 Now, work out this one. So, this is a question using E D means Ellingham diagram find out what is the lowest temperature at which zinc oxide can be reduced to zinc by carbon. Let us say zinc oxide curve is this one, which is the temperature you are going to use or which is the temperature that is required to convert zinc oxide to zinc by using carbon. So, this is the temperature right. So, the carbon monoxide curve is in the green, zinc oxide curve if this is the one, this is the temperature around let us say 1250, this point around 1250 you need to convert zinc oxide to zinc by using carbon right. What is the minimum temperature required for the reduction of magnesium oxide by carbon more than 2000 because this is the green curve, this is the magnesium oxide curve and that is the point where you are having that intersection and that is more than 2000 let us say 2150. So, at that point at this point carbon going to carbon monoxide formation is thermodynamically more favorable compared to magnesium going to magnesium oxide. So, net reaction will be magnesium oxide reacting with carbon giving you carbon monoxide that means carbon monoxide formation is favorable magnesium oxide is converted to magnesium ok. So, there are some web website is given you can play with that or you can look at that. Now, so this is the reaction which is used in thermit process all of you are familiar with thermit process you have heard of ok. So, technically speaking as I was telling you can utilize aluminium to react with chromium to get pure chromium because aluminium this you know diagram from this eligarum diagram you can see aluminium is having more delta G or more negative delta G compared to chromium. So, chromium oxide can be converted to chromium by using aluminium because aluminium oxide formation is more favorable than chromium oxide right. This is this is the standard reaction kind of reaction which you should not forget what when you are looking for elingham diagram you must remember something like this whatever is at the bottom that will be getting oxidized ok not the other way around this chromium is not going to react with aluminium to give you the reverse reaction. So, this is in this case aluminium is going to be your sacrificial reagent it is going to sacrifice itself to give you the pure form of the other one. How is the delta G working out here or more to say since this is almost a two both the reactions are similar aluminium, oxygen the stoichiometric wise everything is exactly same aluminium solid chromium solid oxygen gas oxygen gas and then you are going to get aluminium and chromium or chromium oxide right. So, nothing is changing since nothing is changing in terms of their state like solid liquid and gas or overall stoichiometric are all same if you are taking or thinking about delta delta G or the delta G formation for the whole reaction only thing perhaps you need to worry about the delta H that is the only difference because delta S is going to be similar or exactly you know comparable you can just cut it out. So, delta G equals delta H minus T delta S. So, overall delta delta G the difference between the delta G is going to be equivalent to the difference in their delta H. So, since delta S is similar T delta S term you can kind of delete it ok. So, over here if delta G for this reaction is minus 266 delta H is minus 180 what overall you are having is for this reaction delta G is going to be minus 86 k cal per mole right. Now, this is what we know as thermit process. So, delta H is minus 86 delta G is negative delta S is very small delta G is approximately same at different temperature. So, with respect to temperature you are not going to see any or much difference with respect to delta G whatever temperature you do the reaction you are supposed to get this thermit process. Aluminium plus chromium this alumina formation and there is a chromium oxide you can convert because it is becoming independent of temperature. But what usually you need to do is you need a kinetic you need to cross a kinetic barrier you cannot just mix at room temperature alumina and chromium ok. You need to prime the reaction almost like you know igniting your you know your what your this explosive or any of those you know what are those fireworks yes I can see that ok. Now, this is this is a kinetic. So, you need a prime you need something called priming ok we are done going to be done very soon. So, you need to prime the reaction with magnesium ribbon, but once you initiate the reaction it is a it is a autocatalytic process. Once you initiate one time if you have let us say whole mass of huge mass of this reaction right after initiation you do not have to worry about it because the reaction itself will take care of the later on ignition period ok. So, it is a very you know autocatalytic type of thing once you catalyze it will be catalyzed forever sorry sorry priming is like initiating initiating the reaction ok. You have to initiate the reaction you have to provide enough activation you need to start the reaction activation area barrier is there for the reaction that activation barrier the moment you cross that activation barrier everything is downhill. So, thermodynamically everything is going to work out from there on ok. Now, of course life would perhaps the world would have been a very good place or much better place if we perhaps need not huge charcoal I think one or few of you are having the queries it is going to produce carbon monoxide or carbon dioxide anyway it is a bad thing right. Of course, lot of this greenhouse effect lot of global warming everything is due to let us say blame it on Ellingham diagram ok. It is a charcoal you are using right. Now, the problem is there is no other solution perhaps the best solution could have been this hydrogen right. If hydrogen plus oxygen going to give you water that water has it been gas ultimately it is liquid right hydrogen plus hydrogen plus oxygen 2 gas you are consuming and you are forming something water