 convert into oxide. So if you look at the example I have given in calcination, we have mainly hydroxide and carbonates over there. Metals of hydroxide metals we have, metal hydroxides we have, we have hydrated metal oxide we have, and we have carbonates present. So these compounds, if you look at the reaction, it can easily convert it into oxide. You don't have to put much effort into it. Just you need to heat it in absence of air. You will get oxide easily. But suppose if you have sulphide like HGS, in this that the one that I've given HGS you have, ZNS you have, PBS you have. Okay. So all these sulphides you need to convert it into oxide. Right. But oxygen is not there. We have lead sulphide, zinc sulphide. So you have to supply oxygen from outside. So that metal converts into oxides. Right. So it's basically depends upon the property of the metal impurities that we have, or metal compounds, which is present there in the mixture. If it is hydroxides, carbonates and hydrated form, you heat it simply, it will convert into oxide. But if it is not in the form of carbonate, hydroxides and hydrates of metal oxides, then you have to provide oxygen from outside and then heat it so that it converts into oxides. Yes. Okay. So why do only metals like more reactive metals like calcium, aluminum and all only form the carbonates and not zinc carbon. See, in this, we do not have any logic on this because you see, when we discuss about the extraction of any metals, we are just giving you an overview of what process we use. But in this process, we add certain compounds also for the purification process. Correct. So it totally depends like what compound you are getting in a process. It totally depends upon what process you are following in the previous step. Depending upon that, you will have some chemical reaction and then you'll get the product. Okay. So for calcium, it's very, I cannot give you, even no one can give you the concrete answer on this because you see the metals present in the earth crust as the no, as the compound. In that, what all impurities are there, that is one case. What you are adding from outside for the extraction of metal, that is another condition. In all these conditions, you will have one product in the reaction. Right. So it can be anything. So for calcium, you know, iron, mostly we see we forms hydroxides or carbonates because in the earth crust, we have those compounds present plus condition of reaction is like that. So that it converts into carbonates and all. If condition is different, maybe you'll get calcium oxides also directly. That is also possible. So there is no any concrete answer for this could be anything. Okay. Yes. Yeah. So this is what we have grossed it. Correct. Now, what we get till now, we get metal oxides. Our purpose is to extract metal, eventually. Right. Now, we have got metal oxides. Now, if you are able to remove oxygen from these metals, metal oxides, you will get the metals finally. Correct. So the last, not last, you have second last method. And the last one is just, you know, purification, like last step is just we will enhance the percentage of the metal that you are getting to get the more pure form. Right. So in this step, the next step, what we do, we have metal oxides, we have to remove oxygen. And for that purpose, we do reduction of metal oxide. So introduction process, we extract oxygen from metal oxide, and we get impure metal first. And then the last step will increase the purity of the metal. That's the last step of the reaction. So in the next process, you write down reduction of next process, you write down reduction of roasted or canned ore. Reduction is the process here. Reduction. So this process we call it smelting. Okay. I'll write down the definition. Write down the extraction of metal, the extraction of a metal from a metal from its oxide, its oxide by a process, by a process involving by a process involves melting is called is called smelting. So the roasted or calcined ore that you get, it still contains non fusible impurities. Okay, I'll just in short, I'll just write down the ore that you have roasted ore, right. It contains infuse infusible, I'm sorry. It is non fusible impurities, non fusible impurities, non fusible impurities of like, you know, silica, you know, silicates, or, you know, metallic oxides, these kind of impurities are present to remove this infusible impurities as example of infusible impurities you write down. We can have silica, we can have silicates, metal oxides, all these are infusible impurities. And to remove this impurities, we also add some compounds into this and these compounds are called flux, flux we add to remove these impurities. Okay, so when you add a flux, so already we have impurities present in the, you know, a mixture which we call it as gang. So gang is already there. And you add some compound from the outside in this smelting process that