Hello everyone let’s start lecture 9.And topic of course explanation of corrosion event on the basis ofmixed potential theory and we have discussed 1 particular situation which is 2 cathodicand 1 anodic reaction.So, we discuss 2 cathodic and 1 anodic reaction that is the situation we have consider.We have drawn mixed potential plot this was the plot what we have discussed in lecture8.And then we started discussing 1 solution numerical problem based on that particulardiscussion that when we have a strong oxidizer or a strong cathodic reaction in additionto another cathodic reaction.We have seen that the corrosion rate of that active component increases greatly.Why it increases greatly?Because this is log scale.So, that means if we increase it so it will be in orders of magnitude and that is whatwe could see that the order it was initially 10 to the power minus 8 when there was nooxidizer in the form of Fe plus 3.Now once we have added oxidizer, so because of the mixed potential theory new dissolutionrate which is this has gone into the minus 6.So, that means the there is a 2 order of magnitude increase in the corrosion rate of zinc.Now then we wanted to solve this values okay, so let’s how do we attack this particularproblem, let’s go one by one.The first thing we had to find out this equilibrium potentials and as well as I think we missedone more equilibrium potential which was E. Okay.So, if we consider these many equilibrium, so we have to find out all the equilibriumpotentials.Now here how many equilibrium potential which basically the reversible potential that couldbe present this is one which is E or let me draw it in this colour.So, E zinc equilibrium then this is already I have indicated this one this equilibriumpotential which will be decided by the pH of the solution.And then one more equilibrium potential is this one which is E equilibrium Fe plus 3Fe plus 2, so this is the point and this is the point 3 equilibrium potential.And now we have 2 mixed potentials, 1 mixed potential is this one which is the mixed potentialwhen there is no oxidizer only hydrogen reduction and zinc oxidation are considered.So, this is E mix I simply put it as E mix and this is thepotential which is E mix prime.So, this E mix prime and E mix are nothing but this is nothing but E mix prime and thisis nothing but E mix only E mix as per this diagram.Now one by one first let’s calculate equilibrium potential E equilibrium for hydrogen reaction,we know that E equal to E zero minus 0.059 pH, so where this value is 0 this will beminus 0.059 pH is nothing but E equilibrium H2.And these are the data we have another data weather information what we have is temperatureequal to 25 degree Celsius pressure equal to 1 atmosphere.So, this since we have considered pH to be 3, so E equilibrium H2 equal to minus 0.059pH equal to and pH equal to 3.I can change it into 3, is nothing but minus 0.177 volt and this is all with referenceto a standard hydrogen electrode.The next we have to find out E mix which is nothingbut E mix when Fe plus 3 is not there.Fine.So, this potential would be we have to find out applying this equationi zero, so now here neeta c equal to E mix minus E equilibrium hydrogen.Now here we are considering the polarization for hydrogen evolution reaction on zinc surfaceand this is cathodic polarization because the cathodic reaction is nothing but the hydrogenevolution on zinc surface which should be equal to minus beta c which is nothing but0.1 log of ic hydrogen on zinc surface divided by i zero which is the exchange current densityof hydrogen on zinc surface.So, that means we are considering this line, now what would be ic hydrogen on zinc surfaceat E mix.Since this is the mix potential when Fe plus 3 is not there, so that means at E mix potentialE mix should be ia zinc which is nothing but I corr zinc.So, here i corr zinc is nothing but 10 to the power minus 8 ampere per centimeter square,so E mix minus and E equilibrium is how much we have already found out minus 0.177 equalto minus 0.01 log of 10 to the power minus 8.And what is i zero for zinc hydrogen of evolution reaction on zinc it is 10 to the power minus11 ampere per second, so, this becomes 0.3 with a minus sign.So E mix equal to 0.3 minus 0.177 equal to minus 0.477 volt.Now we have to find out E equilibrium zinc, so third isE equilibrium zinc. let’s find this, what should be this value.Now if we see this plot again we can consider this line and again we can consider polarizationfor the anodic reaction that is zinc dissolution.So, this is my E equilibrium zinc parallel line of that, so this should be my over voltage.So, this is na zinc, so these over voltage would beequal to beta a log ia divided by i zero and here it is i zero zinc over zinc surface andthis is ia of zinc.So, let’s see that those points this is the point which is i zero for the zinc dissolutionand for the zinc at equilibrium zinc potential and this is the E corr where i corr is 10to the power minus 8 that means ia is nothing but 10 to the power minus 8.So, if we put those values beta a is 0.1 log 10 to the power minus 8 divided by 10 to thepower minus 11, here is also it is 10 to the power minus 11, so let’s put that, so youwill get 0.3.And this nothing but E mix minus E equilibrium zinc equal to 0.3.So E mix we have already found out minus 