that is liquid. Has it been something gas ok more of a water has been you know if it is if it is stable form has been gas then what would have happened that instead of that carbon Ellingham diagram that carbon going down you perhaps would have seen hydrogen to water formation hydrogen plus oxygen going to water formation or hydrogen plus metal oxide going to form the water formation that would have been in a negative slope right. And water being very friendly you would not have any problem since most of the cases what you see that this hydrogen going to water or hydrogen reacting with metal oxide going to water is running parallel to that of the metal oxide curve you are not going to use hydrogen efficiently for conversion of metal oxide to metal and that is unfortunate I think and most of the problem was problem would have been really really solved ok. But that is how it is, but of course there are some reaction you have to you have to know relatively where hydrogen is with respect to other metal oxide curve in the Ellingham diagram. So, some reaction you can utilize hydrogen ok for the reduction of corresponding metal oxide to the metal, but it is not a generalized solution. The generalized solution is charcoal that is where everywhere you see the charcoal. Now, the point of these are reduction of metal sulphide. So, of course as you mentioned many metals which are chemically soft you know it is again that hard soft acid base principle oxide is hard. So, metals which are in harder form or hard cation metal oxide will be formed and those are the metal oxide we see in the ore let us say calcium oxide. Calcium 2 plus is hard, oxy oxide 2 minus is hard, calcium oxide is going to be favourable form. Calcium sulphide may not be that favourable that much favourable, but because calcium 2 plus is hard sulphide is softer relatively softer. So, calcium sulphide or let us say magnesium sulphide formation or whatever titanium sulphide if you form these are not going to work out these are not going to be a stable conformation. So, similarly when you have metal which is not hard, but more of a soft on a softer side then metal oxide formation will not be preferable. You will get metal sulphide in the ore you will get metal sulphide. Now, these are the for example, these are the cases copper, mercury, zinc, iron these are the case of course iron can be iron oxide as well, but still it can give you the sulphide form. Now, the reason why we cannot utilize this reaction is very simple the metal sulphide plus carbon this carbon dioxide carbon disulfide formation has no slope almost it is a constant slope in Ellingham diagram. Carbon going to carbon monoxide you have see it is going down running down, but carbon reacting with metal sulphide it is a generalized form we have given metal sulphide is not like just metal and sulphide it is of course sometime it is as you have seen Fe 2 O 3 it is a simplified form we have given metal sulphide. Metal oxide MO it is not necessarily MO it can be M 2 O 3 similarly metal sulphide can be of different form. Anyway this is the problem why we cannot utilize the charcoal for carbon disulfide formation. It is therefore, easier or it is advisable I think you have almost no choice, but to convert metal sulphide to metal oxide one step and then utilize charcoal to reduce those metal oxide to the corresponding metal. So, first metal oxide is roasted or metal sulphide is roasted to metal oxide and then reduce to metal. So, metal sulphide plus oxygen going to metal oxide plus sulphur dioxide plus charcoal giving you the metal and corresponding. Of course, another possibilities are there where your self reduction is a viable technique. What is self reduction? You take copper sulphide you try to do this reaction copper sulphide plus oxygen you try to get copper oxide the moment copper oxide is formed copper sulphide and copper oxide together can give you copper and sulphur dioxide under that high temperature condition. So, this is like a self sorting self reduction. Of course, hydrogen is also a pore reducing agent for metal sulphide. So, you have to see the relative curve if they are you know hydrogen going to water is delta G for that reaction is it less or more negative compared to the metal going to metal sulphide that is all you need to care. So, this is the Ellingham diagram for metal sulphide as you can see these are you know in the metal oxide curve we were seeing minus 2000, 3000 and so on, but you see these metal sulphide these curve are not that much thermodynamically favourable ok. That is a good thing because and then you can convert that into the metal oxide and so on ok. There are also Ellingham diagram possible for metal halides there is no way you need to remember these things most often I think few things you should remember, but most often you will be provided with the Ellingham diagram ok, but you know of course, which is most electro positive metal and thereby what should be the curve for it these are few things you should know, but these are metal halide these are metal halide this is metal halide Ellingham diagram of course. So, we have seen oxide Ellingham diagram sulphide Ellingham diagram and metal halide Ellingham diagram ok. Now, so let me summarize by telling you this part we have seen different techniques as simple as mechanical separation, electromagnetic separation and thermal decomposition you can heat it and decompose it these three things are very easy to understand of course, almost everything is easy to understand, but these are none of these are kind of you know self contained method you cannot just use one method to purify your ores and get the metal in pure form often you have to use the combination of method of course, you can use one metal to reduce another metal oxide right also you can use the electro catalytic reduction where cathode and anode you can have right. The one which is perhaps most important for us is the Ellingham diagram where we take a metal oxide use charcoal reduce the