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Worked example: Balancing a redox equation in acidic solution

AP.Chem:
TRA‑2 (EU)
,
TRA‑2.C (LO)
,
TRA‑2.C.1 (EK)

Video transcript

our goal is to balance this redox reaction in acid and before we get into the steps let's talk about the fact that this is a redox reaction by assigning some oxidation states and so we start over here with the dichromate anion and we know that oxygen has an oxidation state of negative two we have seven oxygens so negative two times seven gives me negative fourteen as our total here we know that the total for the entire anion has to equal negative two which is the charge on the dichromate anion therefore we must have plus twelve for all of our chromium's here so plus 12 and minus 14 give us negative two since we have two chromium's each one must be plus six and so that's the oxidation state for chromium here we go over here to the chloride anion right so the charge is negative one so our oxidation state is negative one chromium ion over here so plus three and then finally chlorine over here so oxidation state of zero so if we look at chlorine chlorine went from an oxidation state of negative one to an oxidation state of zero that's an increase in the oxidation state therefore chlorine was oxidized here look at chromium chromium went from plus six to plus three that's a decrease in the oxidation state or a reduction in the oxidation state therefore chromium was reduced and so this is a redox reaction because something is oxidized and something is reduced in terms of balancing it our first step is to write the different half reactions and so we're going to break those into an oxidation half-reaction and a reduction half-reaction so let's go ahead and get some space down here and let's go ahead and write our write our half reactions and so we had the chloride anion right going to chlorine like that and we said that this was our oxidation half-reaction so I'll put that way over here on the right that's our oxidation our reduction half-reaction involved chromium right so we had we had the chromate anion here so CR 207 to minus right going to chromium 3 plus like that so this is our reduction half-reaction so that's step one write the different half reactions step to balance the atoms other than oxygen and hydrogen and so if we look at our first half reaction right chlorine we have a1 chloride on the left and two chlorines on the right so we need to balance it by putting a 2 over here on the left like that we go down here to the reduction half-reaction and we have two chromium's on the left and only one on the right and so we have to put it to right here to balance it so step two is done step 3 balance the oxygens by adding water so if I look at my oxidation half-reaction right there are no oxygen so I don't need to worry about doing anything to this to this half reaction at the moment I go down to the reduction half-reaction and I do have to balance my oxygens right so if I go over here and I can see that I have seven oxygens on the left side and none on the right and so I need to do that by adding water and since I have seven oxygens on the left I need seven oxygens on the right so I'm gonna go ahead and add seven water molecules and that now gives me seven oxygens on the right of my half reaction so that's this step step three right here step four balance the hydrogen's by adding some protons so let me go ahead and I'll use red for this so step four balance the hydrogen's by adding a protons here once again the oxidation half-reaction we don't have to do anything because we don't have to balance oxygen or hydrogen here but again we go down to our reduction half-reaction and we have the oxygens balanced by adding water but by adding water now we have some hydrogen's on the right side so we can see we have a total of fourteen hydrogen's on the right side so 7 times 2 and so we're going to balance that by adding protons and so we need to add protons to the left side of our half reaction so we need to add 14 so 7 times 2 is 14 so we go ahead and add 14 H+ to the left side of our half reaction so step 4 is done step 5 balance the charges by adding electrons so let's get some more space here first of all okay so we're going to balance the charges by adding electrons so first let's analyze what kinds of charges that we have here so we'll start with the top oxidation half reaction action so we have the chloride anion right which is a negative one charge and we have two of them alright so we have two negative charges so there's two negative charges notice these are not oxidation states so that's what gets people confused sometimes these are charges over here on the right we have no charges all right so that's and that's a neutral chlorine molecule here so we have two negative charges on the left and zero for a charge on the right so we need to figure out how to balance those charges by adding electrons and so it makes sense that we would have to add two electrons to the right over here because that now gives us a total charge of negative 2 on the right so that's one way to think about it just getting these numbers equal here this negative 2 right here and this negative 2 another way to do it would be of course you know the electrons have to go on the right side because this is the oxidation half-reaction and one of the ways to remember that right Leo the Lion all right so loss of electrons is oxidation and so if you're losing electrons they must go in the product side of your half reaction and so our top half reaction our oxidation half reaction is now balanced let's go down to our reduction half-reaction right so Leo the Lion goes ger so