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When the following equation is balanced properly under acidicconditions, what are the coefficients of the species shown?Co + ClO3- Co2+ + Cl2 Water appears in the b...

Question

When the following equation is balanced properly under acidicconditions, what are the coefficients of the species shown?Co + ClO3- Co2+ + Cl2 Water appears in the balanced equation as a ________ fill in theblank 5 (reactant, product, neither) with a coefficient of__________. (Enter 0 for neither.)How many electrons are transferred in this reaction?

When the following equation is balanced properly under acidic conditions, what are the coefficients of the species shown? Co + ClO3- Co2+ + Cl2 Water appears in the balanced equation as a ________ fill in the blank 5 (reactant, product, neither) with a coefficient of __________. (Enter 0 for neither.) How many electrons are transferred in this reaction?



Answers

The following oxidation-reduction reaction occurs in acidic solution. When the equation is balanced using the smallest set of whole-number stoichiometric coefficients possible, what is the stoichiometric coefficient for water? (Hint: In addition to the species shown in the original equation, $\mathrm{H}^{+}(a q)$ and $\mathrm{H}_{2} \mathrm{O}(l)$ can also appear in the balanced eanation ) $$\mathrm{MnO}_{4}^{-}(a q)+\mathrm{CH}_{3} \mathrm{OH}(a q) \longrightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{HCO}_{2} \mathrm{H}(a q)$$ a. 13 b. 19 c. 11 d. 7 e. 4

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.

For this question, we are balancing three oxidation reduction reactions or redox reactions in acidic solution. The first reaction can be separated into these two half reactions, die chrome eight goes to chromium three plus and iron two plus goes to iron three plus. So as you recall, the first step is writing the half reactions. The second step is to balance everything but oxygen and hydrogen, the iron atoms are balanced. The chromium are not. So we need to put a two in front of the chromium here to have the chromium is balanced. Now that the chromium czar balanced, we can balance the oxygen's with water. We have seven oxygen's. So we're going to need seven waters on the right hand side. Now we just introduced 14 hydrogen. So we'll put 14 hydrogen on the left hand side as H plus because it's an acidic solution. Now we need to balance charges. We have a total of six plus on the right hand side We have 14 -2, that's 12. So we need six electrons on the left hand side, so we have six plus on both sides. For the second half reaction, everything is balanced, accept the charges. So we need a one electron on the right hand side so that the charge on both sides is plus two. Now we want to some these half reactions together, making sure that the electrons council For the electrons to cancel this second half reaction needs to be multiplied by six, then the electrons will cancel. And we just add everything up Remembering to multiply this six through to the two irons. We can then verify that it's balanced. We have 14 hydrogen. We have seven times to 14. We have to chromium seems to chromium seven oxygen, seven times 17 oxygen's six irons. Six irons. What about the charges? Three Times six is 18 Plus six more is 24. On this side we have 12 Plus. 14 is 26 -2 is 24. So it's balanced. The next one is vo two plus, goes to V three plus and I minus goes to I two. So first we'll balance the atoms. The vanadium czar balanced. The iodine were not. So we needed to put a two in front of the iodide, then we'll balance the oxygen's with water. So we need one water on the right hand side that introduced to hydrogen. So because it's acidic solution will add to H plus, is to the left hand side. Then we need to balance charges. We have three plus and we have to and two. So that's four plus. So to get three plus, we need to add one electron to the left hand side. Down here, we have two minus. So we need to add two electrons to the right hand side to get the charges balanced. Then we need to multiply the top reaction by two, so the electrons will cancel. And then we just sum everything up. Make sure you don't forget to distribute that to through when you're summing everything up. So we end up with four hydrogen ions to vo 22 pluses and so forth. Then we can verify that it is balanced. To verify that it's balanced. We've got four hydrogen four Hydrogen is to vanadium to vanadium, two, oxygen's to oxygen's two I lines to eye lines. Then what about charges? We've got two times three plus that six plus. No charge, no charge. And we've got to minus And four plus. So that's two plus plus. four. More that six plus on both sides. And for the last one We have no three Goes to n. 0 2. And we've got cobalt goes to cobalt two plus we back up here, the nitrogen czar balanced but we need one more oxygen on the right hand side. So that means we had one water that introduced to hydrogen. So now we need to add to hydrogen ions to the left hand side. Then we balance charges. We have no charge on the right hand side and we have plus one on the left hand side. So to make it no charge, we add one electron. The next half reaction, everything is balanced, accept charges. So we need to add two electrons to the Right hand side to make both sides equal to zero. Then to make sure the electrons cancel, we need to most play the first half reaction by two and the electrons council and then we sum everything up, distributing that to through the top equation. To verify that it's balanced, we have four hydrogen for hydrogen for hydrogen to nitrogen to nitrogen. Six oxygen's 1, 2, 3, 4, 5 or 12342 times two is 456 one Cobalt 1, Cobalt. What about charges? We have a total of two plus, no charge, no charge. So two plus on the right hand side. Then we've got four plus and two minus Gives us two plus on the left hand side also.

