 Now, in this second lecture of module 3, we continue the discussion on electrostatic phenomena. The first topic we will take up is one on diffusion potentials and then we will return to one of the questions which I had posed before you and in order to answer that we may have to look at the theory which allowed us to relate the interfacial potential, the total interfacial potential to the distribution coefficients of the cations and anions. So, the analysis that we had seen last time will briefly review before coming to the question about what if the non-aqueous phase is highly non-puller or just does not permit presence of any ions at all like in case of air. So, first to be taken up is the diffusion potentials. Now, conceptually this is a slightly different idea. Like earlier we imagine that there are two aqueous phases between which a polar oil is sandwiched and we measure the transient potential between the two surfaces or interfaces that is the left hand phase of oil in contact with the first aqueous phase and right hand phase which is in contact with the second aqueous phase. Now, if we have cations and anions getting transported from let us say the left hand phase to the right hand phase, then they would do so with certain ionic mobility through the oil corresponding to that there will be a transport number. If the cations and anions are different in size and appreciably different in properties especially with reference to oil then the different transport numbers of these ions will result in different diffusional fluxes through the oil which would mean a time dependent potential delta phi could be measured across electrodes which are in contact with the two aqueous phases and one would say further if the system is ideal then the opposing phase boundary potentials should be equal as per our equation 18. This is appropriate stage to review the derivation that we had seen last time. The equation we just referred to is this equation 18. So, to have a quick recap we started with the definitions of the distribution cations for cations and anions given by RT ln B plus or B minus equal to W mu plus or mu minus 0 minus O mu plus or mu minus 0. Then we had the electrochemical potentials for the cations and anions in water and oil as related to the chemical potentials W mu plus 0, O mu plus 0, W mu minus 0, O mu minus 0. The concentrations W c plus W c minus O c plus O c minus contributing in this RT ln c kind of terms and then for oil we have a positive phi f adding to the chemical potential and minus phi f for the anions. So, those four definitions of electrochemical potentials for two kinds of ions in two phases and then we had the pair of equilibrium relations where in the oil and water phases the cations and anions would have respectively equal chemical potentials W mu plus bar equal to O mu plus bar and W mu minus bar equal to O mu minus bar. And finally, the electroneutrality conditions that in the bulk of water the cations and anions would have the same concentration in the bulk of oil again cations and anions would have same concentration. So, after plugging in those definitions 1 to 4 in equations phi and 6 equilibrium relations and keeping in mind the electroneutrality conditions we had ultimately arrived at this equation for phi, phi as RT by 2 F ln b plus by b minus. We went on to express this in terms of the distribution coefficient for the salt as such. So, that gives you RT ln b square by b minus square that was what we got after adding those equations 9 and 10 which allowed us to arrive at this result finally. So, phi is RT by 2 F ln b plus by b minus or RT by 2 F ln b square by b minus square. The point to be noted here is that the oil is like nitrobenzene, it is not a completely non-polar oil there is a polarity here. That means, it allows the electrical double layer to form within the oil phase. It is here that we come to the topic of surface and interfacial potentials. As proposed earlier the non-aqueous phase is now chosen such that practically no salt can dissolve in it practically. That means, still it is conceivable that very small concentration of the ions may still be formed. An example could be a parafonic like decay or strictly speaking air. Air will be an extreme example when ions cannot exist there and we presume the oil to be at least about 1 mm thick about 1 mm thick not very much higher. Now, the question is among all the equations that we had over here which ones are still applicable and which ones not. The basic definitions relating the electrochemical potentials of cations and anions in the two phases those would remain unchanged. So, equations 1 to 4 would be unchanged. Equations 5 and 6 are statement of equilibrium. So, in principle this should be fine this pair of equations is also admissible. What about electroneutrality? Equations 7 W c plus equal to W c minus that should be fine. We can have cations and anions and electroneutrality in the aqueous phase. Question is only about this equation 8. The last of the equations we required in establishing a relation between phi B plus and B minus. Remember the constraint of our system is that the non aqueous phase is highly non polar. The tiny concentrations which