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Balance the following in basic solution:(a) $\mathrm{SO}_{3}^{2-}(a q)+\mathrm{Cu}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{SO}_{4}^{2-}(a q)+\mathrm{Cu}(\mathrm{OH})(s)$(b) $\mathrm{O}_{2}(g)+\mathrm{Mn}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{MnO}_{2}(s)$(c) $\mathrm{NO}_{3}-(a q)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{NO}(g)$(d) Alls) $+\mathrm{CrO}_{4}^{2-}(a q) \longrightarrow \mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{Cr}(\mathrm{OH})_{4}^{-}(a q)$

a. $S O_{3}^{2-}(a q)+2 C u(O H)_{2}(s) \rightarrow S O_{4}^{2-}(a q)+2 C u(O H)(s)+H 2 O(l)$b. $O_{2}(g)+4 M n(O H)_{2}(s) \rightarrow 4 M n O_{2}(s)+4 H_{2} O(l)$c. $3 H_{2}(g)+2 N O_{\overline{f}}(a q) \rightarrow 2 N O(g)+2 H_{2} O(l)+2 O H^{-}(a q)$d. $3 A l(s)+C r O_{4}^{2-}(a q)+4 H_{2} O(l) \rightarrow A l(O H)_{3}(s)+C r(O H)_{4}^{-}(a q)+O H^{-}(a q)$

