Balancing Redox Reactions by Ion-Electron Method | Half Reaction Method | Class 11 Chemistry

Hello children, how are you all? Today I will explain to you The second part of the redox reaction Whom we people call ion-electron method This half-reaction method Why do children use this? Those redox reactions you have, right? They are in an unbalanced form. So, whenever they need to balance it, they use this method. By the way, let me inform you that there are two methods of balancing. One is your oxidation method, and the other is your electron. ion-electron method So, the oxidation method, I've already taught you. You can go to my video section and see it in this playlist. And the second part is, Like ion electron method or half-reaction method, I will explain this to you today. How you can balance any redox reaction in seconds Come on, with this trick, let's go, patta patta, some points Let's take a look, because if these points are yours No, if you don't learn, then the difficulty of balancing arises. Write down these points, please, and give them to me. Also keep informing me in between that Sir, these points are being missed. It is okay. What does the first step say? That if Any redox reaction given to you, and you have to balance it, kid. Then you must first write the unbalanced right equation. Write down whatever redox reaction is given to you. And assign the oxidation number to which one? Which one is undergoing oxidation, and which one is undergoing reduction? In which one is electron gain occurring, and in which one is electron loss occurring? Then you have to calculate all of that oxidation number, okay after calculating the oxidation number You will find out in which one oxidation is taking place. After which one is undergoing reduction Write two half-equations separately. Now I will give you a complete reaction I will tell you how you can write a half-cell reaction. These points cover here After that, balance the atom except oxygen and hydrogen All the elements there will be given to you. like manganese, whatever is given will undergo a reaction Balance it excluding hydrogen and oxygen. After that, you need to rebalance the oxygen. by adding H2O to the side containing fewer oxygen atoms Now pay attention to you First of all, you don't need to balance hydrogen and oxygen. Whatever other elements remain must be balanced. But in the very next step, you just need to balance the oxygen. And check for oxygen balance Add H2O to the side where oxygen is less within the reactions. Alright, add H2O, then you need to balance the hydrogen as well. You have to add exactly the opposite side of the H2O here. You can write on the opposite side, opposite side, that's fine. Either you need where hydrogen atoms are fewer, there you go You can add hydrogen atoms inside the reaction, okay? I'll just say the same thing: wherever you add H2O, Add H+ on the exact opposite side, like If you have added H2O product, then H Plus H plus at the reaction side, put it there It will be oxidation side or reduction side, okay, next one Step, which is the final step, says what? Multiply the half-equation by a suitable factor to equalize. the number of electrons means whatever you have You will get the electron current, right? Using cross multiplication. I'll also tell you how to cross it. You will multiply inside the equation So that your half-cell reaction becomes balanced After that, you will add two half-cell equations by canceling the number of electrons on the both sides of the equation you have both Write the half-cell reactions and add them together. I will now tell you about addition. How does it work and how are cancellations handled? Then until you have the final A redox reaction will be obtained that will remain in balanced form. Let's start this question, kids These are tricky points; write them down. What exactly are you planning to do with some of these? I've already told you, shall I start? Just let me know the rules, then let's go. The first question appears here: balance the following unbalanced equation in acidic medium by ion-electron method Or according to the half-reaction method We need to balance this, let's go, first we We will move in step number 1 Step 1, what did he say, write unbalanced equation, write unbalanced equation, I write What is the unbalanced equation here, mine? Mn plus 2 plus ClO₃⁻, you Write aqueous and the like, okay, don't miss it MnO2 plus I have written down this unbalanced equation It's written correctly here, brother. What else would you say now that it's done? Write the unbalanced equation, then assign the oxidation states. Number all atoms in reactants and products for oxidation state You need to assign the number to yourself. Step two oxidation Number, you have oxidation How to find the number This is the next one here Go to find out Mn, minus 2 times oxygen's 2 If it is equal to 0, then x will become what? Two times four and then I'll just go a little this way, okay? This is its, and last If we remove this, then x plus 2 times oxygen's minus 2 is equal to 0 It will be clear, it's clear, it's clear Everyone has now understood; tell us here what you realize. You're about to do it next In step 3, what are you realizing that we are here But let's look at oxidation and reduction—what is the next step? It says that come, you need to submit your two reports on oxidation and Reduction whatever you want to tell and half cell equation Just represent it here. Alright, let me tell you here is the oxidation number Two and the Atom in the pillars And the oxidation number of all products I have It has been revealed that you can see here It's already come out here, and do you know what's going on? It will be that I am here, this here, right here. It is that if you are going here, okay, this is from MN, this is MN. Okay, so it's reaching from plus 2 to plus 4, meaning Here, 2 electrons are being lost, loss of electrons, Loss of electron is occurring, and if I talk about another place Subscribe here That thing Subscribe to the next Write in the step What do I have to do here? I need to write it down, okay, let's go. What is the next step? Let's write the right half-cell reaction. half-cell reaction Listen carefully how to write a half-cell reaction Don't they all appear together like this? So write them all together What will you make with your plus 2? mn o2 is the first half-cell The reaction has