you can imagine for cations and anions there may not be able to build up ever within that 1 mm thickness of oil to give you O c plus equal to O c minus anywhere in the finite system ok. So, that is a constraint now. If O c plus equal to O c minus cannot be taken as a valid relation in the finite extent of the oil, how do we then modify the phi B plus B minus relation? We have water and oil which is non polar in the oil which may let us say have a thickness of about 1 mm. The equality of O c plus equal to O c minus is not valid. Solubility of the cations and anions in this oil is quite low. If this is the case and you could actually at different distances from the interface measure the phi value, how would we expect phi now to depend on B plus and B minus? Earlier relation phi t equal to phi equal to R t by 2 F ln B plus by B minus, we do not expect now to be valid because that last equation 8 is not applicable. Now, this is the starting point for our discussion. I will let you spend a few minutes with the equations 1 to 7, equation 8 is not given to you. Make an attempt to find the relation between phi B plus and B minus on similar lines. And before you do that maybe let me mention the magnitude of the diffusion potentials. Diffusion potentials normally are small about 15 millivolts. And if you think of a very thin film about 1 micron thick, then it is possible that diffusion potential may influence the relative ionic mobilities. And presence of a charged monolayer across the surface of such a thin film may be able to alter the diffusion potentials. Now, you can go into pursuit of relation of phi between phi and B plus and B minus, if we cannot have OC plus equal to OC minus anywhere in this thin film. Perhaps there is no difficulty in arriving at this relation. Last time we calculated potential difference for a slightly Polaroid and the Equus basis. And this time we are trying to calculate it for a very non Polaroid. Non Polaroid. Non Polaroid. Non Polaroid. Right. Now, if you see this, butener said that due to the difference of distribution of ions in that pursuit of resist, they printed them. And to understand that it should be of two different types. Correct. Now, if we have a non Polaroid on one side, then the ions won't be distributed to more is getting dominated. Ok. Ok. Are you able to get what Vijay is trying to indicate? We have deliberately chosen this system such that the oil is highly non polar and therefore, we cannot have now OC plus equal to OC minus. So, this term will remain. Right. But you have gone very far and you said that now we should say bar was right. Right. We have conditions appropriate. Yes and we should be able to prove that. But so far, are you with me? This is ok. Everyone gets this. Since OC plus was equal to OC minus, the second term was absent earlier. Now, it has to be kept. We have to deal with it somehow. All right. What we will do now? We will divide the potential on the oil side of the interface into in general. The contact potential be due to whatever adsorbed or spread layer at the interface. It could be a dipole array or it could be a charge film adsorbed adsorbed or spread at the interface and psi for whatever solubilities they might be for their distribution. And we will show that if the oil solubilities of the interfacial film and of salt which are present there very small, then only delta V will contribute to delta phi when such a film is spread at the interface. So, we are going to prove that now. And the first step in that direction is to obtain this equation which is the result of equations 1 to 7. Four electrochemical definitions, the equilibrium relations and one of the electroneutrality relations right ok. Next and this is something which I had alluded to earlier. I did tell you earlier that we can later think of a situation where only contribution stable contribution can come from delta V. So, we are actually addressing that that particular statement made long time back ok. Now, let us look at this equation 20 closer. What does it tell us? Within the oil we may choose different locations. There is a different distances from the interface and think of what the concentrations of cations and anions are there. Supposing at certain distance the cation concentration is this O c plus. O c minus is the anion concentration. The two are not equal. What does this equation then give you? And this will also address one of the earlier questions. When I did not have the second term, the question which was rightly raised was where do we measure phi? And I had answered that that is the total potential between the bulk of oil and bulk of water. Bulk of water is given arbitrarily 0 potential. Absolute potential anyway we cannot measure. We attribute absolute potential equal to 0 to water. But here we could choose different locations. There is a different distances from the interface. Have the pair of cation anion concentrations they could be measured maybe. Then what will be phi? Phi should be the potential that will measure at that location within oil relative to bulk water right. So, now in order to go further we have to do something about this pair of concentrations O c plus and O c minus. Now supposing in