hi everyone this is our last video on redox reactions we're going to talk about balancing redox reactions using a method called half reactions so if you remember in our discussion about redox reactions we know that there is a transfer of electrons or a change in the oxidation number with one of the species involved in our reaction in order for redox reactions to be balanced we see here that the electrons lost in oxidation must equal the number of electrons gained during reduction and so when we think about balancing redox reactions not only do we need to have the same number of atoms for each element on both sides of the equation but we also need to think about having the same number of electrons on each side of the equation so the method we're going to use to balance these reactions is called the half reaction method these are the five steps that we're going to walk through for balancing simple redox reactions first we're going to write the two individual half reactions we'll write a half reaction for oxidation and another one for reduction then we're going to balance all the elements next we'll balance the charge by adding electrons and then if necessary we're going to multiply each half reaction's coefficients by the smallest possible integers in order to yield equal number of electrons in each and then finally We'll add the two balanced half reactions together and cancel any species that appears on both sides so to start with we're going to balance just some half reactions on their own and we'll only need to use the first three steps in our half reaction method so here we have two half reactions in the first we see we have 10 going from the four plus state to the two plus state so our oxidation number is changing so in order to balance this reaction we first want to make sure that we're balancing all of our elements we can see that the the element is balanced on both sides but we need to balance by adding electrons so we can balance charge so on the left hand side we have the four plus State and on the right hand side we have the two plus state so in order for our charge to be balanced it looks like we need to decrease the charge on the left hand side on the reactant side so that means we'd have to add electrons right in order to reduce that charge since we go from 4 down to two it looks like I need two electrons in order to balance the charge so this is my balanced half reaction I can see that this is a reduction reaction because my oxidation state is being reduced or decreased in my next example we have chromium going from the two plus state to the three plus state and this is already balanced in terms of the elements but we do need to balance in terms of charge so on the left hand side I have a two plus charge and on the right hand side I have a three plus so in order to have the charges balance on both sides of the equation I need to add electrons and I can see that I'll need to add them to the right hand side right so if I have a 3 plus on the right and I reduce it by adding electrons I'll need to add only one electron to get it to equal to two so this is a balanced half reaction and I can see that this is an oxidation reaction because my oxidation state is increasing all right so now let's graduate to balancing a full redox reaction we're going to start by writing the the expected products so the question says complete and balance the following redox reaction which gives the highest possible oxidation state for the oxidized atoms so I have aluminum and I have fluorine as my reactants now I know that fluorine usually likes to have a negative charge associated with it so that is not the species that's being oxidized I expect to generate fluoride ions in this reaction the aluminum is a metal and so this likes to be oxidized to a positive state right so the highest oxidation state of aluminum that we expect would be a three plus state so we have the skeleton for our redox reaction but in order to balance it we need to write our half reactions so I'm going to write a half reaction involving the aluminum species so I have aluminum as my reactant and I have the aluminum ions as my product the other species involved in my my skeleton for the redox reaction is the fluorine F2 and the other species involved with that is f minus foreign so I wrote my half reactions and now I need to go and balance in terms of all the elements so for my aluminum reaction I only have one aluminum on both sides so that's balanced for my fluorine reaction I have F2 on the left which means that on the right I'll also need to have two fluorines all right so we're balanced in terms of the elements next we want to balance charge by adding electrons so if I look at my reaction involving aluminum in order to balance the charge I have a charge of zero on the left for my Elemental aluminum and on the right I have a plus three charge so I need to reduce the charge on the right by adding three electrons and so now that would sum to give a charge of zero for my fluorine reaction I have a charge of zero on the left but I have a charge of 2 minus on the right so I will need to add electrons to the left hand side in order to balance the charge for this reaction so now looking at my half reactions I can see that the reaction involving the aluminum is my oxidation right my oxidation state has increased as I went from 0 to plus 3 for my aluminum and the reaction involving fluorine is the reduction the oxidation state is going from zero in the fluorine Elemental State and to a negative one for the fluoride so continuing to balance this reaction we move to step four step four says if necessary multiply each half reactions coefficients by the smallest possible integers to yield equal numbers of electrons in each so I can see I have three electrons in my oxidation half reaction and I have two in my reduction half reaction I need to have them be equal so I want to find the least common multiple of 3 and 2. so we know that would be six so in order to get six electrons in this half reaction I'll have to multiply by two in order to get six electrons in this half reaction I'll have to multiply by three so I'm going to do that and multiply through all my coefficients by these integers so this gives me two aluminum gives two aluminum ions plus two times three gives me 6 electrons and for my reduction reaction I have three fluorines F2 plus six electrons gives 3 times 2 6 fluoride ions all right so now the