in this video we're going to look at how to balance Redux equations using the half reaction method in acidic solution ready let's do it here's the equation that we're going to balance now balancing a Redux equation is a little bit more challenging than just balancing a regular chemical equation because with a redo equation we have to balance not only the individual atoms but also the charges so there's a stepbystep method we use that guides us through the process kind of to make sure we don't forget anything along the way I'm not going to talk about all the steps of this process right now instead I'm going to introduce them one by one as we use them while we work through balancing this example so here's where we start this is a Redux equation which means that there's oxidation and reduction going on we want to know which elements are getting oxidized and and which are getting reduced we can figure this out by looking at changes in the oxidation numbers for these elements so the first thing we're going to do is assign oxidation numbers to each of the atoms in this equation so our first step is to determine oxidation numbers these are the rules that we're going to use to figure out the oxidation numbers for the elements in this equation now if this stuff is totally new to you I have a couple videos on just determining oxidation number you might want to check that out for some background anyway let's start with ag silver here silver is an element by itself it's not combined with any other elements it's just on its own and so that means that its oxidation number is zero NO3 1 minus here is a polyatomic ion for n for nitrogen there's not a specific rule so we'll have to figure it out oxygen is usually minus two except minus one in peroxide this definitely isn't peroxide here so its oxidation number is minus two now in this compound there are three oxygens 03 so I'm going to multiply this -2 this min-2 * 3 to get min-6 which is the total number the total oxidation number for oxygen in this compound now to figure out nitrogen's oxidation number we're going to use this rule here that the sum of oxidation numbers for a polyatomic ion equals the ion charge the ion charge here is 1 minus and that means that whatever nitrogen's oxidation number is when added with min-6 has to give us -1 so that means nitrogen's oxidation number is + 5 + 5 - 6 = -1 so there we have it now let's look at oxidation numbers on this side of the equation they're not necessarily the same as they were over here because oxidation numbers can change when the element is in different circumstances so here AG now has a charge of plus one it's become an ion it's become a monatomic ion made of just one atom so its oxidation number is the same as its ion charge so it's going to be plus one and then over here we don't know nit oxidation number but we do know that Oxygen's is going to be-2 then there's this rule here that for a neutral compound the oxidation numbers have to add to zero n o is definitely a neutral compound because there's not a charge after it which means that to add together with minus 2 it get zero nitrogen has to have an oxidation number of + 2 + 2 - 2 = 0 so these are the oxidation numbers for the elements in this equation now that we've determined these oxidation numbers we can figure out what's being oxidized and what's being reduced take a look at this chart here reduction is a gain of electrons and it causes the oxidation number to go down oxidation is a loss of electrons and it causes the oxidation number to go up okay silver here starts out is zero on this side of the equation and then over here it becomes plus one so it's oxidation number is going up which means that it is being oxidized from 0 to+ one nitrogen here is + 5 over here but then it's plus two on this other side of the equation so its oxidation number is going down which means that nitrogen is being reduced now just let's take a look at oxygen real quick oxygen is min-2 here and it's min-2 on this side so nothing is happening to oxygen in this equation silver is being oxidized and nitrogen is being reduced now we're going to write half reactions for the oxidation and for the reduction separately we'll start with the oxidation here silver as we said is getting oxidized so we have AG on one side of the equation and then we have AG 1+ on the other side of the equation this is the oxidation half reaction and now the reduction half reaction is going to be NO3 1 minus NO3 1 minus and then on the other side of the equation we're going to have n o now the next thing that we're going to do is we're going to balance each one of these equations separately I'll start with the equation for reduction and then we'll balance the equation for oxidation here's a reduction half reaction we're going to start by balancing the atoms other than o and H here's what that means we have two elements in this equation we have nitrogen and oxygen on both sides right here on this side of the equation we have one nitrogen and we have three oxygen 03 and on this side of the equation we have one nitrogen and one oxygen so the nitrogen is balanced one on both sides so we're ready to move on the next thing that we're going to do is we are going to use H2O to balance oxygen and H+ to balance H here's what that means we have an imbalance right now of oxygen we have more oxygens on this side than on this side in order for this to balance we're going to add H2O which provides oxygen there's one oxygen atom in each H2O molecule and I have three oxygen on this side one on this side so I need two more oxygens so I'm going to put a two in front of this so now I have 1 plus two oxyg which gives me three oxygen now the Oxygen's balance by adding water but I've introduced another element into the chemical equation because now I