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Current time:0:00Total duration:14:02

Worked example: Balancing a redox equation in basic solution

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

Video transcript

in the previous video we saw how to balance redox reactions in acidic solution this video we're going to balance a redox reaction in basic solution and these are a little bit harder but we're going to approach it the same way that we balanced other reactions in the acidic solutions so we're going to once again step four add some protons here and we're going to go ahead and add the half reactions together but after we do that we're actually going to add some hydroxide anions in here as well since this takes place in basic solution the hydroxide ions are necessary to neutralize the protons that we added in step four and so let's let's start off by by starting with oxidation states alright just like we do it in the previous video so if I start with the hypochlorite anion here I know that oxygen has an oxidation state of negative two I know the total has to equal negative one here so that means that chlorine must have an oxidation state of plus one plus one and minus two give you negative one but go to this chromium compound over here the easiest way to approach it is to remember that the hydroxide anion has a charge of negative one and you have four of them giving you a total of negative 4 in terms of oxidation states a total has to add up to equal minus one here the charge on the ion there for chromium must have an oxidation state of plus three plus three and minus 4 give you negative one so chromium's oxidation state is plus three here move over here to the right to the chromate ni n so oxygen is negative two I have four of them for a total of minus eight I need to get a total of minus two the charge on the ion and so for an oxidation state for chromium it would be plus 6 because plus 6 and minus eight give you minus two so chromium is oxidation state is plus six here and then finally for the chloride anion negative one charge therefore an oxidation state of negative 1 in terms of what is being oxidized and was being reduced need to look for an increase in the oxidation state for oxidation so going from an oxidation Zeta plus 3 to an oxidation state of plus 6 for chromium means chromium was oxidized and if we look at the chlorine alright it's going from an oxidation state of plus 1 all the way over here to an oxidation state of negative 1 that's a decrease in the oxidation State or reduction in the oxidation state so chlorine is being reduced and so we know that something is being oxidized and something is being reduced and so we can go ahead and do step one now we're going to write our half reactions so one oxidation half-reaction and one reduction half-reaction this time let's go ahead write the reduction half-reaction first it doesn't really matter which one you do so we'll go ahead and write the one using chlorine since chlorine was reduced here so the hypochlorite anion going to the chloride anion here so this is the reduction half-reaction the oxidation half-reaction is the one that involved chromium so we had this chromium compound c roh for the negative charge all right going to chromate cro4 two - so this is our oxidation half-reaction so let's go to step two balance the atoms other than oxygen and hydrogens the atoms other than oxygen hydrogen so for the reduction half-reaction that applies to chlorine we have one chlorine on the left and we have one on the right so chlorine is balanced in the oxidation half-reaction that applies to chromium right so one chromium on the left and one chromium on the right so in this case step two is already done for us but always be careful because sometimes you actually have to balance these atoms and it will affect your final answer so let's go to step three balance the oxygens by adding water so if we look at our reduction half-reaction we can see there's one oxygen on the left side and we have no Auctions on the right side so we need to add one water molecule to the product side of our reduction half-reaction now we have one oxygen on the right and one oxygen on the left and so oxygen is balanced for our reduction half-reaction let's look at our oxidation half-reaction remember this four here applies to everything in the parenthesis so we have four oxygens on the left we have four oxygens on the right and so oxygen is already balanced next we go to step four balance the hydrogen's by adding protons so if I look at my reduction half-reaction we added a water molecule all right and that introduced two hydrogen's on the right side of my half reaction and so therefore I need two hydrogen's on the left side because right now I don't have any I'm going to balance them by adding them in the form of protons so now I have to add two protons to the left side of my reduction half-reaction and now I have two hydrogen's on the left and two hydrogen's on the right for my oxidation half-reaction remember this four applies to the hydrogen as well so I have four hydrogen's on the left and none on the right side of my oxidation half-reaction so I need four on the right I'm going to add those in the form of protons so for H+ on the right side next we're going to add electrons to balance charge so step five we need to analyze the total charges on both sides of my half reactions so let's look at the reduction half-reaction first I have two protons right so that's a that's two positive charges on the left side here and I also have a negative charge on the hypo chloride anion here so I have two positive charges and one negative charge so of course that's a total of plus one and I have a habit of writing my charges like I have how I write my oxidation states so just remember that these are these are charges not oxidation states and a total of plus one on the left side and on the right side all right I have one negative charge on the chloride anion and that's it in terms of charges and so I have a negative one charge on the right side here so I need to get those charges to be equal and I can only get that by adding electrons which are of course negatively charged so if I want to get both sides of my reduction half-reaction equal the only thing I could do is add two electrons to the left side right here because that gives me that gives me two more negative charges so that takes this positive one to a negative one so now I have a total of negative one charge on the