such a system that is the aqueous and highly non-polar oil or air we think of a certain spread or adsorbed monolayer of film which can offer the contribution to the total potential in the form of say V or delta V ok. And as we go away from the interface the concentrations vary and therefore, we should be able to relate the concentration to the potential which is contact potential at the interface. So, we need basically a relationship between O c plus and O c minus in terms of V or measurable potential. The only measurable potential here will be the contact potential V. So, here we have equations of Boltzmann which are applicable near the interface where there is a contact potential V and these concentrations could be expressed in terms of V as shown in equations 21 and 22. O c plus is equal to O c plus at x equal to infinity at infinite distance into exponential minus epsilon which is electronic charge 4.774 10 to the power minus 10 electrostatic units V minus psi at x equal to infinity upon k t and O c minus is given by O c minus at x equal to infinity exponential plus. If you talk of cations there is a plus sign minus sign if you talk of anions there is a plus sign epsilon V minus psi at x equal to infinity by k t ok. So, the O c plus O c minus here could now be expressed in terms of the measurable quantity V and other entities which are given in this Boltzmann's equations. I would suggest you to plug in equations 21 and 22 in 20 and see what is the consequence. Substitute for O c plus and O c minus in equation 20 yeah you have to know they are certain structure. So, it is not difficult. So, if you substitute for O c plus and O c minus in the expression for phi you can observe that in the hypothetical region at x equal to infinity which is outside the practical region the double layer would be complete and O c plus will be equal to O c minus only at x equal to plus infinity. So, after substituting these equations equations 21 and 22 in 20 and noting that at x equal to infinity O c plus and O c minus are equal we could arrive at this result 24. First term is same as earlier R t by 2 f ln B plus by B minus. In the second term we have minus R t by 2 f ln exponential this by this that will be cancelled that will become minus 2 epsilon V minus psi at x equal to infinity by k t. So, this is what you get ln of exponential minus 2 epsilon V minus psi at x equal to infinity by k t, but what is this R t by 2 f ln B plus by B minus. This is one glance here will tell you at x equal to infinity O c plus equal to O c minus and we get the earlier expression R t by 2 f ln B plus by B minus which was psi, but now this is valid in this case at x equal to infinity right. In the earlier case this phi would be entirely by psi and in this case at x equal to infinity in those terms it will be phi equal to psi at x equal to infinity. So, this is psi at x equal to infinity. Once we make this observation we can write equation 24 I hope all this is clear so far we can substitute for the first term over here psi at x equal to infinity right. So, the first term which is R t by 2 f ln B plus by B minus we can replace by psi at x equal to infinity and therefore, this equation 24 can be written as phi equal to psi at x equal to infinity minus R t by 2 f. Since ln and exponential will annihilate each other now we will have simply result of all this only this bracket minus 2 epsilon bracket B minus psi at x equal to x equal to infinity by k t. So, we have here R t by 2 f into minus 2 epsilon B minus psi at x equal to infinity by k t. Now, we can see that this t cancels with this t ok temperature cancels out Faraday's unit f by electronic charge epsilon is Avogadro's number same way the gas constant R by Boltzmann's constant k will be Avogadro's number. So, basically f by epsilon is same as R by k which is Avogadro's number. So, all this cancels f by epsilon and R by k cancels t cancels. So, we are only left with these terms minus 2 epsilon V minus psi at x equal to infinity. So, when you apply this negative sign we find phi is equal to V right. So, we can prove for a non-puller oil or air in contact with water phi is independent of the two distribution equations B plus and B minus in contrast to the equation 13 where phi was R t by 2 epsilon B plus by B minus. But remember that this is a deliberately chosen system to force conditions that Bohr had visualized. So, there was nothing wrong in that visualization, but just that visualization does not correspond to most of the realistic systems few systems will obey. So, may be broad conclusion could be that serious scientific thought cannot be totally wrong in all context. In reality we can only measure the differences in potential. So, delta phi equal to delta V here the contribution is due to an adsorbed layer. This expectation can be actually blown out by experiments. A capacity method could be used in which an alternating current generated by a small scale amplitude vibration of a gold plate immersed in oil balanced to 0 by a potentiometer connected to the electrode in water shows that if the non aqueous phase is paraffin or air the change in phi is only due to V and change in phi does not depend on B. So, what matters is only