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Balancing Redox Reactions · Acedra
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Balancing Redox Reactions
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in this video we're going to focus on balancing Redux reactions and acidic solution so let's start with an example let's say if we have the reaction between zinc and the chlorate ion and let's say that it produces the zinc plus 2 cation and chloride so what do you think should be the first thing that we should do in order to balance this reaction under acidic conditions so we need to separate this reaction into half reactions let's write a half reaction between zinc and zinc
0:49
2+ now when balanced in a half reaction balance the atoms first and then after that balance the charges so we have one zinc particle on both sides but the charges are imbalanced on the left side the total charge is zero but on the right side the total charge is two so these two numbers differ by two so we need to add two electrons always add the electrons to the side with the higher charge in this case the right
1:23
side by the way what kind of reaction do we have here this half reaction is it ox oxidation or reduction whenever the electrons are on the right side it's an oxidation half reaction zinc is releasing or giving away electrons another way you can look at it is by analyzing the oxidation number whenever the oxidation number goes up as you travel from the left side to the right side of the reaction it's oxidation the zinc is being oxidized in this reaction if the oxidation number decreases it's
2:01
reduction now let's move on to the next half
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reaction so we have chlorate on the left side and chloride on the right side now before we can balance the charges we need to balance the atoms we have one chlorine atom on both sides and we have three oxygen atoms on the left side so we need to add three oxygen atoms to the right side under acidic condition you can balance such reactions by adding H+ or h2o since we need three oxygen atoms on the right side and let's add three water molecules since water contains
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oxygen so now the oxygen atoms are balanced let's move on to hydrogen notice that we have six hydrogen atoms on the right side so we need to put a six on the left side we need to add 6 H+ to that that side so now all of the atoms are balanced on both sides of the reaction now we can balance the charges what is the total charge on the left side on the left side we have six H+ ions so it's 6 * 1 so we have plus six from those ions and we have a negative 1 charge from the
3:25
chlorine so the total charge on the left side is POS 5 on the right side we only have a chloride ion so it's ne1 on the right side what is the difference between 5 and 1 5 -1 is 6 on a number line there are six units apart so we need to add six electrons but to the side with the higher charge since the left side has a higher charge let's add it to the left side
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notice that for the other reaction the electrons are on the right side so for this one they have to be on the left they have to be on different sides of the reaction they can't be on the same side because one of the half reactions has to be oxidation and the other has to be reduction so whenever the electrons are on the right side it's oxidation if they're on the left side it's reduction now the next thing that we need to do is we need to make sure that both of the electrons on each reaction
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are equal so we need to multiply the first half reaction by three so that we can get six electrons so it's going to be three zinc
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atoms Plus or this is going to turn into three zinc plus two
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ions and six electrons
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now the other half reaction we don't really need to change anything I'm going to rewrite it
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though let's get rid of that one
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so in the other half reaction we had six
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electrons a chlorate ion six H+ ions and on the right side it was chloride and three water molecules now that we have the same electrons we can add the two half reactions when you add them the electrons will cancel since they're on opposite sides of the reaction and then everything else you could just bring it down so to the left of the arrow we have zinc chlorate and the H+
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ions and on the right side
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we have the zinc ions chloride and
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water so this reaction is balanced and you can check it we have three zinc atoms on both sides six hydrogen atoms three oxygen atoms and one chlorine atom now the charges must balance as well notice that the total charge on the left side is 6 +1 which is 5 and here it's 3 * 2 which is 6 plus1 and is 5 now let's try another
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example consider the reaction between between iron metal and the permanganate ion go ahead and balance this reaction under acidic conditions using what you know at this
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point so just like before let's separate it into half reactions so let's start with iron metal in this reaction we have one Fe atom on both sides so the number of Fe particles is the same now we can balance the charges the charge on the left is zero on the right it's three these two numbers differ by three so we need to add three electrons to the side with the higher charge so that's it for the first one now let's move on to the next half reaction so we have
8:21
permanganate and it's turning into MN + 2 so the manganese atoms are balanced in its reaction we need four oxygen atoms so let's add four water molecules to the right side now that we have eight hydrogen atoms we got to add eight H+ ions to the left side so now that's balanced now we need to consider the charges the total charge on the right side is two on the left it's 8 +1 which is POS 7 7 - 2 is five so we need to add five electrons to the left side so now the second half reaction is
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balanced now we need to make the number of electrons equal we have three in the first half reaction five in the second so the least common multiple of three and five is 15 so therefore we need to multiply the first reaction by five to get to 15 electrons and the second one by three to get to 15
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so if we multiply the first half reaction by five it's going to be 5 Fe and five fe+ 3 and 15
