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4. You now have mass balance but will need to balance charge. This will be done by adding electrons to the side of the reaction with excess positive charge_ Enter 0...

Question

4. You now have mass balance but will need to balance charge. This will be done by adding electrons to the side of the reaction with excess positive charge_ Enter 0 if no electrons are added, or #e for the number of moles of electrons required to balance the net chargesH20 + HONO3H+ + NO3Submit Answer Tries 0/45This is now a balanced half-reaction: The electrons are aSubmit Answer Tries 0/45This is a half-reaction. Submit Answer Tries 0/45It is also useful to be able to determine the oxidation s

4. You now have mass balance but will need to balance charge. This will be done by adding electrons to the side of the reaction with excess positive charge_ Enter 0 if no electrons are added, or #e for the number of moles of electrons required to balance the net charges H20 + HONO 3H+ + NO3 Submit Answer Tries 0/45 This is now a balanced half-reaction: The electrons are a Submit Answer Tries 0/45 This is a half-reaction. Submit Answer Tries 0/45 It is also useful to be able to determine the oxidation state change of the redox active species, in this case, the N in HONO and NO3 The oxidation state of N in HONO is Submit Answer Tries 0/45 The oxidation state of N in NO3"is Submit Answer Tries 0/45 The difference between these two oxidation states, +3 and +5,should be the same as the number of electrons in the balanced half-reaction, Submit Answer Tries 0/45



Answers

For each of the following balanced oxidation-reduction reactions, (i) identify the oxidation numbers for all the elements in the reactants and products and (ii) state the total number of electrons transferred in each reaction. (a) $14 \mathrm{H}^{+}(a q)+2 \mathrm{Mn}^{2+}(a q)+5 \mathrm{NaBiO}_{3}(s)$ $\quad \longrightarrow 7 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{MnO}_{4}^{-}+5 \mathrm{Bi}^{3+}(a q)+5 \mathrm{Na}^{+}(a q)$ (b) $2 \mathrm{KMnO}_{4}(a q)+3 \mathrm{Na}_{2} \mathrm{SO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$ $\quad \longrightarrow 2 \mathrm{MnO}_{2}(s)+3 \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{KOH}(a q)$ (c) $\mathrm{Cu}(s)+2 \mathrm{AgNO}_{3}(a q) \longrightarrow 2 \mathrm{Ag (s)+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)$

This is the longest super long problem that I have attempted. We're going to be doing eight. Yes, 8/2 reaction method balancing for reduction here I have put a summary for how to write half reactions and a summary of how to balance redox reactions. These are decades old. I don't remember where I got them, but they're handy Dandy. If you need to come in reference this I put these l down on a sheet of paper are on a separate sheet already because this takes so long that I want to make sure it's going quickly. So what I'm gonna do is I'm gonna write half reactions for each part of this. The first half reaction I'm going to write is for it's copper solid and it's going to see you two plus okay, and that's Equus. And now we have to balance this first thing we're gonna look at is bouncing any oxygen's. There are none. Move on, then we do. Hydrogen is there are none. Move on. Then we had electrons to make sure charges are the same on each site. So if I had two electrons to this side than its neutral, I'm done with this part of the reaction. I'm only unhappy that I haven't way over here, so I'm gonna move this closer to the middle, and I'm just going to write down what I just said. Okay, then, for my other half reaction, let's make that in purple. I'm going to consider these two. Let's start with what I have and all three minus. I'm gonna leave the states out this time and Enel Gas. Okay. My first order of business is to balance my oxygen's, and I balance oxygen's by adding waters. I need to waters because that will give me three oxygen's on both sides of this. Now I add hydrogen ions, and I need to add four hydrogen ions on this side to balance this out. We're getting closer as we speak here. Now, the only thing that's not balanced knowledge race this. I have a neck charge of three plus on this side and zero on this site. So I'm going to have to add three electrons to balance that out. Three electrons. I now have my half reactions, and I'm going to add my half reactions together. Hopes that's a very nasty little line. Well, that's it is what it is. And I have to make sure that my electrons match and they don't. So what I need to do now is I'm need to multiply each equation by whatever I need to get the same number of electrons in each side. So it'll be a three and a two Now for this one. I'm gonna write the whole thing down first. I'm going to start with three. See you, and I'm gonna leave the States off. Plus, now I have eight hydrogen and I have to end all threes. Minuses! And I have six electrons. I won't write the electrons every time. Now, on the product side, I have three. See you two. Plus, I have six electrons. I have to Eno's