the properties of the monolayer and not the properties of the non aqueous phase right. In tune with the expectation from the theory which has given you delta phi now only depend on dependent on delta V. The psi came concept came which is the distribution potential unequal solubilities of cationsilence. We have a system analyzed where phi is RT by 2F ln B plus by B minus right. We have chosen now an oil with which has very little solubility for cations and anions, but does not mean it is practically negligible, but there will be certain amount of ions present anyway, but they will take infinite distance to build up OC plus equal to OC minus. If we were to measure the potential at infinity it will still be psi potential at x equal to infinity. You have to understand a fine difference between the visualization and the theory. According to this theory we have a certain adsorb monolayer or spread film which has ions which have very little solubility in the bulk of oil, but given infinite extent of oil and if you were to measure the potential at infinity you could still call it a psi potential ok. So, and theoretically therefore, that RT by RT ln B plus by B minus is equitable to psi at x equal to infinity. We will see later once more that what is strictly a contact potential could be regarded as a distribution potential or the other way round. So, when we understand the physical picture better and we have an idea about the magnitudes of these entities parameters here it is possible to visualize the situations in an equivalent manner. So, experiments do reveal on 100th molar sodium chloride solution at air water or oil water surfaces. If we have this spread monolayer octa desil trimethyl ammonium ions then the expectation from this equation delta phi equal to delta v is actually confirmed and such potentials are completely stable in time. We require however a high boiling paraffin like petroleum ether that is decaying or near homologs for this to be true. We return now to the old picture if one uses instead a fairly polar oil according to our equation 18 there would be no change in potential between a system where the interface is clean or one with monolayer spread at the interface. Only if the distribution questions are changed will there be a permanent potential delta phi in full agreement with the experiment. Otherwise for the fairly polar oil the adsorbed monolayer or whatever is anchored at the interface will always distribute over time leading to contribution only corresponding to delta psi. So, now we have these two extremes appreciably polar oil will have delta phi coming only from delta psi. Highly non-polar oil will have delta phi coming only from delta v. So, that should make a comfortable understanding of the situation in theoretical terms. Now, we come to a slightly different picture about the contact potential originating from the contact potential originating from a monolayer. I did talk about water quite frequently earlier. So, you have an idea what water is like. We also talked about influence of hydrocarbons on structure of water in the vicinity of the interface. We revisit that picture. And now think of what might contribute to delta v in such a situation. In other words, if there is a measurable delta v for a hydrocarbon in contact with water or a monolayer in adsorbed condition at an interface what will constitute components of delta v. In a sense we will go first in the analytical direction, visualize what might be happening and then see what are our limitations and come back maybe to some kind of overall description. So, before we do that let us see something about water again. At clean surface of pure water, the potential psi due to free charge may be taken as 0. And as a consequence in absolute potential phi of water is only this contact potential v. And measurements by different people have led to values of v from minus 0.5 to plus 0.4 volts. And many of this researchers agree on a value between minus 0.1 and minus 0.2 volts. That is a small magnitude. We will anyway avoid the absolute potential in our discussion. This is only to give an idea for a clean surface of pure water, the phi coming from contribution of v will be small minus 0.1 to minus 0.2 volts. Now we imagine in this system a monolayer to be adsorbed at the clean water surface or it could be a thin film of hydrocarbon spread on the surface. We must realize that the kinds of molecules we are dealing with water and monolayer forming chemicals, they will have associated with them dipole moments. Remember the case that influence of hydrocarbon on water viewed in terms of entropies led to a picture that water became slightly oriented in presence of the hydrocarbon bonds leading to semi solidification state for water. The entropy change for formation of a unit surface of water influenced by hydrocarbon led to an entropy value which is in between which is below the complete solidification or that of ice. So, that picture still can be brought in here except that it gets finer. If we have a monolayer at the surface that is how or spread, it is going to create certain semi solid like state within water which means the water gets somehow structured, water HOH with its dipole