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electrons now for the second one if we multiply everything by three we're going to have 15 electrons a * 3 is 24 so 24 H+
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ions three perinate
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ions three of these as well and then 4 * 3 is 12 so we need 12 H2O molecules so now we need to add these two reactions
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as we add the two reactions the only thing that we can cancel are the 15 electrons now these three we need to add them to the left side of the reaction as we add the two reactions so it's going to be 5 Fe plus 24 H+ plus 3 peragon
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and on the right side we have V3 so it's going to be five Fe + 3 on the right side and three manganese ions and 12 water molecules so this is the balance reaction and to check everything notice that we have five Fe atoms on both sides we have 24 hydrogens 12 * 2 is 24 we have three maganese atoms and 12 oxygen atoms on both sides so now the last thing that we need to check is the total charge we need to make sure it's the same on both sides so on the left side we have a charge of 24
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and -3 which adds to 2 21 5 * 3 is 15 3 * 2 is 6 15 + 6 is 21 the total charge is 21 on both sides so the reaction is
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balanced let's try another example try this one iodide reacts with the per bromate ion under acidic conditions to produce iodine and
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bromide so let's start with the iodine ion now the first thing we got to do is balance the atoms we have two iodine atoms on the right and one on the left so let's put a two on the left side now let's balance the tries the total charge of a Pure Element is zero on the left side it's 2 * 1 so we have a total charge of -2 on the left side -2 and Z differs by two which means we need to add two electrons to the side of the higher charge Z is greater than -2 on the number line so let's add it to
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the right side now the next thing we need to do is balance the other half
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reaction now we can do so
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by adding H+ or water so notice that we have one bromine atom on both sides but we have four oxygen atoms on the left so that means we need to add four water molecules to the right side so that's it for oxygen now let's move on to hydrogen 4 * 2 is 8 so we need to add 8 H+ ions to the left side now let's check for the total charge the total charge on the right side is minus1 on the left side it's 8+1 which is POS 7 7 differs from 1 by 8 7 -1 is the same as 7 + 1 which is 8 so we
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need to add eight electrons to the left side
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now the next thing we need to do is make sure the electrons are equal in both half reactions so we need to multiply the first one by four we don't need to change the second one which is good by the way which half reaction is the oxidation half reaction and which one is the reduction half reaction the first one is oxidation because the electrons are on the right side the second one is reduction because the electrons are on the left side that's how you can quickly tell which one is which so now
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let's multiply the first half reaction by four so we're going to have eight iodide
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ions four I2 molecules and eight electrons so let's add these two half reactions so as always the electrons will
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cancel so everything else we just simply need to bring it down so on the left side we have these three
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now for the right side we have everything to the right of the arrows except the eight
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electrons so we have four I2 molecules a bromide ion and four water molecules so this is the balanced reaction as you can see we have eight hydrogen atoms on on both sides one bromine atom and four oxygen atoms and eight iodine atoms now let's check the total charge the total charge on the right side is negative 1 on the left side positive 8 and negative 8 cancels so we just left with this number so on both sides the total charge is Nega 1 so this reaction is
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balanced so here's the last example for this video let's say if you have the reaction between PB and pb2 and under acidic conditions it turns into the PB plus 2 ion how would you balance this reaction so let's start with the first half reaction which is PB turning into the pb2+ ion the total charge on the left side is zero on the right side is two so we have to add two electrons to the right side since it has a higher charge so now let's move on to the next half reaction
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pbo2 that's going to turn to PB plus 2 as well both of these two substances they're going to converge into the same
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ion now before we can balance the charges we need to balance the atoms so we have two oxygen atoms on the left that means we need to add two water molecules to the right
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side so now that we have four hydrogen atoms on the right side so let's add four hydrogens to the left to make it
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equal now the last thing we need to do for this half reaction is that we need to balance the charges the total charge on the right side is pos2 and on the left side it's pos4 so these two numbers differ by two so we only need to add two electrons to the left side since the left side has a higher charge now because the half reactions have the same number of electrons we don't need to modify so we can simply add the two equations so these will cancel on the left side we have PB pb2 and
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H+ so it's going to be lead metal reacting with lead dioxide plus 4 H+ you can call this lead to oxide I mean not lead to oxide I take that back lead for oxide since the charge on PB in this particular compound is plus4 now on the right side we have two PB plus 2 ions and we have two water molecules so this is the final answer for this reaction so we can see that we have two lead atoms on both sides of the reaction we have four hydrogen atoms two oxygen atoms
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and that's about it the total charge on the left side is four and on the right side is four so everything is equal for this reaction so that's it for this video now you know how to balance redo reactions under acidic conditions
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