we got as flux. This combines with the gang, and it forms slag, which comes out from the mixture. And that is how we remove impurities here, right, flux, you add it forms slag and that comes out. One second, guys. Guys, one second, just one second. Yes. Okay, so what we were discussing. Gang plus flux gives you slag. Okay, so slag that comes out and the impurities is removed. Okay. Now, what are the types of flux that we use that is important here, right? So flux we use, we can have acidic flux, we can have basic flux. Okay, anything we can use, depending upon the nature of the gang again, the impurities which is present in the sample. Right. So right down here, next, acidic flux, acidic flux, such as silica. Example, if you see, we have SiO2, it is acidic flux, borax, NO2B407.10H2. Just you need to keep some examples in mind. Okay, NO2B407.10H2. These are acidic flux. Okay. This is added always when the impurities are basic. Added in basic impurities. Added in basic impurities. Okay, basic impurities. Such as we have metallic oxides, metallic oxides are basic. So we have, you know, acidic flux, acidic flux added in basic impurities gives you fusible slag. It was non-fusible initially. Once you add this flux, it becomes fusible. Okay. Look at this reaction, SiO2 plus CaO. See, when we're using borax here, since it has the ability to remove most of the metallic oxides. Okay. So one logic we have that acidic flux we are using against the basic impurities, but it is based on the reaction that we have. What, you know, impurities are present. Based on that, we'll choose the flux here. Right. So NO2B407 also serves as a flux for most of the metallic oxides. That's why we have it over here. Come again. Yeah, tell me. Said it doesn't matter that borax is basic. I see. The structure of borax is what? We have NO2B407, right? Yes, sir. Or we have 2NA plus and B407 minus we have? Yes, sir. So it can take oxygen, sorry, the metallic oxides that you have, it can form a salt with the metallic oxide. It means that metal oxides can replace NA from that. It means reaction is possible. Okay, okay. So then reaction is possible. That's why we are using it. Sir, which acidic character are we talking about? Because if borax is basic and we are using acidic flux, we're telling it as acidic flux. No acidic flux. See acidic flux is this. Okay. So acidic flux, see borax, see it's not like in the solution how it is behaving. The acidic behavior, if you look at the structure of borax we have, the borax structure is this. It depends upon its structure. B407 we have, no. So we have 1, 2, 3, 4, 5, 6. And just a second, let me just check the structure. 1, 2, 3, 4. That's 5bob. Seventh oxygen where it will go. 1, 2, 3, 4, 5, 6. We have 6bob, you know. 1, 2, 3, 4, 5, and 6. Sir, two central bobis are there, sir. I'm not, I'm doubtful of that. And not 100% sure that there's two central bobis. I think it's four sideways bobis, one central bobis, and all those on the, on the other side of the boron, it'll get OHOH maybe. Okay, one second. So we have 10H2, no? Yes, sir. One second. Let me just, let me call it. And it will be 407. We have the structure. We have bobis born. We have four or five bobis born into this. Five bobis. Four border bobis and one bridge bobis. Four, five, no. So I think it would be this, like this. Like that, yes, sir. And we have NaO. Four we have, we have seven here, right? So you are saying this, I think this is possible. And then O with salt, correct, yeah. And this ONA. I think this is the structure we have, no? Yes, sir. Okay. This is the structure we have, correct. So you see, this is salt, correct? So we have this, and first of all, it is a salt. Yes, sir. Basic salt or acidic salt, that's a different thing. But since it is a salt, so it forms O minus here, O minus here. Correct, sir, correct. Which can easily take the metal oxide. You see, if O minus, O minus is present, it can take this calcium or suppose any metal oxide ion if you have over there, correct? So point is here, point is you are confused with, since it is acidic flux. So the compound, the flux that you're using, it must be acidic. That's fine. But this compound, it can show reaction with metallic oxide. And it was in a non-fusible initially, when it combines with borax, this O minus will take these metals, which is there in the non-fusible slag, and then it converts into fusible slag. Since the reaction is possible here, that's why we are using this as an acidic flux. Because it shows reaction with most of the metal oxides. Because Na is a, no, electropositive metal from Na plus Na plus, and these metals will come over here. And it converts into a fusible slag, which comes out as a slag over there. So reaction is possible. Compounds usually, the terms you're using acidic flux, because most of the flux are acidic in nature. And this acid-based behavior is relative, right? You can say this