occurred, that's correct. And this B-containing one has become the second half-cell reaction. After writing here, you will get the balance Expect oxygen and hydrogen Inch, show me and balance oxygen By adding H2O, it's fine, so next Step, now it's our next step Now, what should we do here? adding H2O to balance Add oxygen to H2O to oxygenate We'll balance it, okay, let's see. Look, on the left side there is oxygen, so it's H2O. I'll place it on the exact opposite side, where the oxygen is. Okay, fine. I know now. Listen, kid, I'll add H2O. Look here, why was it added, because there's more oxygen here, isn't there? These two timings won't work anymore. Because there, oxygen gives you only one chance, not two. Balance it, and it'll be fine. So, these two oxygens are balanced for us. Oh, hydrogen has increased; I'll remove the hydrogen now. I'll balance it in the next step, okay? Tell me part B, tell me part B, what is in this? ClO3, there is more oxygen here, less oxygen there. Alright, so I'll add H2O here, I'll add it on this side. Alright, is there a need for balance here? Here, the product contains 2, 1, oxygen There are 3 oxygen atoms in the reactants. So, to balance a few things out, then 30 in your side of the It is added on the exact opposite side. Gaya and not four, this is it Go ahead, move forward a bit more now, next one Step number will come Four, yeah, there's no need to do anything here, yeah, look How much hydrogen is here here? Two two green Four hydrogens are how many hydrogens? Two two none What will you do with four? Balance it with four? And here there are two hydrogens, so now this becomes two. Alright, that's settled. Now the next step comes, your electron. You need to balance it. Look, this is the next step: balance the charge. by adding the appropriate number of electrons to the Half-cell reaction, okay, now let's move on to the next step Our step number five in this, what balance charge now? By adding electrons, it's correct; we're charging here at the balance. We will add electrons here Look here, what's this? Child, look carefully, listen attentively. Look here, what's this? At MN plus 2 plus 2H2O, there are two charges; how much charge loss? It's happening, two, okay, so plus two charges, so you don't need to worry here You'll have to balance two charges, just check it. Show an electron, you'll have to show two electrons, because there are two here. The electron is being lost, like charge, you balance it, three remain, so three Show me the next one. Look here, it's just one. An electron, right? So you have an electron here But what will you do? Add it? Plus two H plus and What will you write now? Will you say, sir, you had plus, so you had If the electron is shown as negative, then the electron is on this side. You've shown a small concept; you must know it now if you're looking at this minus is present, and here an electron needs to be added So, electron directly facing each other in minus It just gets added right here and ends up in loss. Where does the electron arrow point? If you don't understand, just memorize it. if the plus charge remains, then that side of the arrow The more electrons there are, the greater the charge will be. will be negative, then it will have electrons The longer you write, the more it will remain like minus here, then you We write two electrons minus, okay, got it. Come on, now what's the next step, read it Read it, I've also balanced the electrons Gave everything he had, multiply the half equation by suitable factor to equalize the number of two half-cell reactions Look, you need to multiply Whatever electrons you have As you can see, there are 2 electrons here. So in this half cell, you will multiply by 2 Here, there is 1 electron. So in this half cell, you will multiply by 1 Alright, so now I'm moving on to the next step. I'll take it to the next page. Step 4, 5, 6 Step 6 What does it say, cross multiply cross multiply the electron Alright, check how cross multiplication will work. Here there are 2 electrons; multiply by 2 in the sub-... Here, multiply 1 electron by 1 in the sub. Now, if you multiply it by 1 in sub, you'll get the same result. What will come, Mn²⁺, tell me, what is it, come on, tell me Mn²⁺ + 2H₂O → Mn²⁺ + 4H Mn²⁺ + 4H⁺ + 2 electrons, okay, this is 1 Now there are two electrons there, so multiply all of them by two. Then CLO 3 will occur twice. How much time will CLO 3 take? twice, okay minus, plus, one electron Plus, what is H2O? Just look once, H plus 8, two times H plus 8 What is two times, H plus 8, Go to the side of the arrow, plus CLO 3 plus H2O, CLO 3 plus H2O, Now, what was it? Since it was 2, multiply everything by 2. Multiply by 2 here, and multiply by 2 here as well. You will multiply, and here you will multiply by 2. So this will become 2, 2, 4, what will happen, this will become 2, 2, It will become 4; here, multiply by 2, so this becomes 2. 2, 3 will become, and from here, multiply by 2, you will get If you edit it, how will it be canceled, whatever it may be Cancel it; how can you cancel this now? Listen, whatever things are with Aro, they will remain on this side. It will be canceled by the arrow of the second reaction; this side means both of these. And whatever things remain on this side of Aru's, those are what will be seen. It will be canceled from the other side of the previous reaction. I mean, the things on this side get canceled, or those on the other side. The items on the side and the items on the other side of the ARo get canceled. Alright, it's fine on this side, so let's start the cancellation. Look here, you've got 2 electrons from 2 electrons Cancel. Okay, and what else? Yes, here it is 2H2O, right here. 2H2O will cancel out, yes. 4H and 4H will cancel out. What's left here, Mn plus 2 plus 2 ClO3? minus, these two will appear together, arrow MnO2 plus 2 ClO Now this is your entire reaction, balanced, okay. Now your entire reaction has already been balanced here. So with this method using the half-cell method I have this redox reaction Balanced it, okay. So, there are very many such children For unique and very good videos to watch Don't forget to follow up on A to Z Because this is a very hard concept, but I have Explained it to you in such an easy way That I don't think you'll need to watch the video anywhere else again You can solve such a number of questions Thank you, see you in the next video.