are a combination of the analyte molecule plus some an ion present in the buffer. The mass difference of 45 amu fits nicely with the idea that the observed parent ion is the neutral drug molecule plus one formate ion [M + HCO2]- See: www.acdlabs.com/blog/common-adduct-and-fragment-ions-in-mass-spectrometry/

That's a good question and you need to ask it whenever someone reports a value for a formal potential, because people do different things. However, most often people work at conditions other than unit activity for all of the reactants and products for the half reaction of interest. Rarely do people work at pH 0 (unit H+ ion activity), for example. The pH is probably one of the most influential parameters that varies among experiments. Temperature is also frequently varied. Excellent experimentalists are careful to report temperature, ionic strength, pH, reactant concentrations for any measurements that are not at standard state.

Think of a molecule with the formula that you have given containing atoms of only the most abundant isotope of each type. (If you don't have a table of isotopes for elements common to organic molecules, you can find one on-line.) For most non-metallic elements, the most abundant isotope tends to be the lightest. Do this for carbon, hydrogen and chlorine separately. Multiply those isotope masses by the number of atoms of that element that appears in each molecule and sum these numbers for all the elements in the molecule.

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.)

There are these situations when the molecular ion is not the tallest peak in the high mass cluster: When the sample is contaminated and more than one compound is being ionized at the same time. We can usually rule this out, if our separation and/or clean up steps are rigorous. However, when the signal is weak background peaks (trace level contaminants and electronic noise) can complicate interpretation. When the compound of interest breaks apart so easily that no molecular ion survives to give a signal. This happens with long alkyl chains, for example. That is a tough puzzle, but one can get a lot of information by applying the same strategy to the cluster of ions at the highest mass with the caveat that this cluster is due to a fragment of the molecular ion is quite possibly an even mass ion. (See Fred McLafferty, Interpretation of Mass Spectra for an in-depth discussion.) When the molecule has multiple chlorine and/or bromine atoms. In those cases the probability of having a molecule with at least one heavy halogen isotope is greater than a molecule with all light isotopes. But these cases are easy to spot, since the tall peaks are spaced by two mass units from each other. Remember that everything heavier than the molecular ion with all light isotopes must be the result of ions with a combination heavy isotopes (and the same chemical structure, namely, the molecule minus one electron). Other ionization methods that we were not considering here also give a different signal for the molecular ion. Electrospray ionization (ESI) is very important method in LC/MS and in desorption ESI (or DESI). These methods lead to much less fragmentation, but form “adducts” (combinations of the molecule plus a hydrogen ion or sometimes a sodium ion). Here the signals in the high mass cluster must be explainable by combinations of isotopes for the structure of the whole molecule plus a hydrogen, that is, (M+H).

Hi Harshit, That is a very good question. Perhaps, I should have addressed that in the video. Let's try it now. Going back to the table of isotopes (in the video) we see that Si would also contribute to the M+2 peak (3.4 % compared to 4.4% for S, but close enough to consider it. We want to be flexible when the intensities are weak, since the relative uncertainty in real data is often bigger with weak signals.) Now, notice that the Si also contributes 5.1 % to the M+1 peak. We need to correct the intensity of the M+1 peak for the contribution of 1 Si atom before calculating the number of carbon atoms, but we see it is bigger than the signal at M+1. (5.1% > 4.2%) If the M+1 peak is the result of a silicon atom, then there is no room for carbon. So, a silicon atom does not seem likely. Make sense?

Thanks for the explaination. So, sir in the nutshell, do we have to consider only those elements whose M+1 intensity percentage in the table is less than the M+1 intensity given in question? Because if (5.1% of 75.2 =3.8) < 4.1 is consider here, we might not be able to put Carbon in the formula.

@@harshitshukla7208 Yes. I think you have it. I want to apologize for sloppiness in my earlier answer. I should have multiplied 5.1% (the contribution to the M+1 peak that a silicon atom would provide) by 75.2 (the intensity of the M peak) to get 3.8 as you have done. That is the Si contribution to M+! in intensity units. That is the number to compare with the signal at M+1. 3.8 is still smaller than 4.1, but it seems to leave little room for carbon. If we go back to correct our calculation for the number of carbon atoms, we must subtract the Si contribution out first. --> [(4.1-3.8)/75.2]x100/1.1 = 0.398 carbon atoms. Much less than 1.0 does not look good. If the data were really noisy we might round up and consider it, one carbon and one silicon, but we now have to go back and account for 75 amu in the molecular ion again. That would also lead to more inconsistencies. Silicon seems unlikely.

Thanks sir.

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

Your question goes to the heart of the field of chromatography. Basically, when two compounds elute a the same time, you need to change the separation conditions (either the stationary phase or the mobile phase or both) in order to retain one more strongly than the other.

Hi Dawson, The term “slit width” refers to the physical width of the opening used to pass a narrow portion of the spectrum of light to be used in the experiment. The light from the lamp is dispersed by a grating into a rainbow across a wall inside the box that houses the monochromator. The opening in the wall (called the slit) allows only a narrow band (a small range of wavelengths) to pass out of the monochromator and on to the detector. In some contexts, slit width may also refer to the range of wavelengths (usually in nanometers for visible light) that pass through the monochromator (that is, through the physical slit). More often this wavelength range is called the "spectral bandpass". The narrower the slit width the better job the monochromator does in isolating the wavelength for the atomic absorption transition. However, beyond a certain point decreasing the slit width is detrimental because it weakens the signal intensity and, consequently, poorer sensitivity.