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Half reactionEred/VGds+ (aq) + 3e- = Gd(s)-.279PbOz(s) + 4H-(aq) + SO}-(aq) + 2e = PbSO (s) + 2H,O()1.691...

Question

Half reactionEred/VGds+ (aq) + 3e- = Gd(s)-.279PbOz(s) + 4H-(aq) + SO}-(aq) + 2e = PbSO (s) + 2H,O()1.691

Half reaction Ered/V Gds+ (aq) + 3e- = Gd(s) -.279 PbOz(s) + 4H-(aq) + SO}-(aq) + 2e = PbSO (s) + 2H,O() 1.691



Answers

Follow the directions of Question 39 for the following half-reactions: 1. $\mathrm{I}_{2}(s)+2 e^{-} \longrightarrow 2 \mathrm{I}^{-}(a q)$ 2. $\mathrm{Co}^{3+}(a q)+e^{-} \longrightarrow \mathrm{Co}^{2+}(a q)$ 3. $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+14 \mathrm{H}^{+}(a q)+6 e^{-} \longrightarrow$ $$ 2 \mathrm{Cr}^{3+}(a q)+7 \mathrm{H}_{2} \mathrm{O} $$

Electoral chemistry in this podcast, Siris on what we're mostly looking at is how we're able to balance some unbalanced chemical equations and then following this identify the oxidation on reduction part. So in the first example we have, we have our oxidation part of the reaction that is province three p of 34 minus generates three p o for three miners at an electron. So we know that that is an oxidation reaction. The second half of this is M and oh, four minus. And then we need Thio ad for H plus Add three electrons gives us M N 02 to H 20 That is our reduction part. So now we can balance this equation and completed and other altogether you have three p 03 four minus at M N 04 minus at four h plus. That is a good rumors three p both for three miners at, um and to to h 20 Okay, so I'm moving on to the next example, So we'll write out the full equation this time that is balanced. Now we can identify oxidation reduction components where we have our oxidation. That is M G to H 20 generates M g O H two to H plus two electrons and are reduction component that makes up this balanced equation. At the top, we have O C L minus two h plus to the minus. You can see that this thes two oxidation reduction reactions do couple quite nicely on we got C l minus at H 20 now for the next one. What we have is H two c o to H 204 a g NH 33 plus a queer state that is an equilibrium with H c o three minus at five. H plus add four h g at 12 and H three so we can identify oxidation reduction components. We have our oxidation reaction occurring with the H two c o on our reduction, but this material

