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Show all work below each question Balance the follow-g redox reaction in acid solution: points)s2Cr20z 2-2. Balance the following redox reaction in basie solution: ...

Question

Show all work below each question Balance the follow-g redox reaction in acid solution: points)s2Cr20z 2-2. Balance the following redox reaction in basie solution: points)15Mn04BrMnO2BrO3Cr3+

Show all work below each question Balance the follow-g redox reaction in acid solution: points) s2 Cr20z 2- 2. Balance the following redox reaction in basie solution: points) 15 Mn04 Br MnO2 BrO3 Cr3+



Answers

Write a balanced equation for each of the following redox reactions using the oxidation number method: (a) $\mathrm{Br}_{2}(l)+\operatorname{NaI}(a q) \longrightarrow \mathrm{I}_{2}(s)+\operatorname{NaBr}(a q)$ (b) $\mathrm{PbS}(\mathrm{s})+\mathrm{O}_{2}(g) \longrightarrow \mathrm{PbO}(a q)+\mathrm{SO}_{2}(g)$

Problem. 20 asks us to use the oxidation number method to balance the following that ionic Redox Jack Redox reactions. It gives us the reaction cr two of seven qu'est ce plus I minus Aquarius yield CR three plastic queens. Plus I too solid. So the first thing that we have to do is a sign oxidation numbers for each of these Adams, where we're going to start with the CR two of seven. Bob, this is they. This has a two minus charge. There we go. Uh, where the oxygen is a minus two. There are seven of them for a total of minus 14. But in order to make this a hole to minus molecule, the CR two needs to be a plus 12. Ah, and so each CR, because there's two of them is gonna be a plus six. The eye minus is a minus. One charge, the CR three plus is a plus three charge, and the eye, too, is zero. And so our molecules that change throughout this reaction R C C, R, which goes from a plus six two plus three, therefore is produced where it's Delta is a minus three, decreased by 33 electrons and the I, which went from being minus one 20 It was oxidized. And it's Delta is a plus one increased by one oxidation Limburg. And so, in order to balance this chemical equation, we're gonna need Thio uh, increase. Ah, the charge from the iodine by three threefold Thio even out this minds three close one charge differential. And so for that, I want to incorporate cr two the seven to minus Equus waas the high minus a quiz yields Percy are three plus yanquis plus r i to solid. Uh, and we're going to multiply our iodine to get this, uh, by three. Now you'll notice be there's an eye to and then a single iodine. And so if we multiply this eye and I'm by three, we're then gonna have a really hard time balancing belting out the side. So we're gonna put the three here and we're gonna make this 66 I'm minus. Ah, that's gonna change up the total charge center. We have ah six change a six Delta six for the iodine, but we need to multiply. Buy out this minus three. And so to do that, we're going Thio multiply this PR by too. So now we have a plus six change over here. We also have to see ours on this side and to see ours on a reactive side, we now need to balance out these oxygen's eso. We're gonna add water to this side. Um, to her product side, it does tell us it's in an acid solution. And so that indicates that there is ample Ah, hydrogen ions and water available for us to use. And so, at age 20 stands liquid or l Ah, and we're going Thio, we have seven oxygen's on the reactive sides. We're gonna need seven Oxygen's on the product side That gives us 14 Hydra Jin's and so we're going to need to add 14 h plus injurious to the side as well. So our final hi chemical about balance chemical equation is here to a seven to minus plus six I minus plus 14 age plus yields to see our three plus plus three i two plus seven h +20

Problem. 22 asks us to use the oxidation number method to balance the following that ionic redox react Redox equation. Uh, it gives us the equation by minus a Prius, plus a mento four minus equals meals. I too solid Plus, Emma. No. Too solid. So the first thing we're gonna do is assigned the oxidation numbers. It's all right, Those above each of you to be Holden's. The minus is gonna be a minus one. That a man of war will start with the oxygen, which will be a minus two other four of them. So the tool minus eight charge the overall molecule needs to be a minus one. And so that lemon is gonna be a plus. Seven. I chew is gonna be zero. I am a no to the oxygen is gonna be a minus two again. Father to advance. The toll charge is minus four, which means that Emma needs to be a plus four. The, uh, and Aaron went from a plus seven to a plus four, which means that it was reduced and it's delta or it's changed. Isn't minus three because it went from plus seven two plus four. I went from a minus one to a 20 which means it was oxidized. And it's Delta is a plus one. So we're gonna write out this equation and balance both sides, huh? Where I will just write out the chemical equation. We have I minus angriest plus m n 04 minus angriest yields I to solid plus and and about two solid. Okay. And so we need Thio, uh, make these deltas equal, and so we're going to multiply. Be eyes by three. Um, you'll notice that we haven't I minus and I to still need to know by the eye to buy by three, cause if we must find the eye minus by three, we would have, uh I have to put a one and 1/2 here. And that's, uh, not not an option. Hi. So that gives us a total of six iodine. So we're gonna need to put six in front of my minus will then need to balance out because we know how a change of are, uh, 66. Iodine sword. Continue to multiply thes the heaven of four minus and the MMO two minus by two. In order to keep changed the same. And then they tell us that this is in basic solution. So we need to add water and, uh oh h minus molecules in order to balance this equation further. And so we have. Let's see, we have, uh, eight oxygen's on this side on four. Oxygen's on the product side. So R O H minus Equus. It's gonna be on the side and our h 20 liquid gonna be on this side in order to balance out the number of hydrogen Tze and oxygen's, uh, we will need, uh, we currently have for we have eight nine oxygen's on this side and 245 oxygen on the side. And so, in order, Thio, multiply out all of our hundreds of oxygen's we will need to add Wait, oxygen's to the side and then or on this story. So that way we have the eight total hydrogen sze a total hydrogen sze i eight oxygen's plus four oxygen's is 12 total. On the product side, we have four here and ate here also equal £12 of this is our final balanced net Ionic Redox equation