have hydrogen and on this side I now have 2 * 2 four hydrogens and right now I don't have any hydrogen's on this side so in order to balance out the hydrogens I added by introducing water I'm going to add H+ to this other side I've got four hydrogens on this side so I'm going to add four H+ to this side so now I have four hydrogens on both sides of the equation now my nitrogens oxygens and hydrogens are all balanced now that I've balanced the atoms in this equation I now have to balance the charges by adding electrons the concept of balancing charges in an equation might might be new to you so let me walk you through this the first thing I've got to do is determine how much charge I have on each side of the equation I'll start over here I have four H+ each one of these H+ have a charge of plus one so four of those are going to give me plus 4 then I have 1 n 3 1 minus with a minus1 charge so the total charge here I have is plus4 from the hydrogen's minus one from the NO3 1 minus gives me + three now over here on this side of the equation I don't have anything that's charged so the total charge here is going to be zero that's simple enough so now I got to get these to balance I need it to be the same charge number on both sides I need to find a way to cancel out this plus three of charge and I'm going to do that by adding electrons because electrons have a negative charge so to this side of the equation I'm going to add 3 e minus the symbol for electron three electrons Each of which have a charge of 1 minus so I'm going to have + three minus 3 the charge from these electrons here and that's going to give me a charge of zero so now I've got zero on both sides of the equation and the charge is balance so the equation for reduction or I should say the half reaction for reduction is now balanced for atoms and for charge now I have to do the same balancing routine with the half reaction for oxidation here's a half reaction for oxidation we have to balance the atoms other than oxygen and hydrogen but we don't have to worry about that because we have one silver atom on this side and one silver atom on this side so it's already balanced we also don't have to worry about adding water or H+ to balance the o or H because there are none of them in this equation the only thing that we have to keep in mind is the balance of the charges here's how we're going to do that we're going to determine the charge on both sides of the equation over here we just have AG with no charge so the charge is zero over here we have ag1 plus so the charge is + one I need these charges to balance so to cancel out this plus one charge I'm going to add one electron with a one minus charge to this side so now I've got + 1 -1 equal 0 0 0 the charge is Balan now that I've balanced this I'm ready to combine it with a half reaction for reduction now that I've balanced these two equations for both atoms and charge I'm going to start putting them back together let's look at the electrons here there are three electrons down here in the reduction half reaction and there is one electron here in the oxidation half reaction what I want to do is multiply the half reactions to make the number of electrons equal in both since there are three electrons down here and one electron up here I can multiply this whole equation by three to end up with three electrons so they'll be the same so I'm going to do that just as if it's a math equation and I'm going to distribute this three across the whole reaction so that's going to give me 3 AG on that side and then 3 ag+ + 3 e minus now the number of electrons balances in the reduction and oxidation reactions now that the number of electrons is the same in both reactions I can add them together let me show you how I do that we'll start with everything on this side side of the arrow okay I'll start with my 3 a 3 Ag and now I put in everything from this reaction on this side of the arrow 3 A+ 3 e minus plus 4 H+ plus N3 1 minus then I put the arrow in and then I'll put in everything on this side I will have three ag+ plus 3 e minus plus this n o plus 2 H2O and now the step up here says that we will be canceling out stuff that appears on both sides of the arrow let's look we have three electrons on both sides of the arrow so this can be canel out and this can be canceled out and when we rewrite this after canceling out those electrons this is going to be our final equation lastly we'll do a final check to make sure everything balances here the atoms and the charges we'll start with the atoms okay left side AG silver we got three of them hydrogens four nitrogens just one and oxygen 03 we got three over on the right side silver AG we got three hydrogen we have 2 * 2 gives us four nitrogen one and oxygen we have one there plus uh two over here gives us three so Silvers hydrogens nitrogens and oxygens all balance this is good for atoms now let's check the charges okay charges on the left side here we have have 4 H+ so that's going to be a total of plus4 charge and then NO3 1 minus it be minus one it's going to give us + 3 + 4 - one that's for the left side now for the right side the only thing that's charged is silver here ag1 plus and we have three of those so the total charge for this is going to be+ three for those three Silvers so plus three here plus three here the charges are balanced to this equation is perfect we're good to go all right so that is how you balance a Redux equation using the half reaction method in acidic solution here's a list of the steps that we use to do the balancing now you might want to review these but I don't think it's a good use of your time to memorize them instead I'd recommend you try solving a variety of sample problems so you can really get comfortable with the process and it will just become second nature