one is more basic, other one is less basic, that we can discuss. Okay. Overall, what we are doing, we are considering whether the reaction with the non-fusible slag is possible or not. So yes, it is true. When you have basic oxide of metal, mostly we can use acidic flux like SiO2. But some of the oxides of metal also get replaced, all convert into fusible slag when it reacts with borax. Since the structure is this, Na plus will get replaced by the metal ion over here. That's why the reaction is... Got it, sir. Okay. Okay. So we have acidic flux. And the second flux we have is basic flux, you write down. Two types of flux we use. The last one is the basic flux we have. Just opposite, it is basic flux. So we'll add into the acidic impurities which is present. So basic flux, such as we have limestone, the example of this, we can have CSCO3 we can use for this purpose. We can also use the same group. MgCO3 we can use for the same purpose. Okay. This is added to remove, this is added to remove, to remove acidic impurities, acidic impurities. Silica is also like we have seen, it is an acidic flux also present as impurities in the sample. So silica is also the acidic impurities we have. And when you add these basic flux into this, it forms silicate. So you have added calcium here. So it forms CSCO3 calcium silicate, which forms a fusible slag, comes out and CO2 escapes into the atmosphere. So this is, smelting is the process we use to remove the slag, correct? Whether it is to remove the impurities with the help of flux, which can either be acidic or basic, correct? Once you have this impurities you have removed, still you are left with metal oxides, which will use for reduction in order to obtain metal from its oxide. The last process we have here for the purification is chemical reduction. Write down chemical reduction, chemical reduction, the last method. So basically in this we use a various reducing agent, right? Three types of reduction method we have here. The last one is the electro refining method, I think. Electro refining method is more important. We'll discuss that, how it happens and what are the, you know, cathode anode will take, okay? That is important. We'll discuss what important we have into that. But other two are, you know, you already know like chemical reduction is what? You add some reducing agent, it will extract oxygen and will get metal, correct? You see chemical reduction, first method is carbon as the reducing agent we use, as the reducing agent. Just few reactions, you know this already. Suppose we have lead oxide, in general what happens? We are left with metal oxide, metal oxide reduction you can order with carbon. You heat this, it forms CO monoxide and metal will get to you. For example, you see, you have PBO reaction with carbon, heat curvy, it forms Pb plus CO. You have Fe2O3 reduction with carbon. When you heat this, it converts into Fe plus 3CO3 carbon we have here. And like this, we have the reaction with carbon. Apart from this, we can also use carbon monoxide as a reducing agent. So PbO plus CO, Pb plus CO2, carbon monoxide as a reducing agent. Because all are eliminating oxygen from this. So it forms Cu plus CO2. Copy this down. The second method we have, which we use for reduction is, it is auto or self reduction process, auto reduction process or self reduction process, okay? In this process, write down, in this process, we do not require any reducing agent. In this process, we do not require, we do not require a reducing agent. We use this method only for those metals, which are less electropositive, sulfides of those metals, which are less electropositive. So write down sulfides of metals like, sulfides of metals like Hg, copper, lead, sulfides of metals like mercury, copper, lead, mercury, copper, lead, tin are reduced to, are reduced to metal, are reduced to metal when heated in presence of air. Just you need to heat in presence of air, that is it. No other reagent is required. Sir, in? One second. In presence of air. No sir, tin. That, once again, tin are reduced. I don't know what sentence I've written. I tell you what is the, what is the entire thing, because I was framing the sentence, I cannot repeat the exact sentence. Okay. I was saying what the metals like mercury, copper, lead and tin are, no, the sulfides of metals like mercury, copper, lead and tin are reduced to, are reduced to metal in presence of, when heated in presence of air. Once again, Trippan, just a second. I said the sulfides of metals like mercury, copper, lead, tin are reduced to metal when heated in presence of air. Correct? Yeah, very good. Done. Done, sir. Yeah. See, there's a doubt, carbon can be used to reduce any metal oxides or just some. The same logic, Trippan, you can apply electrochemistry to. The reduction of metal is possible if the oxidation potential of carbon is higher than to that of metal. So carbon can reduce