and four Waters. I'm gonna try to stick this together a little closer to make it. Andi, I had six electrons plus two n o plus four h 20 Okay, so now we're gonna cross off anything that we have on each side and you'll notice. First thing I'm going to get rid of is this. You can tell if you're doing it this right, because you'll get rid of your electrons The next thing that I need to check is if I have any waters. I do not any waters or any, um, hydrogen is that I need to get rid of. I don't. So now all they have to do is write this out. Three. See you. That was solid. Plus eight h plus. I was give those in a queue plus two, NL three minus. My products are three seats. See you two plus and to n o. And I believe that's a gas plus four h +20 which is in l first. Problem. Done. Yes, we have seven more. Make sure my cameras on it. Iss okay for this problem, we're going to write our 2/2 reactions and I'm going to do CR 207 and ZR three plus. Well, the first thing I notice is that I don't have the same number of CR, so I'm going to put that there. I have to chrome ian's that premiums on each side. Then they have to add seven waters. And that means I also have to add 14 waters to this site. Ups 14 h plus is so I'm adding 14 age pluses. Now let's check and I'll raise this in a moment. Let's check our arm charges on here. I have a plus six on this side and they have a plus 12 on this site. So I'm going to have to add six electrons to this side to get our charges to cancel up. So there's my first half reaction. And for my second half reaction, I'm dealing very simple. One chlorine and chlorine. First order of business is balance. I've got no oxygen's I've got No, I'd regions. So all I have to do is what's need to add two electrons to this side to balance my charges out. Now, I want to mention if you've done your work correctly, you're gonna have some oxygen's some electrons on the laughter, the arrow and some on the right. That means you've done it correctly, and those indicate your what's being oxidized and once being reduced. Okay, now we're going to combine our reactions. But first, before we combine our reactions, let's make sure we've got the same number. I've got six electrons in two electrons, so all I need to do is multiply this equation by three this time since I've got six electrons in each side. I'm gonna leave this off when I write my equation. So I've got 14 h plus plus CR +207 plus six C o minus on my product site. I have to chromium three ions, seven waters and three chlorine. Okay, if I checked carefully here, I don't think anything is going to cross out. So I guess I'm done. I wish I would have written that a little bit nicer, but there it is. 30 Problem is sort of interesting for this problem. We have PB and P B 02 and they're both turning into r p. B s 04 And that s a poor is a big pain in my life right now. And I know it's going to be an ion in this reaction, so I'm just going to consider this later and put my ions in here right now. These are a little close, but let me move this this equation down a little bit, and I should do a different color. Okay? So, back to our first half reaction, I've got the same number of leads on each side of got no oxygen's. I've got no hydrogen, so All I need to do is add a couple of electrons right there. My 1st 1 is good. Now, my second reaction, I Scott to oxygen's right here. So I'm gonna have to add switch colors, add to waters. And if I had to waters my oxygen's air Good. But I'm gonna have to add for hydrogen is to this site. Now, if I had four, hydrogen is to this side. I can see that. My, um I've got I lost my equation here. Four hydrogen is to this side. I'm gonna have to add two electrons to this side. You look at how nicely that's gonna work out. I've got two electrons and two electrons, so I could just combine my equations. So let's combine these two equations. I'm gonna leave my electrons off. So I've got PB plus four h plus plus P b +02 and this yields to PB two plus and to h 20 Okay, Now we are getting closer here, but we do have to bring back this and our cell fates. So in order to do that, let's just write what we already had here. And I'm going to leave this one off for right now. If I need any hydrogen Zyl, put him right here, plus Peak B O to know I'm adding H two s 04 And on this side, I have PB two p. B s 04 Hope that means I'm gonna need to right there as well. And I still my to age two. All right, here. So let's evaluate what happened. I added four. Hydrogen is in and I had four hydrogen, so I don't need that term anymore, so I'm going to get rid of that term. Now, let me see if I've got all my terms in here correctly, and I'm going to go through and rewrite this one with States. There we go. Third problem. Done. Okay. Lets your fourth problem and let me find out where it is. It's on the reverse of this side. There's Yes, there it is. Okay, so I've got my first reaction. I'm gonna do my men get manganese two plus in m n 04 one minus at my hydrogen in the form of waters. I made four of them that will give me eight. Hydrogen is on this