This is Problem 29 of Chapter 18 Electric mystery. So in this question balance each of the following Redox reactions in acid solution. So in the party of the question we have given, then bless Ferric ion, which is a free three plus that can form then ions plus fairest ions which is every two plus. So in the first step, in order to balance these Redox reaction, we have to find the oxidation number. So the oxidation number of pin in this case it is zero. And for every three plus ion, this is plus three for tin two plus iron. This is plus two. And for every two plus iron, this is plus two. So these are the oxidation numbers. Oxidation number is basically charge assigned of an element according to some set of rule, and that rule are called the oxidation number rules. So now we can clearly see that the tin undergo oxidation because the increasing oxidation number from zero to plus two. So that means this is the case of oxidation. And this is the case of reduction because the charge on after three plus from a free two plus that plus three, two plus two that went degrees in oxidation number. So this is called the A reduction. Yeah. So in the first step, we have to divide this equation into two half. Part one is oxidation half reaction, and the other one is reduction. Half reaction. Okay, so in the oxidation half reaction, we have 10, 2, 10, 2 plus ions, and in case of reduction, half reaction. We have a f e three plus two f e two plus iron, so as we can see that, So that means, in case of oxidation, half reaction. In order to balance this oxidation half reaction, we know that in case of oxidation, loss of electron can take place. So in this case, two electrons are lost, so we can see these, uh, two electrons and in case of reduction, half reaction, a reduction means gain of electrons. So in this case, one electron again so that this can form a few two plus ions. So that means here we can add one electron like this, but this equation. But this situation is not balance until we have to balance the electron on both the side of the equation. So, in order to balance the electron. What we have Do we just multiply this whole equation by two? Okay, so now what we get, we have tend to tend to plus ion plus two electrons. So this is the oxidation half reaction. And in case of reduction, half we have to f e three plus ion plus two electrons that can form to Effy two plus island. So now we just add all of these equations in order to cancel the electrons. So the two electrons no electrons are canceled out and on adding the equation, we get our balanced chemical equation. So which is finally 10 plus two f e three plus ions that can form Tyne ions plus mhm twice off F E two plus ions. So this is the balanced equation. Mhm. Mhm. Yeah. And in the second part of the question. Okay, so similar approach. We can apply in the second part of the question. So we have given this type of equation mhm. So in this situation, first we find out the oxidation number of each element, So the oxidation number of hydrogen, his plus one arsenic is plus three. Oxygen is minus two Irene zero. Oh, this hydrogen in this case, plus one arsenic. In this case, it is plus five oxygen in this case, minus two, and I'll died. Island, this is minus one. Okay, Right. So now so you can easily calculate the oxidation number. Like, for example, if we consider this example like h two, Yes. Or for minors. So if we consider that for arsenic, it is X, and we know hydrogen. The oxidation number of hydrogen, that is, uh, plus one. So we can try to in two plus one plus X plus four and two minus two. This is for the oxygen. Mhm. So this is four into minus two, which is equal to minus one. So we get two plus X minus eight minus one. So the value of, uh, X minus six at its minus one. So the value of X is classified. So this is the just rough calculation in order to find the oxidation number. Okay, so now we we know about the oxidation number. So now first we can see the oxidation half part and a reduction half part. So now this arsenic undergo the observation because, uh, increasing observation number. And this I wouldn't undergo the reduction because decrease in oxidation number. That is called the reduction. Mhm. Okay, so now we can divide this equation into two half part. That means the oxidation half part and reduction. Half part. Yeah. So the oxidation half reaction, and here we write reduction, half reaction. Mhm. Okay, so the oxidation half reactions comes out to be We have edge. Yes. Arsenic. Mhm. Oh, three to minus. That can form H two s or four minus. So the oxidation number that is plus three for arsenic and here, plus five for arsenic. Now we can, in order to balance the oxidation half reaction, we can see that the hydrogen and oxygen are not balanced. But arsenic is balanced. So in order to balance the oxygen atom first. So we add water molecule on that side in which oxygen are deficient. So that means in the left hand side of the oxidation, half reaction so that we can add one water molecule in order to balance the oxygen. So now oxygen is balanced, but hydrogen is not balance. So in order to balance the hydrogen, we can see their total three hydrogen involved in the left hand side and two hydrogen involved in the right hand side. So we just add