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    Welcome to the course corrosion part II.And today we will just recap what we have discussed in part 1 are important aspectsof electrochemical polarization, activation and concentration polarization. So the firstlecture will be on the recap of electrochemical polarizationand we have understood in the first lecture which is this part 1 that there are 2 kindsof polarization we will talk about activation polarizationand concentration polarization.Now when we talked we start a discussion corrosion in part 1 we got into the fundamentals andthe fundamentals are based on thermodynamics of electrochemical reactions as well as kineticof electrochemical reactions. Now later we got to know from polarization aspect thatpolarization actually connects thermodynamics aspect which is basically the potential andkinetic aspects which is basically the current density.So we will just briefly talk about electrochemical polarization, so that our consecutive lecturesbecome easy to understand and I would request you to go back to the lectures in part I.So that it becomes easy. Now we have understood that if we allow 1 metal to be dipped intoon an electrolyte containing that particular metal ion. For example, if I have if you seethat if this is a beaker and there is an electrolyte which is aqueous electrolyte.And here you have metal ion concentration of let say some activity if we mention inthe form an activity, so some activities there in the solution of that particular metal ionand if we dip the same metal in then immediately some potential will be developed, and at thesame time at the interface of the metal there will be double layer formation and those doublelayers you have understood that the double layers are called Helmholtz double layer.And there could be outer double layer outer Helmholtz layer or plane and there could beinner Helmholtz layer or plane, in short we call it OHP and this one we call it IHP. IHPis forming on the metal stuff on the on the metal and just in front of that particularmetal surface in the electrolyte we have OHP and in this case this forms OHP and this formsIHP. Now once we introduce this metal into that electrolyte containing metal ion youwill experience 2 reactions.One reaction is in M n plus plus n e which forming M at the same time opposite reactionwould also happen M plus and we we should realize that this particular arrangement thatmetal inserted or dipped in that electrolyte containing metal ion at particular temperatureand pressure. So this temperature and pressure both are constant for this particular experimentation.Now we see that one particular reaction is cathodic in nature.Cathodic or reduction and another reaction is anodic or oxidation, now every reactionwould try to reach equilibrium for a particular temperature pressure condition. These 2 reactionswould also like to go to equilibrium. Now if they would like to go to equilibrium, sothat means the rate of forward reaction if we consider this is to be forward reaction,and if this is to be a backward reaction. So then this rate of forward and backwardreaction at a particular temperature pressure condition should be same should be equal.So that means at equilibrium we will have M so that means forward rate if this is basicallyrate of forward reaction and this is rate of backward reaction for equilibrium conditionr f equal to r b. Now interestingly for a steady state that time also we have the situationlike the rates are equal. But in this case we will see that we are not having any netreaction, later on we would understand that corrosion event is basically a steady stateprocess.If there are nothing external agents added, so that time it steady state the rate of reductionreaction is equal to rate of rate of oxidation reactions will see later, but that time thereis a constant change in the consent in the amount of a species we will see later, butin this case whenever the situation arises we will see that rate of forward and backwardreaction are same. But at the same time there is no net change in product or reactant.So that means this and this they will not have any net change, so the net change ifI consider net should be 0, so this condition we call it as equilibrium condition. Now atthe same time interestingly if I consider r f, r f is basically reduction process orcathode process. And in the cathodic process we see that electrons are accepted by metalions and they are actually going to deposit on the metal object.Whereas in the backward reaction I can see the electrons are taken away from metal atomand they are forming metal ions. So once they form metal ions they will be part of thatelectrolyte. Hence if we take that rate of electron flow, then that can be consider inthe form of current. Because here I could see the charge which is electron in case ofcathodic we are introducing electron. So that forward reaction is possible.And in case of backward reaction we are taking away electron and if we consider time scale,so they can be considered as a current flow. When the forward reaction if we consider therate of charge flow per unit time there would be current and if we consider the overallarea of the surface just to normalize it with respect to area and we know why we normalizefrom corrosion part 1.So then that can be considered as current density which is ‘i’. Now this ‘i' whichis current density it is basically nothing but ‘i by a’, ‘a’ is basically thearea. Now here ‘i' is ampere which is current which is basically the rate of charge flowper unit time. And then per unit area it becomes current density. And if we see the kineticsof corrosion we have realized that rate of any reaction or electrochemical reaction.Because we have seen that these reaction is electrochemical in nature rate of electrochemicalreaction can be expressed in the form of current density and let’s say in this case the forwardreaction is deposition, and the backward reaction is ion formation.So then I can express either deposition rate or the dissolution rate in the form of massper unit time per unit area which is written as M by t by area.It can be related to i A, A is nothing but the atomic weight n F what are those terms,i of course is the current density A is atomic weight in gram, n is equal to the number ofelectrons participating in unit reduction process or oxidation process and this n inthis case it is easy to understand that this is nothing but oxidation number of that particularmetal electrons participating and F equal to 1 Faraday.And this can be calculated from charge of an electron which is 1.602 into 10 to thepower -19 coulomb into Avogadro number which is nothing but 6.023 into 10 to the power23 because 1 faraday is nothing but charge of 1 mole of electron. So that becomes 96488which approximately consider as 96500 coulomb. Now you could see that the rate of depositionor rate of dissolution I can term it as cathodic or reduction and in this case this is oxidationor anodic. Both can be