In each part of this problem were given an unbalanced Redox reaction. And for each one of the given reactions, we need to balance it using the half reaction method. So starting with the reaction given in part A, you first need to split up the unbalanced total reaction into 2/2 reactions. The way that we do that it was, is that we split up the oxidation and reduction reactions so chromium is oxidized in nitrogen is reduced. So that's how we know that these are the two half reactions. And now, in order to balance each one of the half reactions and then add them together, we first need to balance all of the atoms that are not oxygen or hydrogen. And then we need to balance the oxygen's with water and the hydrogen is with H plus. Since we're in acidic solution, then after that we need to balance the charges and then we can add the 2/2 reactions together to get the total balanced reaction. So starting with the oxidation half reaction above, we see that the chromium zehr balanced and we have no oxygen's or hydrogen, so we just need to balance the charges the reactive side has a charge of zero. The product side has a charge of three plus, so we need to add three electrons to get a charge of zero on each side. And now for the other half reaction, we see that we have one nitrogen on each side. So those air balanced and we need to balance the oxygen's with water. So if we add to liquid water on the product side, we can see that we now have a total of two oxygen, a total of three oxygen atoms on each side of this reaction. In essence, we're in acidic solution. We need to balance out the hydrogen is that we introduced from water with H plus. So we have four. Hydrogen is from water, so we need for equals H plus ions, and now we need to balance out the charges. On the left side, we have a total charge of plus four plus negative one from an +03 so total of plus three. And on the product side, we have a total charge of zero. So we need to add three electrons to get the charge on the left side from plus 3 to 0. And now that we have balanced both of the half reactions, we can add them together to get the total balanced Redox reaction we can cancel out. There's three electrons on either side of the reaction arrow. We can get our final balanced reaction to be for each plus yanquis. Plus I know three minus equally ists plus chromium solid goes to chromium three plus Aquarius, plus a no guess plus two liquid water. Now we can see that all of the Adams and the charges are not are now balanced. So this is the final answer for the balanced Redox reaction. Now we use that same method to get work through Parts B and C for Part B. We split up the 2/2 reactions again, and we start by bouncing the top one, and we see that we have one carbon on each side, so that is balanced. And now we need to bounce out. The oxygen's with water and we can add liquid water to the left side, and that will give us a total of two oxygen atoms on each side. So now the oxygen's air balanced, and now we have a total of two plus three plus one. So six hydrogen ins on the left side. So we need six h plus against. Since we're in acidic solution, we need to balance the hydrogen is with H plus and now we need to balance out the charges and we see that on the left side there is a charge of zero in on the right side gave a charge of six plus. So we need to add six electrons to the product side to get aid charge of zero on each side. And now for the other half reaction, we just have CE and so the Adams air balanced. And now, in order to balance the charges, we need to add an electron to the four plus charge to get a charge of plus three on each side. And now we add these two reactions together. But notice that we need to cancel out the electrons and we have one electron on the reactant side from the second reaction, and we have six electrons on the product side from the first reaction. So we need to multiply the second reaction by six, and then we can add these two together. So if we multiply that second reaction by six. We now have six elect electrons that we can cancel out on each side, and we can begin to write out our overall balanced reaction each to oh Liquid was ch three. Ohh, Equus plus And remember, Since we multiply that second reaction by a factor of six, we now have six C four plus acquis and for the products we have again from multiplying the second reaction by six, we have six C three plus Equus Plus Co. Two Equus plus six H plus Equis. So that's the final balanced Redox reaction for the given reaction in Part B, and now we do the same thing for part C. We have our reaction given in the problem, and we split it up into the 2/2 reactions. And so, starting with the the first reaction, we can see that the sulfur zehr balanced and we need to balance the oxygen's with water. If we add water to the react inside, we have a total of four oxygen's now on on each side of the reaction, and we have to hydrogen on the left side that we need a balance out with to H plus ions on the right side and now we can balance out the charges. We have a total charge of minus two on the left side and a total charge of minus of of zero on the product side. And so we need to add two electrons to the product side to get a charge of minus two on each side. And now for the other half reaction. See that man Younis is balanced and we need to bounce out those four oxygen's with four waters and now we have 888 Hydrogen is from the four. Water is that we need to bounce out with eight h plus ions and now we bounce out the charges we have on the left side eight plus and minus one. So a total charge of plus seven on the reactant side, On the product side, we have a total charge of plus two should get from plus seven to plus two. We need to add five electrons. And now when we add these two together, we need to cancel out the electrons so we can multiply the first reaction by five and the second reaction by two so that we have 10 electrons cancelling out on both sides. And so now we can go through and write out the final reaction. So we want to play that first reaction by five. So we have five s 03 tu minus acquis. And for the H plus ions, we see that we have a total of eight on the reactant side, times two so 16 and to times five So 10 on the product side. And so all of the products will be used up and we'll be left with a net of six h plus on the react inside and from the second re actually multiplied by to sue two mn 04 minus Equus. And then on the product side, we have to and then two plus after multiplying that second half reaction by two. And for the waters, we see that we have four waters times two so total of eight on the product side and one water times five. So a total of five waters on the reactant side. So we have eight waters on the product side and five on the reactive side. So all of the reactant waters are gone and we're left with three waters for a product. So plus three each to a liquid and then was five s 04 tu minus a quiz. And that's from that first reaction, multiplying it by five. So that is a balanced Redox reaction. Were they given reaction and part C.