gain of electrons is reduction so we already know we're going to have to add electrons to the reactants side of this half reaction let's see if we can figure out how many electrons we're going to need to use that we have 14 positive charges from the protons and then we have two negative charges from the dichromate anion here so we have 14 positive charges and two negative charges which gives us a total of 12 positive charges on the left side on the right side we have a chromium ion right so this is a charge of 3 plus and I have two of them so 2 times positive 3 gives me positive 6 so I have positive 6 on the right side of my half reaction so I have positive 12 on the Left positive 6 on the right I need to add some electrons to balance out that charge I already know I'm going to add them to the reactant side right I know that from Leo the Lion goes ger or I can just think about the fact that if I have 12 sort of charges I would need to add six negative charges to get me to a total charge of plus six so I need to add six electrons to the left side over here so I'm going to go plus six electrons like that and now we have we have the charges balanced and so this step is done so Step five is done we move on to step 6 so make the number of electrons equal so what does that mean well let's focus in here on the electrons that we just added to our half reactions right so if I go back up to here all right we added two electrons to the oxidation half-reaction and we had six for the reduction half-reaction but we know that that number has to be the exact same number because the electrons that are lost right from our oxidation half-reaction are the exact same electrons that are gained in our reduction half-reaction so we that is why we have to make these electrons equal in terms of the number and so the way to do that would of course be to multiply my first half reaction by three because if I multiply my first half reaction by three that would give me a total of six electrons which is what we're looking for making the number of electrons equal so let's go ahead and do that I'm going to rewrite our first half reaction I'm going to multiply everything in our half reaction through by three so everything in parentheses so if I take three and multiply that by two chloride anions I would of course get six all right so we have six chloride anions like that and then the three would go in front of the chlorine so I have three CL two and then three times three times two electrons right gives me six electrons like that okay let's go ahead and rewrite our reduction half-reaction because we have a lot of stuff going on here so I'm just going to rewrite exactly we have six electrons plus 14 protons 14 protons here plus the dichromate anion like that alright and then I have two two chromium ions and seven waters okay so now I have my two half reactions I've made the number of electrons equal and already for the last step we just take our two half-reactions and we're going to add them back together and that's going to give us our overall balanced redox reactions let's go ahead and we'll go ahead and say we did this and let's go ahead and get some more room here so we can add those two half reactions alright so I'm going to do is just take everything on the reactant side alright taking everything on the reactant side here right and add that together so let's go ahead and just rewrite everything on our reactant sides we have six cl- right and then six electrons 14 protons all right we have the dichromate anion here and then I'm going to take everything on my product side all right so I'm going to take all of this stuff right here all right and put that on my product side so I have 3 CL 2 plus 6 electrons alright plus 2 chromium ions plus 7 water molecules like that and so now it's just a little bit easier I find for students to see that your electrons are going to cancel right there you have these six electrons and then reactant side you have these six electrons on the product side so you can go ahead and take those out and we're left with our final answer alright so you could go ahead and rewrite it if you want to make it look better we have 6 CL minus plus 14 h plus plus CR 207 2 minus yields 3 CL 2 plus 2 CR 3 + +7 h2o and this should be our final answer so I always like to box my final answer so it just makes it easier for your instructor to grade and the nice thing about redox reactions is when you're finished you can always check yourself because you need you know that you need to balance both the atoms and the charge so let's go ahead and check that real fast let's let's let's first start with chlorine right so we have 6 chlorines on the left then over here on the right we have 3 times 2 which is 6 so chlorines balanced hydrogen we have 14 on the left and then seven times two gives us 14 on the right the chromium is we have two on the left and then over here we have two on the right oxygens seven oxygens on the left and then over here right this seven applies to this oxygen and so the atoms are balanced properly let's next check charge because you have to have the correct charge we have we have let's see we have six negatives right so six negative charges almost put minus six here and then I have 14 positives right so I have positive 14 and then I have negative 2 right here right so when I add all those up I get a total of plus 6 on the left side right so about 14 minus 8 over here on the right side the only charge is this chromium ion it's 3 plus and I have two of them so 2 times 3 gives me plus 6 and so the charge balances as well and so you know this is the correct answer it balances both in terms of atoms and in terms of charge