site. And then to make my charges work. Take a look at what I need to do to make my charges work on this. I will need to add five electrons on this side. My second equation I've got and I'm gonna leave my n a office. You'll see I don't have any. Sodium is on this side, so I I'll add this I on in when we're finished. B i 03 and B I three plus. So for this one, I need to add three waters over here. Six hydrogen over here, and that'll leave me two electrons. So now this, I have two and five. So I'm going to multiply each equation by a number to get the same number of electrons. So multiply this equation by two. And this equation by five. I'm gonna leave my electrons off cause I'll have 10 of each And let's write our equation. Don't forget to distribute your to and your five over each term. Eight h 20 plus two mn two plus plus. Now look at this. 30 h plus is plus five b I 03 And that should be one minus. Okay. And I should have five of those, okay? And I'm gonna spread this out just a little bit and I'll just go ahead and write on the next line I m in plus six. 30 h plus plus five B I, 03 minus. And my products are going to be to Emma. No. 41 minus and 16. H plus is five. B, I three plus and 15 waters. Okay, let me make sure I wrote all those down correctly. That's a lot to remember. I believe I did. Now we're going to combine any terms that we need to combine here. And I see I have 16 age pluses and 30 h plus is So that leaves me 14. H plus is on this site. No, I'm done with those. And I have eight waters and 15 waters, eight waters and 15 waters. So that is going to leave me seven waters. I'm just gonna put I've got seven waters to deal with over here. Then I'll start rewriting the rest. So still got two mn two. Plus, I've got five b i 03 minus and from my products, I'm gonna have to m n 04 minuses plus five b I three plus plus seven waters Okay. This is a quiz. Sequus, and this is liquid. Let me see if I'm done. Almost done. I do have to add my, um, sodium is back to this. And here's how we'll do that. Let me write this one more time. I'm gonna try to get it all in one line. I'm just gonna add my sodium ion back here, and I need five of them. And is that everything on this side? Yes. Now, on the other side, I'm gonna have, um, two mn 04 minus. I can't get this on the other side. And then I will have five b i three plus and they will have five in a plus and seven waters. How exciting. That was exciting. Okay. And then I gotta turn my computer back on because it went dead. It's been sitting there for so long. Okay, Half done. Let's do our next problem. And this one, I'm just going to go a little faster. I'm not going to include states here, and I can see that for my first equation. What I need to do is I need to add four waters under this site, which means I need to add eight hydrogen onto this side. None of these had any charges, so I also need to add eight electrons onto the side. My second equation is ink to think, and all I need to do is at a couple electrons right there. Check that. I've got a whoopsie, an eight and a two. So all I need to do is multiply this equation by four. And I'm gonna combine them eight to go a little darker color eight h plus plus h three a s 04 plus four z and then produces S H three plus four h +20 plus four the n two plus. And I believe that 814 So I'm just going to write this with states real quick. And I have a S H three, which is a gas, and four waters, which are a liquid and four sinks. Think ions, which are a quiz. There we go. See how fast you did that one. Okay, that was E. Let me find F. Okay, so are six problem even faster. Let's do this. We've got a as to 03 and h three a s 04 So I'm gonna have to have a two right here to get my, um, are cynics, which means that I'm going to have to add Got eight. So I'm gonna have to had some waters to this side. I got eight on this. I'm gonna have to have my waters to this side. Give me 10 of those which needs and me. Four more waters over former H plus is over here, and also ill have four electrons on this side for my second equation. And we're starting with my mental three and my panel. I feel like we already did This one. Let me make sure I got everything here. I did. Okay, if we'll check with four electrons here and three electrons, we need to multiply this by three. And this by four. So this is gonna be a long one. Let me just write this out quickly. Don't forget to distribute this term for each thing. My electrons, which are 12 and 12. I'm gonna cross off. I forgot to distribute 16. And now I'm just going to change colors a little bit so we can look at this side. So I've got six, then. I've got 12 hydrogen electrons. I got rid of. I have four annals and eight waters. So make sure you look at that and know what you're doing on here. Because that's what's important to look at. Now I'm gonna combine my h Plus is and I'm couldn't combine my waters. If you look at these, I'm just gonna jot this down quickly. I've got seven waters on this side and I've got with 12 and 16 I have four h plus is I've got three as two or three. And