How to Balance Redox Equations in Basic Solution

let's talk about how to balance redox equations using the half reaction method in basic solution we can balance redox equations in acidic solution or in basic solution and a question will always say balance this in acidic solution or balance this in basic solutions so that's how you'll know which method to use doing this in basic solution just means that we're going to be adding Oh H - is here to help balance the atoms and that's because OAH - hydroxide makes things more basic okay now the process for balancing in basic solution is really similar to the process for balancing in acidic solution except the process for balancing and basic solution has a couple extra steps at the end in this video I'm going to go through one basic solution problem from start to finish so you can see all the steps and you don't need any prior knowledge about redox balancing now if you're already a rockstar at balancing redox equations in acidic solutions feel free to skip through most of this video and just focus on the end part where you'll see some extra steps used for balancing in basic solution that you don't use for balancing in acidic solution got it let's get started this is the equation that we're going to be balancing and we're going to start just like we do with acidic solution by determining the oxidation numbers of elements in the equation okay so starting here ZN zinc is an element by itself it's not combined with any other elements so it has an oxidation number of zero no.3 one - is a polyatomic ion we don't know what nitrogens oxidation number is so we'll have to figure it out oxygens is usually minus two we have three oxygens here so we'll do minus two times three gives us minus six now whatever nitrogens oxidation number is has to add with minus six to make minus one because that's a charge on this polyatomic ion in all of the oxidation numbers to add up to this charge so nitrogens oxidation number is going to be plus five plus five minus six equals minus 1 so plus five for nitrogen over here on this side of the equation now over here on this side zinc has become a monatomic ion with a charge of two plus so it's oxidation number is just going to be the same as its ionic charge okay so plus 2 is zinc's oxidation number on this side and then we got the same thing here with nitrogen we got to figure out what it is okay oxygen is going to be minus two we have two oxygens so minus two times two gives us a total of minus four no2 here is a neutral compound now so nitrogens and oxygens oxidation numbers are going to add to zero so that means that nitrogens is going to be plus four plus four minus four equals 0 so plus four for nitrogen on this side of the equation so what's being oxidized and what's being reduced we'll look at changes in oxidation number okay zinc here is zero on the left side of the equation and then it's plus two on the right side of the equation so zinc so oxidation number is going up which means it is undergoing oxidation so zinc here is oxidized from zero up to plus two nitrogen here is plus five on this side and then it's plus four on this side so it's oxidation number is going down it is undergoing reduction nitrogen is getting reduced from plus five to plus four oxygen is on both sides of the equation it's minus two here and minus two here so it's neither oxidized or reduced so we don't have to worry about it at all so zinc is oxidized nitrogen is reduced now let's write some half reactions for the oxidation and the reduction the oxidation half-reaction is going to look like zinc Zn and then Z n 2 plus on the right side nitrogen is getting reduced so on the left side of the reduction half reaction is going to be no.3 one - then we got the arrow and then the right side no.2 okay now let's balance each one of these will start by balancing the reduction half-reaction what's the Sun my got some marker on the finger it's an occupational hazard sometimes I'll go out with my friends in the evenings and they'll be like you're doing a video today right I'll be like how did you know it's because my hands are just covered with marker anyway we're going to balance the reduction half-reaction we have two elements right now nitrogen and oxygen okay so on this side we have one nitrogen we have oh three we have three oxygens then on this side we have one nitrogen and we have o to we have two oxygens so our first step is to balance the atoms other than oxygen and hydrogen that would be nitrogen here and we're all set we have one nitrogen on this side and one nitrogen on this side so let's move on to the the next step here add h2o to balance o and H+ to balance H okay so right now let's focus in on oxygen there is an imbalance in the number of oxygens we have three on this side and we have two on this side so I need to add more oxygens to this side okay so I'm going to add h2o here now how many h2o is do I want each h2o molecule has one oxygen atom I have two I need three so they balanced so all I need is one oxygen atom okay so by adding just one h2o I get that one oxygen I need so I can go from two to three on this side okay all set nitrogens and oxygens balanced but by adding this h2o I introduced a new element into the equation and that's hydrogen okay so now we got it now we got to put that in here okay h2 there are two hydrogen's in a - oh so I've got two on this side right now I have zero on this side okay so that's a problem but I can balance out my hydrogen's by adding H+ okay so here's my H+ how many H pluses do I need I have two hydrogen's on this side so I'm going to make two H pluses on this side so now that's going to give me two hydrogen's here on the left now real quick about these H pluses adding H pluses to something makes it more acidic we're supposed to be balancing this equation in basic solution but we're going to worry about the basic part later okay so right now we add the H plusses they temporarily make it acidic but then in some of the end steps we add additional Oh H minuses to cancel out the H+ so don't worry about this for right now even though it does make the equation a sick okay so the atoms all balanced nitrogen oxygen and hydrogen the last thing we want to do for the reduction half-reaction here is to balance the charges by adding electrons so let's take a look at the charge situation in the equation okay on the left side we have two h plus so that's going to be plus two and then we have one no.3 one - that's going to be minus one plus two minus one equals plus one and then here on the right side we don't have anything that's charged so we just have a charge of zero okay so what I need to do in order to balance these is to lower the charge on the the left side here and I can do that by adding one electrons here it is e minus this electron has one negative charge so adding to the left side plus one minus one for that one electron gives me zero now I got zero on this side let's line that up now I got zero on this side and I got zero on this side the charges are balanced the atoms are balanced and now I'm going to move on and balance atoms and charges for the oxidation