only those metal sulfides, like the, for which if you compare the reduction potential of the metal should be more than to that of carbon. Or if you talk about the oxidation potential, carbon should have more oxidation potential than the metal. So all we cannot say. Okay, sir. So basically like lesser reactive. Yes, yes, correct. Okay. So it's the choice, no, see, these are the process we are doing in the industry. So it's the choice, whatever metal we have, accordingly, we'll choose the reagent to react and gives the desired result. Okay. Yes, sir. So, yes, so metal sulfide in this auto or self reduction, we don't require any external agent. So the reaction is you see two HGS when heated in presence of air. So we have three O two, it gives HGO and SO two first sulfur dioxide and oxide, which further reacts with HGS, then it converts into metal and sorry, converts into metal plus will get SO two, SO two. Okay. So some part of it converts into oxide, which then again reacts with the sulfide itself and converts into metal. Same thing goes with copper also, you see this reaction. Cu two S two Cu two S reacts with three O two converts into two Cu two O two Cu two O plus two S O two. We get this. Okay. Then this again copper oxide that you have over here, oxides of copper, it combines with the sulfide, which is left and converts into four into two six, six Cu plus SO two. So this kind of reaction is there. Okay. The next one we have that is electrolytic reduction method. Electrolytic reduction method among all is more important. Okay. And this I'll tell you what we need to understand. Electrolytic reduction method that we call it as, write down electrolytic reduction method for the reduction process. This process we also call it as electro refining process. We use the, we use the concept or the method of electrolysis in this purpose. So write down the third type of reduction process that we use is electrolytic reduction, electrolytic reduction and we also call it as electro refining process, electro refining process. Write down this method is used for the purification of this method is used for the purification of copper, zinc, tin, silver, gold, nickel, lead, aluminum, zinc, tin, silver, gold, nickel, lead, and aluminum for all this. Okay. So this one you have to memorize the one that I'm giving you now. Cathode. Obviously we have electrolytic process. You should know what is cathode and what is anode here. Cathode here is made up of pure metal, the metal which is used to be, which is there to refine. Okay. The metal which we need to extract for the same metal we have the cathode electrode. Okay. We'll have the cathode electrode of the same metal in which when you're talking about this one, electrolytic reduction. Yes, electrolytic. It depends here. See, it depends here if you need to extract metal because we have condition, you will understand. Give me some time. Let me explain this, you will understand what we should use over here. So cathode what I'm telling you, cathode is made up of, we'll have a thin strip of it made up of, made up of pure metal, pure metal, what metal. Same as that to be refined. Okay. Suppose copper you need to extract, then copper cut thin strip you have. Okay. Same as that to be refined. Okay. And we'll have a thin strip of it, thin strip cathode on this. The last question is information. Now, if you look at anode, anode is made up of, made up of impure metal. Same metal, which is to be refined. Same impure metal, made up of impure metal, but we have large slab of it, thick and large slab of it. So this one is thin, this one is large slab of the same metal, which is to be refined. Okay. Electrolyte that we use in this purpose, in this process, obviously we have the electrolyte of same metal, which is to be refined. Okay. Electrolyte aqua solution of, aqua solution of suitable salt of same metal, which is to be refined. Now you see with this diagram, you will understand. So it's not like Venkat, we need very pure or very active metals. It depends upon what metal is to be refined. Based on that, we'll select the cathode and anode. Okay, sir. Okay. Okay. So we have this, suppose the vessel we have container in which the electrolysis is taking place. So anode will have on the left-hand side or the right-hand side, anode on the left or on the right. Anode will have on the left. Correct. Anode oxidation, oxidation on the left. So we'll have a large, thick slab for anode, but for cathode, we have relatively thin, thin strip of cathode. Obviously, whatever is required for reduction for electrolysis, that will definitely do. Okay. These things will be there. And here we have the solution, the electrolyte, aqua solution of the same metal. Suppose metal M, you need to refine. So M-ca solution you need to take here. The same concept of half cell. So we have on this. Okay. Now this is cathode and it is pure, right? Cathode, pure metal. This