that's a solid and four and all three minuses, Which is are a quiz. Okay, those are my react. It's let me check something for 347434 Yes. Now, on my product side, I've got six h three s 04 That's a Quist. And my age is air gone. I got for a nose, and I believe that's it. There's the answer for this one. Okay, we're making it two more. Make sure my camera still on. Okay, let's do our half reactions, bro. Me and bro Bean to and plus two electrons. Okay, Second half reaction is all we have done. This one before, Emma. No, for minus and mn two plus. So I have to had four waters, eight hydrogen. And where are my electrons going to go on this one? I believe I have to add five electrons right here. Okay, so I've got five and two. I'm gonna have to multiply each side. I didn't leave much room, so I'm gonna write it here times five and times two. So let's combine these leading off our electrons. So I've got 10 bromide ions and 10 electrons are leaving up 16 hydrogen ions and to permanganate ions. Okay, Now, on the product side, we have five bro means to manganese, two ions and eight waters. I mean, see if there's anything else there isn't. So I wish I would have put states on here. I think I'm just gonna put him underneath a que a que a que L a Q and l? Okay, we're on the last one. Hooray! Hooray! So let's do our c h. 30 h and R C h two. Oh, Okay. So those air good. The only things I'm gonna have to worry about here are adding my hydrogen. So I got 400 years in this side. I'm gonna have to have two h plus is and that will require two electrons. Okay, now, on the next part, we've got CR Joel seven to minus and see her three. Plus, Here's another one. I'm sure we've done so. You've got seven waters, 14 Hydra jin's. And that will leave six electrons. So all I have to do on this is multiply this one by three. And let's write this out. Three C h 30 h. Plus, I'm leaving my these off 14 h plus plus here, too. 07 I forgot to put it to right there. I think I just needed it to right there. Yep. It's good. Okay, there's my product or my reactant and my products are going to be ch two. Oh, plus, I believe six h plus is And these cancel out C h 206 h pluses. And then I have to see our three plus and seven waters. Okay, So our terms that air could going to combine her right here. Now. All right. This I'll have a little more room cause I'm gonna leave out one of my hydrogen ions, so I'm gonna end up with eight. H plus is three of these and one C R. 207 to minus those air. My reactant and my products are three C H 20 two chromium, three ions and seven. These were both a quick as I think. What was that? Yep. That's a quiz And seven liquid waters. We did it. If you watch this far, Congratulations.

So now we'll work on problem 18 from chapter 20. This problem, we're told for each of the following balance Oxidation, ria production reactions Identify the oxidation numbers for all the elements of the reactors and products and state the total number of electrons transferred in each reaction. So let's go ahead and write down our reactant here. So in the first reaction, we have ah to mn for minus plus three s two minus plus water will produce three neutral suffers two men Guineas oxides in 80 H my in ants. So now we're asked to write down the oxidation number of each of the reactant. So we know that for oxygen here, it will be negative too. So if we multiply that by four, we get negative eight. And since the ion here is negative one, we know that Megane should have a charter plus seven here. Silver should be negative too. Hydrogen will usually always b plus one and oxygen negative too. And so silver here is zero because it's neutral element and then oxygen is negative. Two here in manganese is a positive force, since we now have it a neutral element. And, uh, may give two for oxygen, plus one for hydration. So now in part do they asked us the total number of electrons transfer, and we can see that many in the electron. The redox reaction that's occurring here is we have three electrons being transferred from one manganese of from sulphur two manganese. Manganese is gaining three electrons, and that happens for two guineas. So there are a total of six electrons transferred from sulfur, two manganese. So for part B, the reaction is for hydrogen peroxide h 202 plus sil O C. L +207 plus two hydroxide. And this will produce to Cee Lo to minus five waters and for oxygen molecules so we can go ahead and look at the oxidation numbers of each captains. Ah ha jin will be plus one and peroxide Zahra unusual because the oxidation number of auction in peroxide is negative one instead of negative too. So that's cool. Here, seal Oh, to seal 207 Oxygen will be a negative to oxidation of her. So that means negative 14. Which means that each chlorine has to be positive. Seven oxygen in the hydroxide. His negative two plus one interesting for curing Cielo to We have negative too, for the oxygen. Which means that the Lorene has to be positive. Three and for water plus one for hydrogen Negative too, for oxygen and for oxygen molecule zero. So now we need