one. Uh huh. H plus iron in the right hand, side of direction. So now finally, we can balance the electron so as we know the oxidation that when loss of electrons or how many electrons are lost, so see, the change is only for two so that maybe two electrons are lost in this. Okay, so we can add two electrons in the right hand side of the equation. So Okay, so So this is the balance, the equation balanced reduction, oxidation, half reaction. So now we can reduction half reaction. That is I two to I minus. So here, oxidation. Number of Children zero. And for all right. And this is minus one. So first we can balance the island. So we just multiplied by this documentary coefficient to in the right hand side of the reduction half reaction. So as we know, reduction. So that means the loss in case of reduction, gain of electrons. So how many electrons are so here? Two electrons are again in this case because we're not two Children are involved in this case. That's okay. So now both oxidation, half and reduction. Half reaction are better. So now just we can do just add both this situation in order to get the final balance Paradox the action. So simply we can, right? Yeah, mhm. And the other regulation I Odin plus two locked on that can form too ru minus. So just simply add both the equation in order to get the final balanced Redox reaction. So the two electrons are canceled out on what they said and what we get. So we get finally. Yeah, on adding both the equation, we get balanced Redox reaction, and this is in acidic solution. Okay, And now, in the third part of the question, in the third part of the ocean, we have Cooper plus silver ion that can form called part two. Plus, I own Let's silver. So again, the oxidation number of copper zero. So, in the native form, free form the oxidation number of the element that is zero according to the oxidation number rules in case of Silvera I own that is possible in case of, uh, corporate two. Plus, I own the oxidation number plus two and for silver at +80 So, now see, this is the oxidation because observation number increased from zero to plus two. And this is the reduction because the oxidation number decrees from plus 1 to 0 so that cooper and rocky oxidation and silver plus and silver ion undergo reduction. Okay, so now again, same battered follow divide the population into cool parts. One is oxidation, half reaction. And another one is the reduction half reaction. So the oxidation half reaction, that is copper too copper, two plus iron. Mhm. So, in order to balance this oxidation half reaction, we can see the change in changing the oxidation number, which is plus two. So that means the two electrons are lost in this case because oxidation wins philosophy electrons. So we just lost these two electrons and in case of reduction, half reaction. We have silver plus I own two from silver. So here oxidation number of silver iron plus one. And here it is zero. So now you can see that the change in oxidation number and in case of reduction that when the gain of electrons So how many electrons are gained? So it is only one electron are gained in this case. Okay, so now this occasion both the equation oxidation, half and reduction half reaction are balanced. But you can see the electrons are not balanced until now. So in order to balance and cancel the electrons, what we can do, we just multiply this whole equation by two so that the electrons are cancel out on the side of the equation. So in the final equation, we don't need to write the electrons. So we get copper to copper, two plus ion plus two electrons. And here we get twice of silver, plus ion plus two electron that can form twice of Sullivan. Okay, so if we add about this situation, yeah, two electrons are canceled out on both side of the equation and what we are left with. So we are left with copper, plus those silver ions that can form copper ions less the very soft silver. So this is the balance paradox reactions. So I hope this answer your question. So thank you so much for watching this video

And start by balancing The guy died Adams there. Then I'm gonna track electrons. Aluminum goes to aluminum three. Plus, it lost three electrons. The two I dine, Adam's each gain one electron for a total of two electrons transferred there. Now I've gotta multiply those by some factor so that I have the equal number of electrons lost and gained. And so then I'm just gonna transfer those numbers and green in his coefficients on the I died I on because it was too iodide ions. Now, two times three. That's where that six comes from. I'm going to examine the half reactions here. I'm gonna sign oxidation numbers and then on the right side is two plus on the left side. It's plus four. So I know I'm gonna be adding two electrons on the react inside. Now I gotta balance out the charge by adding hydrogen I'll and then I gotta balance out the oxygen biting to water to the right side of that equation. So then I've got the bromide. I am becoming bro. Mean assumes liquid. So I need to bromide ions. So then we're losing two electrons Now the electrons there are balanced, gained two and lost two, so I can simply add those 2/2 reactions together to get the balanced chemical equation.


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