noc18-mm14 Lecture 28-Estimation of corrosion Rate-III

[Music] [Music] today we'll have lecture 28 and we'll continue our discussion on kinetics of corrosion and in the last lecture we just started looking at the corrosion rate in terms of current flow since for corrosion that means the metal oil formation we have to have electron generation and per unit time if we see that electron generation we can get some idea about the corrosion rate and some correlation with the current flow during corrosion now while doing that we started looking at laws of electrochemistry and the first law says that W is proportional to Q and where we finally we got W equal to Z I T this should be capitalized which is the current in ampere so this is electrochemical equivalent and I is current in ampere ampere T's time in second now there we saw that Z is nothing but the value of W when I equal to 1 T equal to 1 now then in order to find out Jed we started looking at second law it says that e which is the chemical equivalent is proportional to said why because we saw that we had taken two examples one was electrolysis of aqueous initial solution another one is the electrolysis of aqueous eg no.3 and we saw that in case of a NaCl we had chlorine gas on the positive electrode that is the anode and on the negative electrode we had hydrogen generation that was basically h2o reduction of water reduction and then in case of hno3 a Agni no.3 we had silver deposition on cathode that means the negative electrode and oxygen evolution on the anode and this oxygen evolution is coming because of the h minus oxidation now when we saw that and then finally we realized that if we send one faraday of electricity or the charge chemical reaction or the deposition or the evolution of gases on the cathode or anode they're all equal to chemical equivalent or the gram equivalent of that particular substance for example in case of AG no.3 we had one one eight point where around 107 point eight eight gram of silver deposition because of the because of that one faraday of charge flow in that particular circuit as well as around eight gram of oxygen on the anode so we see that every time this one faraday of electricity or the charge is nothing but equal to the chemical equivalent or the gram equivalent of that particular substance now if we say that E is proportional to Z that then e equal to s said so F is basically the one Faraday and Z is electrochemical equivalent now this equation can enable us and is nothing but the chemical equivalent as we have said this equation will enable us to find out the Z value so in case of silver that we can find out Z is equal to e divided by F which is e is nothing but the one gram equivalent which is same as the molar atomic weight of that particular silver because it has got the atomic this oxidation number to be 1 so hundred zero seven point eight eight divided by nine six five zero zero and gram and this is coming out to be zero zero zero point zero zero one one eight rather one one one eight if we check this yes it's zero point zero zero one one one eight gram per Coulomb now so with the help of second law we can find out the value of Z now we try to understand this importance of these two equations on corrosion rate now whenever we talk about M going to em n plus plus any that time its gram equivalent would be a M which is the atomic weight of that particular metal divided by M this is equal to gram equivalent fine now when you see that then we could see that whether it's this way or this way if I send one faraday of electricity or the one faraday of charge that means nine six five hundred Coulomb of charge it will always liberate when it transforms to metal i'm or well it will also deposit the metal of this amount which is a M divided by n that means this is the e value of metal or chemical equivalent of this matter now if the deposition is in the chemical equivalent if I send one faraday of electricity on the other way the dissolution of metal ion would be same as the deposition that means one faraday would be equivalent to this gram of metal dissolution isn't it so that much gram of metal dissolution would take place if I send one faraday of electricity no electricity or one faraday of charge now if I take this particular concept to understand if we send Q amount of charge what would be my the amount of mass that will be lost in the form of ions into the solution so that we can easily find out now let's see if I try to put it in the form of differential mode let's say one faraday of electricity is equivalent to a M divided by n this is basically amount of metal that will be dissolving now if I send an F amount of charge or Coulomb so it will be equivalent to one mole of that substance why because this is nothing but a gram of methyl or the the weight the atomic weight of that particular matter so one particular atomic weight so this much of metal ion if it forms then we would definitely have one mole of metal dissolution so if we send Q DQ amount so it will be simply 1 by n F D cube so this much of mole that dissolves so now you see if we send DQ amount of charge we get this much this much of more loss of mol of that particular metal that is dissolving in the solution now if this DQ charge is sent for Delta time or DT time which is a very small miniscule time then we can find out what is the number of mole dissolved parney time so let's say this value is the number of mole would be then doubleu or this if I note it down as in prime which is the number of moles that is dissolving so this is mole it is equivalent to n minus small so n minus divided by T is nothing but the rate of number of moles of metal dissolve per unit but anytime which is same as because n prime is nothing but D cubed R by n F now n F is a constant n is nothing but oxidation number for this particular reaction now F is 1 Faraday which is also constant so now I can write this now if I see this quantity this quantity is nothing but I so I can write this one equal to I divided by n F now when I write this it says the rate of I can rate of number of moles of metal dissolved or instead in in that case I don't have to mention this part any time because when I am using this rate time is taken care of now this particular is also indicating the corrosion rate only in the form of number of moles dissolved or anytime but it's not taking care of the area because area is very important as we have discussed in our last lecture because whenever corrosion happens it's basically surface phenomena so the exposed area on top of it this reaction will take place not in the bulk