reflected by this reaction by this particular equation. Now, If I considerthis particular process that is in this case when I am considering reduction process.I have to term it to some current density, and since the cathodic process is taking place,I term this particular current density what is flowing into the circuit I can term itas i c and c should subscript is not nothing but the cathodic and whereas the backwardreaction which is anodic reaction, the current density I can term it as i a, now interestinglythis i c is actually flowing opposite to i a that is the flow of current for the anodicreaction anodic rate rather is from positive to negative whereas in case of i c is negativeto positive.Now in order to indicate the direction convention is we put negative sign in front of i c, sothat is it is not a negative current it is basically the direction of that particularcurrent and remember if we do not know explicitly express that its current most of the situationsthis i small i means current density. Even if I say current for a small i take it asa current density.Now for this case if they become equal. So that means i A c divided by n F for the forwardreaction which is nothing but r f should be equal to i a A by nF, because for both thecases A are same because the same area of the metal that is taking part in the reactionn is the number of electron participating in for unit reaction and F is 1 Faraday. Sothat means this part and here this part both are same for both the rates and this is rb which is backward reaction.So that means i a equal to i c, now since we have to indicate the direction so thatwhat we put a negative sign, and when this situation arises we call it as i zero andremember this i zero is with respect to non deposition with respect to the situation wherethere is no net change in that particular process that reversible reaction we do nothave net change. So that means there is no net dissolution there is no net deposition.So we call it as i zero is exchange current densityand with respect to non-corroding equilibrium situation, fine, now at when thesystem reaches this equilibrium that means the flow of current or the current densityfor both these backward and forward reactions they are same which is indicated by i zerowhich is exchange current density area in the under non-corroding equilibrium situation.That time the system must have attained some equilibrium potential, and that potentialwe express in the form of n plus M. So this isreduction potential and remember our consideration will be all throughout our 40 lectures whereverwe consider potential this potential is in the form of reduction potential. Now thiscan be written in the form of E zero M n plus m plus RT nF ln activity of ox divided bythat is activity of oxidant by activity of reductant.And here oxidant is M n plus and activity of M, now this is a generalized form for Nernstequation where we have considered in the form of activity of that particular metal ion andhere E zero n plus M is nothing but standard reduction potential.Now if we write it again, so that means M n plus M is equal to E zero M plus n plusRT nF ln a M plus a M. Now this activity as per thermodynamics we can write in the formof concentration with concentration and activity coefficient in the form of concentration andactivity coefficient and this concentration we can write in the form of molarity. So Mn plus activity coefficient of M n plus. I can write this way. Now if the system is notbehaving ideally then it must have some finite value which is different than unity.And when this goes to 1 we can call it as ideal solution, so that means in that caseM n plus equal to molarity of that particular metal ion concentration in the solution. Nowinterestingly most of our situations unless it is specifically mentioned will considerin the form of molarity assuming that this activity coefficient is 1. And now here weare considering a pure metal though we have not mentioned in the beginning.Let us considered that it is a pure metal that is dipped into the solution so that caseactivity of M goes to 1. So that means I can write this equation in the form of molarityn plus M is equal to E zero M n plus M plus RT n F ln. Fine. So this is my final equation.Now interestingly these value this is the actual potential that is been developed onthe surface on the metal. Now if this is 1 then of course E n plus M equals to E zeron plus M but otherwise if this has got value other than 1 then this would be my equilibriumpotential when i equals to i zero in the system that is in the non corroding system.So we could see that whenever we have a situation of equilibrium setting up between a metaland metal ion in an electrolyte at particular temperature and pressure, we always will developan equilibrium potential corresponding to that condition. Now if we can express thisone in the form of energetics then that energetics will look like this. So that means we havedouble layers and this is M n plus layer.This is M and then at this condition that means equilibrium potential and when i equalsto i zero reversible condition, if I draw the energetics the energetics will look likethis. So that means this is M side, this is metal ion side and if this process is takingplace that means M minus ne equal to M n plus and if this process is taking place M n plusplus n e equal to M. So the reduction is the thermally activated jump of metal ion fromOHP to IHP crossing the barrier of if I consider this barrier to be this which is del G star.This much barrier is crossed by metal ion to go to the metals surface from OHP to IHPafter taking n number of electrons. Similarly, if metal is planning to go to the other sidefrom IHP to OHP it has to release any number of electrons to form M n plus but that timealso it is crossing the same area size area height. So as per the thermally activatedjump theory this 2 rates 2 jump situations are actually having the same barrier samebarrier height.If they have same barrier height their rates will be same, so that means r f equal to rbackward or I can write I c equivalently or maybe absolute sense I can writeequal to I a A n f equal to i zero A n f or now we can see equivalently i c equal to ia equal to i zero. So this energetics is valid for equilibrium condition which is the noncorroding situation and when the equilibrium potential is this one and whenever we mentionthis that means it is not that condition for M n plus equal to 1 the molarity of the particularmetal ion is 1 in that system.In some other conditions and when exchange current density is experienced. So this isenergetics for the equilibrium situation. Now we will talk more on this and we willget into the polarization concept and then after that will slowly get into this mix potentialtheory and understanding of corrosion events via the mix potential theory. So let me stophere and in the next lecture will continue our discussion on this energy distributionin a situation when polarization takes place. Thank you.