number of electrons is the same in both of my half reactions now step five says to add the balanced half reactions together and cancel any species that appears on both sides so I'm going to write my reactions again so that I can add them up easily and so now I want to add them together I notice that my electrons can cancel because I have the same number of electrons on both sides of the reaction and then I simply add together what I see on the left hand sides of my two half reactions and add together what I see on the right hand side so we have fluorine plus aluminum giving aluminum ions and fluoride ions we can notice in this that the products here right could form an ionic compound if we take the charge on the aluminum and make that our subscript on the the fluoride right we would get three there if we take the charge on the fluoride and make that our subscript on the aluminum that would give us the ionic compound aluminum fluoride alf3 and it looks like we would have two of them based upon the balanced chemical reaction so let's do a few more examples to get some practice so here we have another unbalanced redox reaction and if we look at this it and just look at the numbers of atoms we have on the left and the right it almost appears to be balanced right I have one tin atom on the left and I have one on the right I have one copper Ion on the left I have one copper Ion on the right and so this appears to be balanced but when we consider the charge it is not so if I look at the charge on the left I have two plus two plus so I have plus 4 on the left and on the right I have plus 4 and plus one so that would be plus five on the right so obviously our charges are not balanced and so we need to go through the steps of balancing this so I'm going to first write my half reactions I have the tin on the left and the two plus State and this is going to form 10 in the 4 plus state and then I'll write my half reaction for the copper as well as it goes from the two plus state to 2 plus 1 state all right so there's my two half reactions that I'm going to work on I'll notice that they are already balanced in terms of the elements so now let's balance by chart for charge by adding in electrons so for my reaction involving 10 I'm going from the two plus state to the 4 plus state so in order to have the charge balance on both sides it looks like I'll need to add in electrons onto the right as a product right and in order to have the charges equal I need to add in two electrons right so 4 minus 2 would give me the two plus charge I have on the left when I go to my copper reaction I have a plus two charge on the left which needs to be reduced by 1 in order to equal the charge on the right so I'll have to add in one electron on the left for this reaction now thinking through my change in oxidation number I can identify that this is my balanced half reaction for the oxidation right because I can see that the oxidation number of the tin has increased and increase means oxidation and when I look at the half reaction for the copper my oxidation state has been reduced or decreased so this is my reduction half reaction so now we've done step three so next is step four which says to multiply each half reactions coefficients by the smallest possible integer so that we have the same number of electrons in each so I have two electrons in this half reaction and I have only one electron in my copper half reaction so it looks like I'll need to multiply that reaction by 2 in order to have my electrons cancel so when I do that multiplying through by 2 for all my coefficients I end up with 2 copper two plus plus two electrons gives me two copper Plus and now I want to rewrite my reactions so that I can add them together and I'll notice that the electrons should now cancel and then I write um what I have left on both sides all right now I can see that it is balanced both in terms of the number of atoms or ions right I have two copper ions on the left and two on the right I have only one ion of 10 on the left and the right but now it's also balanced in terms of charge so on the left hand side I have 2 times plus 2 plus 4 from the copper and then another plus two from the tin so a total charge of plus six on the left hand side and when I look at the right hand side I have 10 and a 4 plus State and I add on another 2 plus so that gives me a total charge of plus 6 on the right as well all right let's just do one last example to really drive home how we balance these redox reactions so I have another skeletal reaction that needs to be balanced we can see that this one is not balanced in terms of the atoms or the charge so we're going to go through and follow our steps by first writing the half reactions so I'm going to write a reaction based on iron and another reaction based on iodine all right my first step after this is to balance in terms of the atoms I only have one iron ion on both sides for that half reaction so that one is okay for my iodine equation I have I2 on the right hand side which tells me that I'll need two I minus on the left hand side so now we are balanced in terms of the atoms or ions and so now we can move to balance in terms of the charge by adding in electrons I can see that for my iron half reaction I have a three plus on the left so that looks like it needs to be reduced by adding in one electron on the left hand side for my iodine reaction I have a charge of 2 minus on the left hand side so it looks I will need to add in two electrons to the right hand side so these are my balanced half reactions I can see that the iron's oxidation number is decreasing right so decrease is a reduction and the iodides oxidation number is increasing as it goes from negative one to zero in I2 so this is my oxidation half reaction the next step in my process for balancing these would be to make sure that I have the same number of electrons on both sides so I have 2 for this half reaction I have only one for my iron half reaction so it looks like I'll need to multiply this one by two so let's do that we multiply through our coefficients by 2 which gives me two iron three plus two electrons giving rise to two iron two plus and then I'll also write in my half reaction for the oxidation and then I'll need to add these together my electrons cancel and I can add up my reactants and products to give my balanced redox reaction all right so now you are all experts on balancing simple redox reactions