of the system or the bulk of the metal so until unless we normalize it with area we will not be able to get the proper indication or proper expression for the corrosion rate so in order to do that we have to divide it with area so let's say small is the basically area this is area so I can put it as n prime a DT equal to AI n F n so now it becomes number of moles of metal dissolve per unit time per unit area so this is a proper indications now it's in the number of moles now we can convert this into a grams per unit area about any time so in order to do that n prime is basically the number of moles so this W if W prime is the amount of metal that has dissolved divided by a which is basically the a.m rather is a m is the atomic weight divided by am a DT is equal to AI NF a I can write this way so where W prime is the weight of the metal or the mass of that metal coming into solution formed and then W and a M is nothing but equal to n prime or the number of moles then W prime a DT equal to I M and F so finally we are getting the corrosion rate is equal to gram or I could say in the form of weight divided by area divided by time equal to I am this is the expression what we are getting now we can simplify this one am i a NF and I divided by a is written in the form of I small I which is nothing but ampere per centimeter square I can mention in this fashion ampere per centimeter square which is nothing but the current density so corrosion rate is equal to i NF now am for a particular metal in and f for m equal to m n plus plus ne they are all constant so now corrosion rate I would say is proportional to I or the current density interestingly it's not proportional to the current flow rather it's the proportional to the current density because if we take current then we are actually not considering the area that is exposed to the solution for the corrosion reactions to happen on the on the metal surface but current density will look when we consider it actually takes care of this area that is actually on top of which the reaction the corrosion reaction is taking this so hence the corrosion rate is directly proportional to the current density not I but we get this particular information by using this electrochemistry of the laws of electrochemistry now once we get this idea we can see that it's basically in the form of weight loss or I can say weight loss per unit area per unit time now this we have seen that in those cases where the dissolution pattern is localized we have to use a different way different way of expressing corrosion rate which is nothing but the depth penetration depth divided by time so we can convert this particular expression in the form of the depth penetrated per unit time by simply dividing this corrosion rate with density of that particular metal so the corrosion rate in the form of depth per unit time we can put it as I am n F Rho and where Rho is basically density so we get another expression like this which is equivalent to weight loss area time density now we see that we have a very good idea that corrosion rate is electrochemical in nature and by using laws of electrochemistry we can relate the current flow in the form of current density with the corrosion rate now before we get to the details of this current density part because we have to now get to the exchange current density which is exchange current density is basically I is zero which is a very very fundamental aspect in electrochemistry as well as in corrosion because while explaining corrosion in more scientific fashion we have to use make use of mix potential theory and there this isida will be extremely needed and this is a fundamental parameter fundamental parameter so before we get to this I 0 which is exchange current density let's use this particular formulas and let's see whether we can find out the corrosion rate in different units and express them in different units so let us solve some new miracles we are just giving some examples of those miracles tears will give you many more problems on these two formulas and then you can solve them and great and practice so the first problem let's let's say our teen immerse in seawater Souls what is the rate of corrosion in MDD and mpy now in this particular problem we have used a corrosion current density now if we go back and see this particular formula the here the corrosion current density is nothing but the current density at which the dissolution is taking place so this is nothing but the corrosion current density this is nothing but the corrosion current density so we can start expressing this or mass loss divided by area and time which is nothing but I a SN teen is nothing but the s n divided by n is two here and F nine six five zero zero Coulomb and where e s n is nothing but one one eight point seven one gram and s n is going into the solution as SN plus two plus two e so that means n becomes 2 and I is nothing but in ampere per centimeter square e s n is nothing but in gram one one eight gram / and here I I can put this value 2.4 5 into 10 to the power minus 6 n is equal to 2 into 9 6 5 0 0 Coulomb now this quantity I can write in this way 2.45 into 10 to the power minus 6 Coulomb power in it time power in it area because I is now I is nothing but I divided by area or Coulomb per per second per unit centimeter square so then I can write this time as second area in centimeter square one one eight point seven one gram and this becomes 2 into nine six five zero zero Coulomb so this Coulomb Coulomb would get cancelled and then two point four five into ten to the power minus 6 divided by 2 into nine six five zero zero and now I will convert the unit's because it's in MD D which is milligram per decimeter square per day so this becomes gram so I have to convert this gram into milligram so thousand milligram divided by second I have to convert in two days so ii becomes 1 divided by because one day is equal to 24 hours 60 into 60 second so 1/24 into 60 into 60 day / centimeter should be in decimeter so one centimeter equal to 10 to the power minus 1 decimeter so it should be 10 to the power minus 2 decimeter square so final expression would come as 2 point 4 5 into 10 to the power minus 6 2 into 9 6 5 0 0 into 1 1 sorry I miss this amount of gram 8.7 1 in 2000 into 100 into 24 into 60 into 60 this much milligram per decimeter square part a so finally this this value would be 13 point zero 1 M DT so this is the corrosion rate so we'll solve the other part which is in the mpy in our next lecture let's stop here we'll continue our discussion on kinetics of corrosion thank you [Music] [Music]