half-reaction balancing the atoms for the oxidation half-reaction is going to be pretty easy because all I have is one element here and I have one atom of each on both sides one zinc here one zinc here so I don't have to worry about balancing the atoms other than oxygen and hydrogen and I don't have to do any of this stuff with oxygen and hydrogen because I don't have those so I can skip right ahead to adding electrons to balance the charges we are going to have to do this let's take a look at the charge here on this side I don't have anything that's charged so I have zero charge and then over here I have one Zn 2 plus so I have plus 2 of charge here in order to get these two balance I got to lower this side by 2 so I'm going to add 2 e minus 2 electrons that have a negative charge so I'm gonna have plus 2 minus 2 gives me 0 now I got zero charge on both sides and these charges balance now I'm going to start to put the oxidation and reduction half-reactions together but in order to do that I've got to have the same number of electrons in both of them and right now I don't have that I've got two electrons up here and I have one electron down here so to change that I'm going to multiply this whole reduction half reaction by two just kind of like it's a math equation okay so now I've got 2 times all of this I'm going to distribute the 2 across this reaction okay so 2 times e minus is going to give me 2 electrons perfect now the electrons are the same in both okay two electrons plus two times two h plus is going to be 4 h plus plus 2 times that is 2 no.3 one - then we got the arrow 2 times no.2 2 no2 plus 2 times H 2 L 2 h2o ok so now the electrons are equal and this is my new reduction half-reaction so now that they have the same number of electrons I can add them together go add them together and then cancel stuff out that appears on both sides of the arrow so let's do that to add these together I look at the arrow and I choose everything from both equations on the left side of that disturbin okay so left side we start with ZN ZN and now i move down here for everything to the left of the arrow in this reaction ZN + - e - plus 4 h + + - no.31 - then i have the arrow in both of them so put the arrow here and now everything on the right side of the arrow for both the equations so I have Zn 2 + ZN 2 + + - e - and now down to here + 2 no2 plus finally - h2o a little bit of a tight squeeze + 2 h2o ok now what i want to do is i want to cancel out stuff that appears on both sides of the arrow and for this equation here that is these 2 electrons - E - here - e - here okay so get rid of those get rid of those and now when I rewrite this to get rid of the electrons the equation is going to look like this just the same thing with everything else except for the electrons so there it is for you now if our goal was to balance this in acidic solution we'd be all set everything balances now and we're left with H pluses here which make the equation acidic but we've got to balance this in basic solution which means we're going to have to go through a couple more steps to get rid of these H pluses and replace them with OAH minuses which make things basic so let's start to turn this into a basic equation here's the first step that we use only for balancing in basic solution that's why I put this little basic above here this is an extra step at the end if you're balancing in basic solution okay as I said we got to get rid of these H pluses because the H pluses makes something acidic so to get rid of these for each H+ add one Oh H minus to both sides so I have four H pluses here so to this side I will add four Oh H minus and then to this side I will add four Oh H minus okay the next thing that I'm going to do is I'm going to combine the H+ and OH H minus to make h2o so I have this h plus or these H pluses and these Oh H minuses when H pluses and OS minuses come together they make water okay so here is the equation that we're going to get when we combine those to make water you see on this side I have the 4o h and the four H plus or I should say the four Oh h minus and the four H plus they came together to make four h2o okay nothing happened on this side because there weren't any H pluses for the O H minus to combine with so the four o H - are just sitting over here okay so now our last step for basic solution here is that I'm going to subtract h2o from both sides if possible okay think of this sort of like a math problem okay I have h2o here and I have h2o here so I can subtract something if it appears on both sides so I have two h2o here for there so I can do minus 2 h2o and that will cancel out out I can do minus 2 h2o and that will leave me with 2 h2o on the side because 4 minus 2 is to so this will be my final equation I have zinc plus 2 h2o because I subtracted 2 from here plus 2 no.3 one - scuse me zinc 2 plus 2 no2 then I subtracted away my water my h2o from this side so there's none left + 4 o H - so I got rid of these H pluses and I'm left with Oh H minuses so this is going to be in basic solution the last thing that I'm going to do is I'm going to do a final check to make sure that atoms and charge balance just because I want to have the confidence that I did it correctly all right our final check I discover mistakes all the time when I'm doing a final check of this kind of equation balancing so I highly recommend that you do this - just to make sure that your answer is right before you hand in your test or you hand in your homework or whatever ok Zn we have one of them hydrogen - x - 4 oxygen I have 2 here 2 plus 2 times 3 which is 6 gives me 8 and then I have nitrogen I have 2 of them ok over on this side zinc I have one hydrogen 4 times 1 from the Oh H - oh I got 4 hydrogen's oxygen I have 2 times 2 4 plus the 4 oxygens here 4 plus 4 equals 8 and nitrogen - from the 2 no2 1 for 8 - awesome the atoms balance alright let's take a look at the charges on the 2 sides of the equation here I have no charge no charge - no.3 one - that's going to be minus 2 of charge here and then I have Zn 2 plus so I have plus 2 and then I have four Oh H - so that's going to be minus 4 because I have four of these and they're each minus 1 plus 2 minus 4 equals -2 minus 2 here minus 2 here the charges balance so I did every thing right so these are the steps for balancing a redox reaction in basic solution most of these steps are used for balancing in both acidic solution and basic solution except for these three steps which are used only for balancing and base balancing something in acidic solution gives us a bunch of H pluses and so these steps for balancing and base use Oh H minus to get rid of those H pluses and then make it a basic solution

Channel: Tyler DeWitt

Views: 2,108,827

Keywords: Redox (Literature Subject), Chemistry (Field Of Study), Acid (Chemical Classification), base, oxidation, reduction, Khan Academy (Nonprofit Organization), mcat, sat, ap, ib, dat, oat, Electron (Subatomic Particle), Chemical Reaction (Literature Subject), chemical equation, lesson, tutorial, help, lecture

Id: v5sDNmYCaqo

Length: 17min 59sec (1079 seconds)

Published: Mon Jun 29 2015