0.477 minus E equilibrium zinc equal to 0.3, sowe are getting E equilibrium zinc equal to minus 0.777, fine, this is the value.Now then let’s find out what should be the E equilibrium plus 2, so in order to findthat if we would like to find this we have to find first we have to find E mix primewhich is nothing but let’s so instead of this let’s get to E mix prime which is nothingbut E mix when Fe plus 3 is present.So, to solve that we can consider na zinc up to the potential of E mix prime, so herewe consider, so if we take this extend this values, so we have to consider this polarizationwhich is na prime.So, here na zinc prime equal to beta a log of ia zinc divided by i zero zinc over zinc,so here this is E mix prime minus E equilibrium zinc equal to 0.1 log.Now here corrosion rate i corr zinc equal to 10 to the power minus 6 at E mix primewhich is nothing but ia zinc.So, this is 10 to the power minus 6 and this value is 10 to the power minus 11 so it becomes0.5 and this is minus minus 0.777 and this is E mix prime, so E mix prime equal to 0.5minus 0.777 minus 0.277 volt.So, once we know E mix prime then we can find out this value we can find out, how, so ifwe extend this further this particular point.So, if we extend this particular E mix prime and then we have to see the polarization thisparticular, so this is nc of Fe plus 3 Fe plus 2.So, this polarization if we consider we can find out this particular quantity, how, sonow we have this dotted line, this dotted line we have if we consider polarization alongthis line I can fit this particular equation in this form.So, 5 finding E equilibrium Fe plus 3 Fe plus 2 we have to find that, so that case I canconsider nc equal to neeta c equal to beta c log ic by i zero here neeta c equal to Emix prime minus E equilibrium Fe plus 3 Fe plus 2, ic equal to nothing but ic Fe plus3 Fe plus 2 and i zero equal to i zero Fe plus 3 Fe plus 2 over zinc surface equal to10 to the power minus 9 ampere centimeter square.Now when we try to solve this we have to see this particular data until as I here 1 coordinateis known which is E mix prime that other coordinator is not known which is nothing but this isthe value of ic Fe plus 3 Fe plus 2 at E mix this value is not known.And here also i zero is known E equilibrium Fe plus 3 by Fe plus 2 is not known.So, we have in this equation we have this is unknown as well as this is unknown.So, we need to find at least 1 known parameter, so in order to find that we have to find outic hydrogen on zinc at E mix prime.So, we can find that by using this particular equation hydrogen over zinc surface minussorry we missed the minus sign here.So, the beta c log of ic hydrogen on zinc surface i zero of hydrogen on zinc surface.So, here we know this particular voltage because we know E mix prime, so here this is the polarization,so let me put it in a different colour, so let me put it this colour, so this is thepolarization we are considering, this is the polarization.So, this is neeta c hydrogen at E mix prime, so E mix prime minus E equilibrium hydrogenequal to 0.1 log ic hydrogen on zinc surface and this is known 10 to the power minus 11.And this is also known I can replace it with minus 0.177.So this is also known I can replace it with minus 0.277.So, this value becomes minus 100 minus 0.1 equal to minus 0.1 log of 10 to the powersorry, so this is ic hydrogen over zinc surface divided by 10 to the power minus 11.So, then ic hydrogen over zinc surface at E mix prime equal to 10 to the power into10 to the power minus 11 this becomes 10 to the power minus 10 ampere centimeter square.So, once we know this current that means this current if we know then I will be able toknow the current corresponding to this.Why?Because ic hydrogen over zinc surface plus ic Fe plus 3 Fe plus 2 over zinc surface atE mix prime equal to ia zinc at E mix prime which is nothing but 10 to the power minus6.So, if that is true, so then ic hydrogen over zinc surface equal to 10 to the power minus10 ampere per centimeter square and ic ia zinc equal to 10 to the power minus 6 ampereper centimeter square at E mix prime.So, ic Fe plus 3 Fe plus 2 would be equal to 10 to the power minus 6 minus 10 the powerminus 10 which is nearly 10 to the power minus 6 ampere per centimeter square.So, once we know that then E mix prime minus E equilibrium Fe plus 3 by Fe plus 2 equalto minus 0.1 log of ic Fe plus 3 Fe plus 2 divided by i zero Fe plus 3 Fe plus 2 overzinc surface this is also on zinc surface.So, now this value is known now which is 10 to the power minus 6 this value is 10 to thepower minus 9, so this becomes 0.1 log of 10 to the power minus 6 by 10 to the powerminus 9 and this is minus 0.277 minus equilibrium Fe plus 3 Fe plus 2.So, E equilibrium Fe plus 3 Fe plus 2 equal to this is this becomes 0.3 and this shouldbe added up 0.3 minus 0.277, so this volt equal to so, 0.023 volt.So, now we got all the things what we wanted to have this value is obtained, this is alsoobtained, this is obtained, this is obtained, this is obtained.Now even we have found out these 2 okay, so this is the entire solution of this particularnumerical please go through this we will have other problems given to you on the basis ofthis during your tutorial section.So, let’s stop here we will continue our discussion in our next lecture, thank you.