moments will reorient near this interface. And when the water dipoles get reoriented near the surface, they will contribute a certain dipole which is denoted by this later mu sub 1. Secondly, this has happened because you have some kind of long chain molecule anchored at the interface. So, they must be a dipole which is caused behind the orientation of water to give rise to this dipole moment mu 1. So, if you think of something like a long chain amine, then the terminal C NH 2 groups will have their own dipole moment that those dipole moments are responsible for creating this dipole moment within water due to the reorientation of water molecules. So, that is our second entity mu 1, mu 2 coming from this terminal head group C NH 2 like in long chain amine contributing the group dipole moment mu 2. Third we are thinking of electrically neutral films here. At the terminal limit or upper limit of the monolayer, there would still be another dipole moment mu 3 associated with this C H bond. So, if you think of this long chain amine, the head group which is submerged in water is orienting water dipoles, that head group has its own dipole moment. These are respectively mu 1 and mu 2. At the top the C H bond which exists that will have a dipole moment mu 3. So, the picture will be somewhat like this. You take a long chain amine here C H 3, C H 2 and so on that backbone of carbon chains and terminal NH 2 group. This group C NH 2 group has dipole moment mu 2 that orient water molecules, they will have dipole moment mu 1 and this top end here will correspond to a third dipole mu 3. So, we recognize that conceptually delta V should have these components mu 1, mu 2, mu 3 although mu 1 is a consequence of presence of this monolayer. The extent to which ordering or reorientation will occur for water molecules will depend on this mu 2. Coupled with this fact, it is also hard to measure mu 3 sorry mu 1. Therefore, we first write what would be the theoretical contribution. Here Helmholtz formula comes handy. It gives you for n dipoles per area per centimeter square. If all these dipoles are vectorially additive in vertical direction, we will have delta V equal to terms of this type 4 pi n mu 1 plus 4 pi n mu 2 plus 4 pi n mu 3. We recognize mu 1 is hard to measure, but is the result of mu 2. So, we may want to actually combine this and it turns out that mu 3, mu 3 is a constant because these hydrocarbon chains are very common in surfactant like molecules. So, while mu 1 cannot be measured, but depends on mu 2, we may be able to combine them and mu 3 normally being a constant because in surface active molecules we have this paraffinic chains. We may say after visualizing that those will be the three principal components. We could say that we could add up mu 1, mu 2, mu 3 and say it is some kind of overall dipole movement mu D. So, delta V is now 4 pi n mu D and this should be a correct expression for contact potential delta V in terms of the observed or overall dipole moment for any electrically neutral film which only can contribute dipoles. So, if there is a monolayer that should mu 2 and mu 3 represented by a single quantity because since a molecule will have a single dipole moment, then the combined paraffinic chain plus the image should give us one dipole moment. That is what we have done mu 3 because mu 1 is induced by mu 2 and mu 2 and mu 3 can be combined together with mu 1 we get mu D. So, what we are doing is we are looking at what might be the components, but we ultimately add up all these. So, if we add them up beforehand, we would not realize that in a long chain there are these dipole moments at the top different and at the bottom different. There will be always this distribution of electrons leading to dipole moments at the terminal ends is one end which is reorienting water. So, that is dependent we combine everything in mu D now. So, if this picture is adequate it should give us some consequences. For example, I told you about the surface molecules. We should be able to see some changes in measured dipole moments depending on the state of compression of the surface and depending also on the symmetry or the molecules which are anchored the head groups which are anchored in water. So, let us see whether we can have something like that. Before we do that let me mention here that these terms like mu D whatever is the overall dipole moment for such a system it will be measured in milli device m D. And for a head group like of ethyl permeate the head group has a moment of 535 milli device experimentally calculated one from such analysis will be 525 milli device. We chosen ethyl permeate here because this would not be a very symmetric kind of head group. If compressed we might be able to alter the magnitude of the dipole moment. If it were symmetrical it should not matter because then there will be no change of orientation. So, dipoles would not be able to contribute differently. So, if the area available for such single molecule is greater than 72 angstrom square we will have a picture of this sort. If we compress it then we will get a different value. I guess we can conclude lecture this lecture today and we will take up charge monolayers next time.