to look at the total number of electrons transferred not to do that here. Um, what we need to see is it's hard to track because we have some electrons coming from oxygen. But we have high Shiprock side. We have Cielo seal to a seven. We have oxygen and each of these reactant ce. So it's hard to see exactly where the electrons air coming from. How many electrons are coming from oxygen? But what we can look at is how many electrons the chlorine is gaining. So we have two Korans plus seven, and on the right, we have two. Chlorine is plus three. So four electrons for each our gained here to here. We have two times for electrons. So that means there's a total of eight electrons involved in this process. In part. See, the react in all reaction which were given is very, um, two plus two hydroxide, uh, passion peroxide edge to two into Cielo to minus kilo two, and then that produces a salt of barium C l O too minus ceelo, too. And there's two of those. Then we have two waters and an oxygen molecule, so here we can put in hour sedation numbers. Barium is positive, too, as the superscript indicates, in oxygen for hydroxide will be negative two plus one for hydration, plus one for I shouldn't hydrogen peroxide and negative one because it's a peroxide for oxygen in the cielo. To we have negative too, for oxygen, which means that chlorine has to be positive for On barium, we said again have it as a positive, too. Corning here is a negative two, which means Sorry, Thea Austin is negative too, which means that the chlorine has to be the same as before. In this case, it's not the same because we over here we had a neutral molecule. Here we have a two plus here, which means that we have to have a negative here. So in this case, the corn will only be positive. Three. Then Hussian is plus one negative too, and syrup. Now we need to look for the total electrons transferred in this reaction so we can look and see that once again, from the sea alot to compound, we see that we haven't addition of one electron here on there's two so two times one electron and that is coming from thes this oxygen here. So there are two electrons transfer.

This next reaction will use the half reaction method to balance it. Sophia, change in oxidation numbers for the first half reaction as the Taschen goes from zero to plus one and then we see water having a change in the oxidation number of hydrogen as it goes from +120 Well, then, make sure that the charges are balanced by adding electrons. You did for this first oxidation half reaction, and then this must be a reduction half reaction where we're gonna have electrons now on the left hand side because we can't have an oxidation without a reduction. So we'll add two electrons so that both sides have a to minus charge. The one will supply the first half reaction by two so that our electrons will cancel when we sum everything up and this ends up being are balanced. Redox reaction

So now we'll work on problem 26 from chapter 20. In this problem, we're asked to complete and bounce the following equations and identify the oxidizing and reducing agents so we can start with part A where we have no to reacting with micro mate in acidic solution. So the best way to approach these problems is the first identified, the reduction and oxidation, half reactions and then work from there. Yeah. So for part, a R reduction is occurring at this by crow, mate. Because we can see that the crow mate Bikram made the chromium on the left side has an oxidation state positive six. And on the right side, it's positive. Three. So it's being oxidized. No. So we have Bye, chrome eight cr 207 two minus. And where an acidic solution. So we know we're going to react with protons and with electrons, it's and produce, uh, two chromium because of the reaction and water. So now we can balance the equation, so we know that we have seven oxygen over here, so we need to have seven waters over here. And since we have 14 height should over here, we need to have 14. Hydrogen is over here and we have two Chromium is going for plus six six two plus three. So a total of six electrons are transferred. Now we can work on the oxidation half reaction, which is nitrate going to nitrate. So you have no too minus h 20 In this case, it's not enough to react simply with the hatch and protons because they're the protons. Were will not react with these and nitrate or nitrate molecules because they would form the acids which would just ionized again. And also we need an additional source of oxygen to add the oxygen to go from nitrite nitrate. So in this reaction, nitrogen is going from an oxidation number of positive three to an oxidation number of positive five. So we have a transfer of two electrons and we also have two protons. So too, add these reactions up. We need to multiply the bottom reaction, the oxidation reaction times three and then we can go ahead and have a thought. We get si r 272 minus plus 14 protons plus three and oh, to Linus. Our water will cancel with this water here so we don't have any water here in our electrons. Cancel. Then