Chemistry 102

Chapter 17

Electrochemistry

Carleton College

Drexel University

University of Toronto

Lectures

00:44

In chemistry, electrochemi…

03:11

In chemistry, a redox reac…

19:16

Balance the following oxid…

02:03

Balance the following equa…

06:04

01:55

03:30

22:11

Balance these equations f…

21:16

Balance the following redo…

22:32

Balance these equations fo…

33:38

11:05

Balance the equations for …

05:26

10:41

22:17

31:26

01:51

Complete and balance the f…

07:01

03:12

29:53

02:32

11:21

Okay. So never going to balance Redox reactions in basic solution, which is slightly different. An acidic solution. Look at the steps here. We can see that Step eight is not on the checklist for Russian ex solutions. So and basically, all that we're doing is for every age plus that we haven't set for we neutralize it or cancel it up with the hydroxide. I'm because it's basically anyone you don't really see. H plus in the basic solution blooding around it automatically reacts in solution to form water. We'll just see us as we go through. I'm not gonna talk too much more about it right now, so we can split up or half reactions are split up Our unbalance. Redox reaction is to half reactions. The green is oxidation and the blue is reduction. And check the oxidation. Say, if you want. We're not gonna do that right now, though. And now we'll start balancing. So we have one s one s. So that's good. You're three O's and 40 so we need one more. Oh, on the left side in the form water. Now, I just added two hydrogen za donors on the right side so we'll have those in the form of each plus. Now we have a start of zero on this side and negative two on the side. So you add two electrons to balance charge and move on to the reduction reaction. The coppers are equal. Already speaking from the oxygen, we have two on the side and one on the side. Silly ad and H 20 And then now we have three hydrogen sze on this side and two on this side. So we add one each plus. And this is the only charge. So you should cancel that charge up. Not withdrawn now. 00 This place it looks like tease, but you get the picture. We have one electron, this equation and two electrons on this equation so wouldn't multiply this whole thing by two. Add the coefficients. Yikes! Two. Uh, it's too close to the edge two. And now we'll add these together. We're going to see that our two age plus and two electrons cancel out. And we also have one water on this side and to water on this side. The road disappeared. So two minus one is net of one water on the side now, we actually didn't have in the remaining H plus in this stage to cancel out. So we are all good here for this reaction is gonna check to make sure all the atoms are equal on both sides of the charges. So you could do that on your own. Okay, so this one, because there's only one product which looks a little bit different than what we're used to. We're knocking the split it into half reactions. You can check, though, and there is changing and changing oxidation. States, we have water zero here, and it's negative to here. Then Mangin eases. Plus two here and plus four here. So Megan is being oxidized and auctions being reduced. Um, but I think it's actually gonna be easier for us just to balances on our own, not spitting up in the half reactions. You can definitely try if you want, but I think it just makes it way more complicated. So, seeing one point, I that's my cue. Maybe we should just balance it as is and skip that half reaction step. So we have equal Megan these right now we have that lets balance oxygen. We have 1234 on this side and two on this side. So let's add to age to O. Now it's Fallon's are Hydra Jin's. We have four on this side and two on this side so we could add to H plus. But actually, let's just try and see if we can balance it intrinsically. I know little Hardwick's. I tried. It doesn't work out with, like, some of your charges. And then if you do tried that but to each plus, you can try that yourself. But we're gonna try just doubling this here. So now we have two times 24 how you're Jen's for hydrogen. But the issue is that we have to Ming and it's not. We just come over here, the two there, much double check. So now we have to making these two manganese. We know our hydrants are going. Let's take our oxygen's one too. 3456 six on the side, two for six on the side. So we're all balanced. We didn't have to add in the H plus are charges. Look skeptically on having here, and we don't need to do this Clunes. But in the half reactions or any of that. So this is actually final reaction. All right of it again, we only have one product, so you could try to balance this, just like we didn't know. Ask problem. But I'm actually gonna split them up this time. Really? Trial and error, I think is is that's how it is. Um, Okay, so there's something a little bit tricky about this. Let's go over the oxidation states. So hydrogen is zero here? Oh, no. Zero. And then we have oxygen is negative. Two times three negative. Six. They want a total charge of negative one. So measure does, plus five because negative six plus five equals negative one. And then over here, oxygen cylinder be negative too. And then nitrogen is going to be plus two to make to charge of this molecule. Zero. We don't have anything for Hadron. So there is definitely missing some kind of reacted. It seems like over here that includes hydrogen. This could be water. Um, potentially so we'll see what happens. But we can think about the fact that in solution, hydrogen goes from age too. Stage plus acquiesce environment, even in the basic environment. So I know I told you we gotta neutralize later. That's fine. So he balanced this and then we have extra charge, or so we add to it. That's the oxidation reaction that goes from 0 to 1. And then for the nitrate We are going Thio, We have the Nigerians balance. We need to add some water. It looks like cane. Our oxygen balance. We have 300 side. What? We have four. Hydrogen is over here. I need to add on the side. And then now we have four minus one plus three over here. So we need the ring. Uh, three legged runs here. Okay? And we need to multiply these. So they have a common the same number of electrodes, electrons so will multiply this one by three years. So three and then these will both become six. Okay. And then this will be multiplied by two. That we will get six electrons. Um, okay, Four times two is eight and three times two is six. No, some mess. Okay, a six two thio. And okay, now we will add these together and cancel the eight minus six. Get to H plus on the side. The electrons cancel out everything else comes down. Okay, Now we're gonna see this. Step eight action happen that we have to age plus in this equation. So for each age plus at it a wage minus, I'm both sides out This warm in black H plus plus o H minus makes each to O. We have to. It is not one. Let me. Right. Well, it's too. Oh, okay. That I I don't know. That's h 20 this cancels. And then so does the H 20 Because there's some more on this side. So the foot ah, the four becomes a two. Okay, So our final our final equation would be this. Plus two each two and +20 H. Myers. And we can double check to make sure everything adds up. We have six hydrogen sze. We have four hydrogen plus two hydrogen. That's six. Perfect to end to end three O's. 123 four, five It's Oh, sorry. Six owes. And then because three times, too. And then four plus two plus two, six. Great. Then we have negative too. And negative too. So it all adds up okay. And that we have one more example. So we'll split this into half reactions. This is oxidation. This production ales are equal. We need Thio Ad three h 20 to balance oxygen. And now we just added six hydrogen Sze, We have three on this side so we need three age plus more. And then now we have charged on the side though I need to cancel. Okay. And then here chromium are equal. Oxygen's are also equal Hydrants are not what we have four on the side and on the side, four inch less. And then now we have four minus two is plus two and we have a negative one. Swimming toe, three electrons on the side. Okay, Now we have three and three we can add combined these into one equation. So form on this three We have one each plus here everything else just comes straight down. Now we have to do this. Step Step eight again. We have one h plaster we at 10 h minus both sides. He's cancel and create a TSH to o can't write down there. But basically this three comes before because he made one more each other. There's nothing else to cancel over here. So now all we have to do is check our work. It looks good. All right. That's how you balance Redox reaction

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