So this problem wants us to balanced several redox reactions occurring an acidic Aquarius solutions. And so we begin with We have I'm minus aquarius plus an O two minus a quiz forms I too solid plus and oh, gas. And so the first thing we can do is stick a two in front of this iodine so that we have to on both sides. And then we want to split into half reactions. So half reactions here, too. I minus Aquarius forms I too solid and an O two minus Aquarius forms and oh, gas. And so now we have to add water molecules to balance out oxygen and then h plus to balance out the hydrogen tze from those water molecules. And since there's no oxygen in the iodine half reaction, we can ignore that for now. But with the, uh, nitrogen nitric oxide half reaction, we're going to need to add one water molecule over here, and we'LL label that age to a liquid. And then when we add H Plus to balance that out, we're going to need add to age plus and that's Aquarius over here to balance out the two hydrogen sze fromthe one water molecule. So now we want to do is add electrons to these to balance out the charges on each side of the equation of each half reaction. So if we start with our first, we have two eye minus Aquarius forms. I too solid. And, sir, we're going to want to add two electrons to this side so that both sides of the equation are now minus two. And with their second one, we have to age plus Aquarius plus an O two minus aquarius forms and oh, gas plus H two o liquid. And our right side of the equation is neutral. However, are left side is a plus one. And so we'LL just need to add a singular electron to the left side of this equation. And I've done that over here kind of scrunched in there. So now we have two sides of our second half reaction are both neutral. Now we want to scale these so that we have the same amount of electrons for each half reactions that we can cancel them so the top one has to and the bottom one has one. So we need to multiply this entire bottom equation by two so that we have two electrons and I'll show that on the next page here. So we're still on letter, eh? But when we do that, we end up with our same to eye minus Aquarius forms I too solid, plus two electrons. And then our second half reaction becomes too e minus plus two enoh to minus aquarius plus for age plus Aquarius forms to and no gas plus two age to a liquid. And so now we have two electrons in each half reaction. We can cancel those out and combine them. T get our final answer, which will be to my minus aquarius plus two enoh to minus a quiz plus for age plus agree. A CE forms I too solid plus two an o gas plus two h to a liquid and we can double check that. We've done this right as the right side of our equation is neutral. The left side we have two minus charges from the iodine two minus charges from the n O two and four h plus. So that is also neutral. And we can also check the number off Adam's on each side. We have four, um, Oxygen's on the left. Four oxygen's on the right to iodine is on both sides to nitrogen is on both sides and four hydrogen is on both sides. So we know that this is our final answer for letter, eh? So now moving on to letter B. We have the half reaction off c l o four minus Aquarius plus C l minus Aquarius Form's C l O three minus Aquarius plus C L to gas. And so again, the first thing we can do is put a too right here in front of the C L to balance it with C L, too, and split this into two half reactions. So we have C l o for minus Adria's form's Cielo three minus a quiz and to see l minus Aquarius form C L to gas. And so the first thing we want to do is add water to balance out oxygen's, which only affects our first half reaction. And so we just need to add one water molecule over here to balance that out with the extra oxygen Adam on our left side. And now we need to add a tch plus over here to balance out the hydrogen tze from our waters so and that is a curious And so now we have balanced our oxygen's and hydrogen Tze And again that didn't affect our second half reaction. Now we need to add electrons to balance the charges between these reactions that is between each half reaction the chargers on both sides need to match. So if we take our first one, we can see that it has a plus one charge on the left. That's too from the age plus minus one. So plus one and on the right it has a minus one. So we want to add two electrons to the left in order to get that minus one on both sides and with our second half reaction, we just want to add two electrons to the right to balance it out with the minus two that is on the left. And now normally this is the part of the process where you scale these reactions. However, since we already have the same amount of electrons weaken, Just cancel those and write our final reaction, which is C l o four minus Aquarius plus two h plus agree a CE plus two c l minus Aquarius form's C L O three minus Aquarius plus H to a liquid plus C L to gas. And so this will be our final balanced equation, and we can double check to make sure that the charges of the same on the left we have a minus one charge and on the right, we also minus one charge, and the atoms balance out as well. So this is our final answer for letter B. So now, moving into letter C Our equation we need to balance is n o three minus agree a CE plus s end two plus I agree. A CE form's s and four plus agree a CE plus N o gas. And so, since our number of atoms, except for oxygen are the same, our first step is to split thes into two half reactions. So we have n O three minus aquarius forms and oh, gas and s end to plus Aquarius forms as and four plus Aquarius. And so our first step is to add water to balance out our oxygen. So we'll have to add to h two a liquid over here so that we now have three oxygen's on the right side and three oxygen on the left side. We also have to add R H plus over on the left side so that we balance out with our hydrogen atoms from our water. And so now we can add electrons so that both each individual half reaction is balanced with itself and sew the bottom one is a lot easier. We just had two electrons over here. Now we have two plus on both sides for the top one on the left. We currently have a plus three and on the right, we have a neutral. So we just want to add three electrons over here. Now we need to scale these so that the electrons are the same for each half reaction. So to do that, we're multiplied this one by two and this one by three. And then we'LL have six electrons on each. And so when we do that, we end up with sixty minus plus eight age plus a Prius plus two and oh three minus aquarius forms to and no gas plus four age to a liquid. And our second half reaction becomes three s and two plus a prius forms three s and four plus Aquarius plus six e minus now that are electrons of the same week and cancel and then combine and all combined on the next page for clarity's sake. So now our final balanced equation is to Anil three minus Aquarius plus eight age plus agree a CE plus three s and two plus a Prius forms two and oh gas plus for age to a liquid plus three s and four plus Aquarius. And so this is our final balanced equation for letter C.


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