information you must remember. Okay. This one is anode. It is impure metal. So impure metal means what? We need to extract metal from this only because they are the impurity we have. So if you can extract metal from anode, right, that is what we want. This is to pure only. So we have some impurities present over here, right? Cathode, we have only one reaction possible. The reaction of cathode is what? Suppose the metal M we have, so M, N plus plus N electron reduction gives you M-solid. So no doubt about whatever M ion we have over here, it goes here, get, you know, M plus from the electrolyte will go here, N plus, MN plus will go here, get reduced and deposit on the cathode. So we wanted to take this M plus from the solution. That's what the purpose. We need to extract, correct? What we also wanted, we also want here to take the metal from this impure slab. If this M plus comes here into the solution, right? And the same thing will get reduced over here. So we are extracting metal ion from this impure electrode and that is getting deposited on this and that is how the purification is done. Are you getting my point? Correct. So could you repeat my internet got disconnected? I said what? You need to impure, we need to extract metal from this impure slab that you have. No, I don't. Correct? So yes, in order to extract metal from this, suppose if here the oxidation of metal takes place, M converts into MN plus, it comes out into the solution and the same metal ion gets reduced at cathode. So eventually what happens? We extract MN plus from this slab and then this only will get reduced at cathode and get deposited over here. So that is how the refining process is taking place. Correct? So at anode, we basically have two types of reaction possible. Anode, we have two types of reaction possible. One is the reaction that we want to happen, which is nothing but M metal converts into MN plus plus N electron. This is the oxidation of metal atom, which is there in this impure electrode. But here we have some negative ion also, right? That is X, suppose N minus. This also wants to get oxidized like this. This reaction we want to happen. This reaction we do not want. So it is an unwanted reaction we have. We want this to happen so that we can extract metal from this slab. So what is required here? Logically you can understand. See, whatever is there to mug up that you need to memorize, but this you can understand logically. We want this to happen. We want this not to happen. This reaction we do not want. So we will choose the electrolyte in such a way that the anionic part of the electrolyte that you have is the reduction potential should be should be more than to that of metal. Or if you talk about in terms of oxidation potential, the oxidation potential of metal should be more than to that of the anionic part of the electrolyte. Because if this oxidation potential is more, it will get oxidized first relative to this, right? In comparison to this. Correct. So we will choose the electrolyte in such a way that the oxidation potential of metal is more than to that of the anionic part that we have. Are you getting it? So we will choose the anionic part in such a way that this reaction does not take place. Only this will be there. And this is only possible when the oxidation potential of metal is more than to this because this note you write down the anionic part of the electrolyte. The anionic part of the electrolyte is to be is to be chosen in such a way in such a way that reaction to this is the first and the second reaction. First and second reaction. The anionic part of the electrolyte is to be chosen in such a way that anionic reaction 2 does not take place. In such a way that. In such a way that anionic reaction 2 when I am opening the numbering 1 and 2, anionic reaction 2 does not take place. Thank you sir. Okay. One last point in this. Write down the metallic impurities having lower oxidation potential. This one. Lower oxidation potential. The metallic impurities having lower oxidation potential having lower oxidation potential then the metal then the metal gets separate in the form of anode mud gets separate in the form of anode mud. I am repeating metallic impurities having lower oxidation potential then the metal gets separate in the form of anode mud. Mud simply we say or anode mud we call it as because it is coming from anode. So anode mud we represent like this. We will have some impurity that accumulates over here. So this is the anode mud. So this is it. This is the process of refining. Now after this we have to discuss Ellingham diagram and then some purification process of this thing. Purification process of some metals like aluminium, zinc, copper we will discuss. One we will discuss and that is it in this chapter we have. So we will take a break now. After the break we will start Ellingham diagram. So we will resume at 6.55. Take a break.