we get to screw me. Um, three plus, uh, four waters three and three. And so we actually have less than 14 protons because we have six protons coming from this reaction and that will subtract from the 14 to give eight protons here. Now, the question asked us to identify the oxidation and production agents. And so we know chromium, This bike Romany compound is being reduced, so it means it's the oxidizing agent, and this nitrate is being oxidized. So it is the reducing agent. No. And part be, the reaction which were given is between sulfur and nitric acid in acidic solution. So first, we have, uh, nitric acid h n 03 as our reduction reaction, and it's in acidic media. So they have a class, and this is the reduction. So we have an electron has to react, and and we're going to end to oh, plus water. So if we're not sure about the oxidation numbers, we can see that this nitrogen here is losing a lot of oxygen. So it makes sense that it's reduced as opposed to oxidize. So now we can balance this equation. Uh, we need to go from this. Nice action on the left. Here. Excuse me. Nitrogen on the left. We'll have the oxidation number we can figure out. So the I am that it's a part of his n 03 minus. So n 03 minus with three additional oxygen with three oxygen. We have a charge of negative six, So this charge over here will be positive. Five. Over here, we have a charge of positive one. And so this is happening for two electrons. So that means we have a total of two. Adams, we have a total of eight electrons being transferred here. And since, uh, we need to keep charge neutral. That means there should be eight protons as well. And we need to put two here so we could have two naturally ins on this side. So now, in order to balance the number of oxygen, we put a five here. So we have six oxygen on the side, six oxygen on this side, and we have 1,000 on both sides. Now we can do the oxidation reaction where we have sulphur plus water producing. Uh, this H two s three and protons and electrons because it's the oxidation reaction. So now, just to figure out how many electrons we're going to need over here we have zero as the charge oxidation number for sulfur. On the right side, we have s 03 to minus, which means that the oxygen will be negative. Six. So sulphur should have an oxidation number of four. So we have the transfer of for electrons, which means they should before protons on the right balance the charge, which means that we need to add a subscript ofthree for the water. I'm not a right left side. So now we can add up the reactions we need to multiply this reaction times to sew. The electrons will cancel out. So we have to h and, uh 03 plus eight protons, actually not a prude. The protons here will cancel out completely, plus two sulfur plus h 20 produces too. H two s 03 plus for into Oh, So the oxidizing agent here is the reaction, the reactant which involves the production. So that's H and R. Three snitches in here will be our oxidizing agent. And that leaves the silver to be be reducing agent. So now we can move on the part, see where we have the bike remit on Diana. Reacted with methanol form formic acid and chromium three plus an acidic solution. So first hour oxidation reaction will be the methanol to form a casted. The reduction reaction would be by crow made to the chromium three. So we can split those 2/2 reactions up right now. So we have this bike roommate CR 207 two minus was protons less electrons producing chromium three plus and water. Yeah, so you know right away that there should be two chromium tze on the right because we have the two chromium sze left. Now we need to check in, see our we need to check and see how many electrons are being transferred here. So we're going from chromium six on the left, two chromium, three on the ray. So that means for two chromium Adam's six electrons will need to be transferred, which means that to maintain the charge on both sides of the reaction will need to have 14 protons, which means that we're going to have seven waters remain for the oxidation reaction. We have this methanol ch 30 age going reacting with water oxidized to form H c o h or formic acid Less protons and electrons. So here way have the oxygen for the carbon, the balance, this reaction. We'd see that there are four electrons for our sorry six editions on the left. So we need six high positions on the right. We have to already. So that means we need for protons. And to neutralize this charge we need for electrons. So here we can add up these reactions. Since we have six and four elections respectively, we can multiply by two and by three. So we can have 12 electrons being transferred total and we can catch them. Cancel our so we get to see our 207 to minus. Then we have three methanol ch three a wage. Now, if we look at our protons, we have 28 protons on top 12 on the bottom. So when we tried to get 16 h plus, we have four chromium three three h, c o h. And if we subtract 14 waters from three waters, we get 11 age too. So our oxidizing agent in this case is thiss agent. This species has been