Let's balance the following oxidation reduction reactions that occur in acidic solution. Using the half reaction method to use the half reaction method, we have to separate into two half reactions a reduction, half reaction and not reaction and used the major ohh method for balancing For a we're gonna split 7 to 2 half reactions. Balance the major and we have minus three on the left, minus one on the right. Add two electrons to belts charge and we'll have C O minus two C minus. The major chlorine is balanced. Add H 20 to balance the oxygen and to H plus to balance the electrons and a two electrons to balance charge. Uh, two electrons will cancel and we will get three. I minus. Add Cielo minus two h plus to I three minus at C l minus at H 20 And there is our balanced half reaction in acidic solution for be one half reaction with a S 203 two h three e s 04 Arsenic is there a major Adam will put a to hear develops that That gives us two arsenic on each side at eight. Oxygen on the right hand side three oxygen on the left. So I need five h 20 for eight. Oxygen on each side. 10 hydrogen on the left, six hydrogen on the right. So I need four h plus to balance that to balance the charge at four electrons, second half reaction and all three minus two an O major nitrogen is balanced. Three oxygen on the left, one oxygen on the right. So two H 20 to balance. Have the oxygen for hydrogen on the right. So I need four hydrogen on the left and charge here plus three. So I need three electrons to balance the charge. Now we need to multiply by a common factor of the electrons on. I'm gonna multiply the oxidation. Half reaction common multiple is 12. So times three times four Let's rewrite thes here multiplying by three. I get 15 h 20 three a s, 203 26 h three a s, four. 12 h plus 12 electrons and multiplying by four would give me four times three is 12 electrons, 16 h plus four and all three minus four and no h 20 cancel out the electrons 12 on each side. I can simplify the hydrogen. Za sees me with four here, simplify the waters issues us seven here. And my final balanced equation is three s 203 at 403 minus add four each plus at seven. H 20 26 h three e s 04 and for And Oh, and there is my balanced Redox reaction. Percy. First half reaction would be B R minus to be our to balance the major balance. The charge second half reaction is Emma No for minus two mn two plus Mangin eases balance on both sides. Four oxygen's on the left. So four h 20 on the right. Eight hydrogen is on the left. So eight h plus on the right. So left there. Charge is seven plus two. So five electrons on the left would give me a charge of plus two plus two rates now will multiply electrons common multiple is 10, sometimes five and times to let's rewrite thes. Here I'm gonna have 10 b r minus 25 B R. Two at 10 electrons and doubling. This one is 10 electrons 16 age plus to M N 04 minus two mn two plus and eight h 20 Cancel out the electrons and my balance equations 10 br minus to elemental four minus at 16 h plus going to five b R two two mn two plus and eight h 20 on theirs are balanced equation for C for D first half reaction of ch three O h two C h 20 carbon is balanced on each side. I've got oxygen one on each side, hydrogen four on the left to on the right. So I need to h plus balance is charged two electrons. The next half reaction is dead crow mates here are 207 to minus two chromium three. Plus I needed to hear it amounts to chromium seven h 20 on the right, 14 age plus on the left and I have plus six on the rights plus 12. So I need six electrons to balance the charge here. I need to multiply. Common multiple is six groups, so we're gonna multiply the first half reaction by 33 ch 30 h 23 c h 20 and six h plus and six electrons. The second half reaction is multiplied by one. So as is electrons will cancel. Uh H plus will simplify to leave me with eight there, and our balance equation is three c h +30 H at crow Mates at eight h plus two, three ch, 20 two chromium three plus and seven h two hole and there is my balanced redox reaction for D.

So here we are, given this chemical reaction, and we will need to complete it and balance it using the half reactions. So if we look up our half reactions for each of our reactant for the chromium part are half reaction is CR 207 tu minus plus 14 h plus plus six electrons forms to see our three plus plus seven h +20 and those do all have their forms with them. But just to save space. I've left those off and then the second part is for iron and that half reaction is iron two plus forms iron three plus bless on electron. And then one thing to note is that there are six electrons in our blue equation, but there's only one in our red equation. So because of that, we're going to multiply everything in that red equation Bus six. So we're going to give a coefficient of six to all parts of that, Um so we can then go in and we know that those electrons are on opposite sides, so therefore they will cancel out, and then we can write our full equation. So the left side is CR 27 less that 14 hydrogen from the top blue equation. And then our other reactant is that six every two plus from our red equation, and then those will form are three products which were to see our three plus plus seven h 20 plus six f e three plus. So that whole thing in green is now our new balanced.


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