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11 ofDeterqine what occurs , each olectrode Ihe electrolysis molten sodlum chlordo NaCl a5 shown part A: Match the words the Ieft column with the appropriate blank...

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11 ofDeterqine what occurs , each olectrode Ihe electrolysis molten sodlum chlordo NaCl a5 shown part A: Match the words the Ieft column with the appropriate blanks the sentences on the naht Make certaln cach sentence complete before submitting your anewerView Available Hintis)Reset HelpSodlum metal /consumed Ine anodeChlorine gas produced a1 the AnneChloride Ions areconsumed ho cainodeSodlum Ionsproduced a" tho Cathodehtc i0 scarchDE

11 of Deterqine what occurs , each olectrode Ihe electrolysis molten sodlum chlordo NaCl a5 shown part A: Match the words the Ieft column with the appropriate blanks the sentences on the naht Make certaln cach sentence complete before submitting your anewer View Available Hintis) Reset Help Sodlum metal / consumed Ine anode Chlorine gas produced a1 the Anne Chloride Ions are consumed ho cainode Sodlum Ions produced a" tho Cathode htc i0 scarch DE



Answers

Suggest what products are discharged at the anode and cathode during the electrolysis of (a) molten $\mathrm{KBr} ;(\mathrm{b})$ fused $\mathrm{CaCl}_{2} ;(\mathrm{c})$ dilute aqueous $\mathrm{NaCl}$ (d) concentrated aqueous $\mathrm{NaCl}$ (brine); (e) aqueous CuSO_using Cu electrodes; (f) dilute $\mathrm{H}_{2} \mathrm{SO}_{4}$ using Pt electrodes.

They're here. We have to determine the products off the electrolysis off several compounds. The first one is molten petition, bro might, when it's being electoral ized, bra Might is oxidized on the an odd forming Brahman, which is likely kazoos and on the cathode. But I assume irons are reused to petition metal, which is likely liquid. Next one is a fused calcium chloride, so on the cathode costume is reduced toe custom metal and on the adult chloride is oxidized toe chlorine. Next question is about, um, dilute sodium chloride. So here it's an adequate solution of sodium chloride. That's why on the cathode, hydrogen irons will be reduced to hydrogen and on the energy. We have two competing reactions, which are, um, oxidation off glory chloride to chlorine or oxidation off water to oxygen. So next question is about, um uh, sodium chloride Breen. So here again, sodium cannot be cannot be reduced. That's why on the cathode hydrogen will reform. So H plus is produced toe hydrogen gays. But on the n it most likely ah, chlorine is being oxidized to Colorado is being oxidized to chlorine gays. And the last one is Ah, sorry. Not the last one to more. Next one is Ah, copper sulfate. When ah, to copper electrolysis I used Let's visualized the system. So we have ah, best with cup liquid copper sulfate and to couple elections here, copper toe plus and well yeah, Here, couple two plus will be oxidized tonic, method forming copper metal. But on the energy Ah, copper metal will be reduced. Copper metal will be oxidized, so copper will become copper. Two plus was to elections. So basically, it means that here copper will move from cassette from there, not to the cathode. So let's say this is couple Ryan's will move from They're not to the cathode and new copper will be deposited on the cathode. So in the last question yes, about the electoral A. Jeez off! Dilute off the value to sue for cases on platinum electrodes on the cathode ah hydrogen IRS will be reused to form hydrogen gays. And on the Senate, water will be oxidized to form oxygen. Yep. Now we've completed this question. Thank you very much for your attention by

In his electrolytic cells. We found that at a mood, bromine is easily excised. Then clue Ryan, because promoted has less electoral negativity buoyant at a food. How soon? Postive, too. Here. The higher ability to reduce, then sodium boasted because cats impulsive to Maybe we said it has high organization energy. Thank you.