left side of my half reaction and that of course is now equal to the negative one on the right side of my half reaction so I have balanced the charges by adding electrons let's do the oxidation half-reaction now so I have have only one of these ions so I have Oneg one charge on the left side on the right side I have one chromate but each chromate has two negative charges like that and then I also have four positive charges for my proton so negative 2 plus 4 gives me a total of plus 2 so I have plus 2 on the right side and negative 1 on the left side so once again my goal is to balance those charges by adding electrons only and if I add 3 electrons to the right side of this half reaction right I would change this total charge from A plus 2 to a negative 1 and so now I have negative 1 on the right and I have negative 1 on the left and so my charges are balanced another way to to to think about where those electrons go right is to remember that Leo the Lion goes ger right so oxidation Leo loss of electrons is oxidation so that's why you have to show your electrons on your oxidation half-reaction being lost or given off on the product side if you will and then Leo the Lion goes ger so gain of electrons is reduction so so you would have to show the electrons for your reduction half-reaction being on your reactant side of your half reaction because they're being gained here so that's another way just to double-check yourself or think about it so we're on to step 6 now make the number of electrons equal so for step 6 make the number of electrons equal so the reason you have to do that is because the electrons that are being lost in the oxidation half-reaction are the exact same electrons that are being gained in the reduction half-reaction so we had 3 electrons being lost but only 2 electrons being gained that doesn't make any sense right the same electrons they should be the same number and so we need to make those number of electrons equal and the way to do that of course is think about lowest common denominator so we have a 2 and a 3 so we can make those 6 electrons so we need to figure out a way to make both of our half reactions have 6 electrons and we can do that by multiplying our first half reaction by a factor of 3 and multiplying our second half reaction by a factor of 2 and that will give us equal numbers of electrons let's go ahead and do that math here so let's let's go ahead and take 3 times 2 electrons right that gives me 6 electrons and then three times two protons Ryan gives me six protons three times the one the coefficient in front of the hypo chloride anion gives me three CL o minus the three times the one in front of the chloride anion gives me three CL minus and then finally there's a one in front of the water as a coefficient as well so three times 1 gives me three waters so let's go ahead and do the oxidation half-reaction now let's get some more room here so I take the two I multiply it through this coefficient of 1 in front of here in front of this ion so now I would have to see our parentheses Oh h4 minus the two applies to the 1 coefficient in front of the chromate as well right so I would have to see our o 4 2 minus the 2 applies right 2 times 4 gives me 8 protons so I have 8 protons here and then finally 2 times the 3 electrons gives me those 6 electrons and so now we have 6 electrons for my reduction half-reaction and 6 electrons for my oxidation half-reaction next we're going to add our half reactions together so once again we're still pretending like we are in acid here so let's go ahead and add our half reactions together we're going to take all of our reactants alright and put them on one side so let's go ahead and do that first so we have six electrons plus six protons plus 3 CL Oh minus plus 2 C our parenthesis Oh h4 minus and so on the right side so we take all of this alright and we add it all together for our product side alright so let's see what we have we have 3 CL - plus 3 H 2 O plus 2 CR 0 for 2 - plus h h+ + 6 electrons so finally all right now that we've written that out I think it's easier to see that you have 6 electrons on the left and 6 electrons on the right so you can cancel those out you also have some protons on both sides right you have six protons here and eight protons here so you could cancel out these six protons and you can make this two protons on the right side so let's go ahead and rewrite what we have once again so let's let's see what we have now now that we've simplified it a little bit we have three CL o minus plus two C our parentheses o H four minus yields three CL minus plus 2 h2o sorry three H 2 O plus two C are 0 4 2 minus plus 2 h plus all right so now we have to remember that this reaction was actually done in base all right so we're going to add in some hydroxide anions to neutralize the protons so if we look at the number of protons we have right we have 2 protons over here on the right so we need to add two hydroxide anions so we're going to add two hydroxide anions to neutralize those protons and since it's a balanced equation what we do the right side we have to do to the left side as well so since I added two hydroxides to the right I need to add in two hydroxides to the left now let's think about what adding those hydroxides to those protons will do right so they will neutralize each other if you take H+ and OH H - you will get h2o and so we're going to get water molecules so we're going to get two of them since we're since we're neutralizing two protons with two hydroxide anions and so actually on the right side we're going to get 2 h2o s here so two water molecules in the right side so let's let's go ahead and and write the final answer here including everything that we've that we've done so on the left side we have 3 CL o - plus 2 C our parenthesis o h4 - and now we added in those two hydroxides right so plus 2 o H - and then on the right side we had 3 CL - plus well over here we have three waters and we just made two more and so we really have a total of five so five h2o and then once again we have the two chromates over here so two cro4 two - and that should take care of everything so we can go ahead and box our final answer so we got there it took us a long time but we have our final answer and remember you can always check your final answer in terms of number of atoms and charge so both atoms in charge should balance and so you go ahead and and check this one on your own just like I showed you how to do it in the previous video