reduced by crow, mate. And the reducing agent is the methanol. So now we can move on to a party where we have a reaction between a roommate Ion and dine Nature shin touch and to age four. Okay, so this prom it will be a reduced to bromide. We see the loss of the oxygen's there So we can do that. Electric, that half reaction first and where it's again. We're in acidic medium. So we're goingto react with each plus in electrons to produce br minus and water. So on the left side, our bro Ming has an oxidation number of positive. Five should be positive. Five on the right and on the left, it's negative one. So that means six electrons are added to this bro me. Which means that we have six protons and three waters. No, no, we can do our oxidation reaction, which is into age four. Yeah, reacting to produce end too good protons and electrons so we can see that this is being oxidized we have, and two with an oxidation number of negative too going to zero. And for two nitrogen atoms, that means we should have four elections here and we should have for protons. So, Tio, have these add up? We have six and four again. So you multiply times two and times three to have 12 total electrons sewing it to B R 03 minus plus three and two h. For now, the protons will cancel as well as the electrons. So we don't have pro tones. You have to be our minus plus, uh, three and two in six h 20 Now, are you oxidizing agent? Is thie species being reduced, which is this promise. So this is oxidizing agent, and then the nitrogen from the into each floor will be our reducing agent. So now we can move on our e where the reaction is between nice right and aluminum to produce ammonium and H for plus and the aluminum oxide and ion so we can see easily that aluminum is being oxidized, so we can work on the reduction reaction first with ah nitrite to ammonium. So we have an 02 minus and also we're in basic solution this time. So we're going to be reacting with O h instead of protons. And we're also reaction with the right Sean. Since it's the reduction half reaction and we produce ammonium. And because of the need to reduce this year, we're producing the, uh, oxygen and I am 02 to minors. So Okay, so, hear what we have is first to balance the reaction, we see that we're going to need four hydrogen on Be right. So we put in for drug sides, which means that there is now four plus 26 oxygen molecules. This is not oh too. It's just go to my home menace, too. So we should have six oxygen's on the left on the right. And to neutralize the reaction or to make the charges equal on both sides, we need to add six electrons that makes 67 our six ten 10 11 negative charges on left and on the right. We have 11 negative charges as well, So if we do the oxidation reaction, it's aluminum again, reacting with hydroxide to produce aluminum plus protons, plus these electrons. Now the aluminum here is going from an oxidation state of zero too positive three. So we have three electrons we have to neutralize. That means we're going to have to oxygen's on the right left, and we're going to need to have two protons to balance the additions. So here, too, make the electrons cancel out. We just need to multiply the bottom reaction Times two and then when we had them up, we get n o to minus plus two aluminum plus eight hydroxide producing an age for plus ammonium plus 2 +02 minus. These hydrogen sze here, which are produced here, will react with, uh, the water that we have the oxygen that we have here to produce water. So two of these actions are used to make two waters in the rest of the oxygen is left as the n ion. Yeah, so that's the reaction. And now the reducing agent will be the aluminum because it's being oxidized and the oxidizing agent will be no too, because it's being reduced. So in part f, we have Htoo too, uh, hi shin peroxide plus Cielo to quarry producing Cielo to minus and 02 in basic solution. So our reduction reaction pretty apparently will involve the Cielo to going to see low to minus. Since it's the fame, all of you, simply with an addition of a negative charge so we can see the electron being added. So we have seo to wass. Electrons will produce yellow to minus and that is the reaction They're finished. Since there's no additional Adams. Then we have Htoo to hydrogen peroxide reacting with drug science. Since we're in basic media and this will produce a molecular oxygen, water and electrons, we have two oxygen's going from plus 2 1 to 0. So we're ready. So we are. We have the oxygen going from negative 120 Sawyer taking away two electrons. We have to have visions on the left for actions on the left. So we need four on the right. So we have two waters and we need two hydroxide, so we need to multiply on the top. Reaction times choose so the electrons will cancel out. You get to see l. O, too, plus H 202 wass to drug side producing feel 02 minus times two oxygen and two waters are reducing. Agent is the H 202 because it's being oxidized and it's reducing Cielo to which is the oxidizing agent


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