metal oxide to corresponding metal and then we form carbon in oxidized form carbon monoxide or carbon dioxide right. What we try to also say that similar things are possible for sulphides and halides metal sulphides and metal halide, but those are not of really that great importance compared to oxide metal sulphide almost invariably is converted to metal oxide first and then that is converted to the corresponding metal. Why metal sulphides are converted because the carbon disulfide curve is not having a negative slope as we are seeing carbon monoxide case that is all. Now, there are other techniques lot of other techniques which are utilized quite routinely these are fusion distillation crystallization ok fusion is something let us say you have a you have a rock you have a rock. Now, if you fuse that rock what will be happening inside that rock some dissolved or some amount of gas will be there. So, if you melt it those gas will be evaporated right. So, the trapped gases will be evaporated of course, that is a purification technique. You can distill you can distill means let us say you start with a liquid you heat it at a particular temperature let us say 200 degree centigrade. So, those which are having boiling point suitable less than 200 degree C selectively some of those or all of those which are having boiling point less than 200 degree C that will be selectively coming out rest of the material will be in the container where you started with. So, you can both collect the material that is left off and material that is going to get boiled and you collect it. You can collect it in fraction wise one fraction two fraction and you can increase or decrease the temperature adjust the temperature slowly you can increase the temperature. So, that one by one organic or compound can come out. Let us say you heat it at 1000 degree C one metal oxide comes out ok. You heat it further 1500 degree C another metal comes out. So, by heating at 1000 degree C you get that in pure form one over in pure form by heating just let us say 1500 or 1400 you get another one. So, this is another technique. So, what essentially this extraction technique tells you is it is a combination of things you need to have knowledge both of course engineering knowledge when you are going to do it practically engineering knowledge is essential because you cannot just heat without knowing what is going to happen ok. You have to have those knowledge in addition of course you can have let us say crystallization technique. You have let us say you have a five things five solid material combined you try to add some solvent. Solvent let us say could be water solvent could be something else and selectively one of those material get dissolved in that solvent. So, you take it out you try to grow crystal or you know that that is what is in pure form. So, solubility can be utilized either to grow the crystal or to wash it off or take it off take off particularly one of the metal or metal oxide. So, there are other lot of other techniques it is not necessarily you have to read in detail, but it is important that you do understand that those things exist. So, thermal decomposition of course thermal decomposition we have seen zone refining this is the last topic two minutes I will take what is zone refining? Well zone refining is very simply it is it is a it is a purification done. So, that you get a high level of material let us say you have 99 percent of titanium you want to make it 99.99 percent titanium. How do you do it? Because that is going to be you know in your lot of these metal all the electronic devices you use it require that your metal be pure in that form without that impurity can cost you dearly right. So, you need to really purify the material. So, this is what is zone refining one of the technique what you have let us say this is your 99 percent or 90 percent pure rod of metal let us say iron bar or titanium bar or whatever metal you have. Now, this is this is a circular heater now circular heater if you have. So, this is a let us say we are having a impure germanium rod let us say this is 90 percent impure germanium rod. Now, it is a circular heater it melts selectively at one position at this position wherever it is heating it is melting temporarily. Let us say it is heating here momentarily it heats over there and it melts only that region it does not melt this region on the top or on the bottom it is a very powerful heater. It melts only a selective zone and from top to bottom this circular heater you are kind of rotating and bringing it down again putting it back bring it down. So, by doing that what is happening is you you are melting it and then cooling it cooling means the moment the circular heater is going out of that zone it is going to cool down. So, impurities when it is going to get solidified impurities will be coming to a particular. So, let us say it is a small zone the top material in that small zone the top region will be pure metal and the impurity will cool down and come at the bottom. That way if you because the cooling profile is going to be different for pure metal and impure material whatever is there. So, that way impurities are going to be on one of the end either it is on the top or on the bottom does not matter. If you keep on doing this circular heat if you keep on heating it what is happening is one end of this metallic bar one end of it is going to have the impurity concentrated. And thereby you can just cut it off you can just cut it off that impure part and get almost let us say 99.99 percent pure metal. You keep on doing let us say if first time is not giving you very high purity you can keep on keep on doing this process after 4, 5 times from 90 percent material pure material you will get let us say something like 99.99 percent pure material. So, it is a zone refining you you just zone wise very selective area wise you keep on doing this process. So, that is how impurities get collected I think I am done. Thank you for your hospitality.