In each part of this problem. We're told that we have one more of a solution of the given compound at standard conditions. And we want to We want to know which reactions will occur in the cath food and the an ode. So since these air all Equis solutions, if we start with the compound given in part A which is an I b r two, we know that that means that we have an I two plus acquiesce ions in solution the ar minus acquis ions in solution. And because this is acquiesced, we also have water. And we know that in the cathode reduction occurs and in the an ode oxidation occurs. So we need to use the information of those initial ions that we have in order to determine once we electoral eyes this one Moeller solution which of these half reactions will occur in the cathode in which one will occur in the an ode. So we know that we have an I two plus and so, based on the standard reduction potentials table with the provided half reactions, we know that and I two plus, if we start with that, then we have to add two electrons to reduce it. You saw the nickel so and I two plus can be reduced. And for br minus. If we look at the half reactions table, we see that we produce BR minus from the reduction of BR to So therefore we have to oxidize br minus to B R two and we also know that water h 20 can either be reduced or oxidized. So we have two reactions in each case that can either be reduced or oxidized, and so we need to know which one will occur. So again, the 2/2 reactions that air above this red line are both reduction reactions in the to be low. It are both oxidation reactions. And so when we look at the standard reduction potentials, you know that the top to Europe recent reductions and therefore the reaction with the higher value and in this case, the less negative value for the standard reduction potential will have a greater potential to be reduced and therefore will undergo reduction as opposed to the other one. And we can see that negative 0.223 is greater than negative 0.83 and so therefore we can conclude that the first reduction reaction will occur since it has a higher potential to be reduced than the reduction of water. And since that is the reduction reaction, it will take police in the cathode. Now the bottom two reactions are the two possible oxidation is that can occur. We have flipped the signs of the standard reduction potentials after we changed both of the reduction half reactions into oxidation reactions. And so that means that both of these potentials correspond to oxidation potentials. If we write them out as they're given in the table, then those come out to a value of positive 1.23 and positive 1.9 And those would represent the reduction of 02 and b R two in each case. And we know that if those were both positive as they're given in the table than 1.23 is greater than positive 1.9 And so that top reaction. The reduction of 02 to form water would would occur over the reduction of br two to BR minus. Because it has a greater standard reduction potential. It has a greater potential to be reduced. But that means that the other reaction which is the reduction of BR to to be ar minus since it has a lower positive value for the standard reduction potential, would have a smaller potential to be reduced or a greater potential to be oxidized. So when we flip over those reduction half reactions to oxidation ins, we see that we can compare them now, since they are both negative now, when represent the oxidation reactions to see which one is larger, and that will correspond to which one has a greater oxidation potential and therefore will be oxidized. So we see that negative 1.9 is greater than negative 1.23 So either based on the fact that the oxidation of BR minus has a greater potential to be oxidized than water does, or based on the fact that water or that oxygen has a greater potential to be reduced into water than BR two has a potential to be reduced and to be ar minus ions, we can conclude that this is the oxidation reaction that will occur, and we know the oxidation takes place in the Anna. So we use that same logic to work through the other parts of this problem, Part B. The acquiesced compound we have is a left three. So that means we have a l three plus which will be reduced. And we have F minus ions, which will be oxidized and again. Water can either be reduced or oxidized, so examining those 1st 2/2 reactions, which are the two possible reductions that can occur. We see that the reduction of water has a greater state of reduction potential in the reduction of a L three plus. So that means it has a higher potential to be oxidized. Then the reduction of a 03 plus That s how your potential to be reduced and so it will be reduced. So this is the reaction that will take place in the cathode and undergo reduction. Now, the bottom 2/2 reactions both represent oxidation. So if we compare the two standard reduction potentials which are both now corresponding to oxidation potentials, we see that the oxidation of water has a higher potential than the oxidation of F minus. And so therefore the oxidation of water will occur in oxidation. Takes police in the an ode. Finally, for part C, the compound is M and I, too. So we have mn two plus and I minus ions and water hers to reactions represent the reduction half reactions. If we compare their two standard reduction potentials, then we see that the reduction of water has a higher potential to be reduced than the reduction of mn two plus. And so therefore water will be reduced in reduction occurs in the cathode, so that reaction takes place in the cath food. In the bottom, 2/2 reactions represent the oxidation that that will occur. And so, in order to determine which one we'll proceed, we need to compare their potentials to be oxidized. We see that negative 0.54 is greater than negative 1.23 So therefore I minus has a greater potential to be oxidized and h 20 And so it will be oxidized and oxidation occurs in the an ode

In each part of this problem were given the molten form of a compound and we want to determine what reaction will occur in the cathode and the an ode when we electoral eyes this solution import A We have an I b r two. So that means that we have an eye to plus and br minus ions and acquia solution and so we can look up on the standard reduction potentials people list Of those half reactions we're in, I two plus is being reduced and where br minus is being reduced. But we know that BR minus can only be oxidized to produce BR too. So, really, the half reaction for the reduction is br to being reduced to BR minus. But since we're starting with br minus as part of that a quiz compound in part A, we have to reverse that half reaction from the table soon it so that it now becomes an oxidation reaction and we conceive from the standard reduction potentials after we flip the sign of the standard reduction potential for B r two. Then we see that it corresponds to a negative value of 1.9 which is less than the standard reduction potential for the reduction of an I two plus insist, since the standard reduction potential of an I two plus is greater than that for the oxidation of BR minus. Then we can see that the first half reaction has a greater potential to be reduced. Therefore, the second reaction has a smaller potential to be reducing the greater potential to be oxidized. So that's how we know that the first reaction is a reduction that occurs in the second half. Reaction is the oxidation that occurs again. We know that reduction occurs in the cathode in that oxidation occurs in the an ode. So that means that this top half reaction corresponding to the reduction of any two plus ions will occur in the cath food, while the oxidation of BR minus ions will occur in the antidote. The compound that we're electoral izing in part B is a L F three. We have a all three plus ions and F minus ions. We look up the half reactions from the Stayner reduction potentials table. We see that we have a l three plus, which is what we start with being a reduced. We have f minus being oxidized after we reverse a reaction of F to being reduced so therefore similar to part a. The talk reaction will be the one undergoing a reduction and therefore occur. And the calf food when we electoral eyes this compound and that must mean that the bottom reaction occurs in the Anna because oxidation is occurring. And finally, in part C, we have m and I to So we have mn two plus in I minus ions. So from the standard reduction potentials list of the half reactions in the table, we can find these 2/2 reactions that correspond to and then two plus being reduced an i minus being oxidized, where we had to again flip the reaction of I to being reduced. And so again, the first reaction is undergoing reduction and will therefore therefore occur and the cathode are the second reaction will again occur in the an ode and undergo oxidation


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