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Question1 ptsAn electrochemical cell has the following cell reaction: PbOz(s) + |-(aq) - Pb2t(aq) Iz(s)Which is the half-reaction occurring at the anode?2 I-(aa) Iz...

Question

Question1 ptsAn electrochemical cell has the following cell reaction: PbOz(s) + |-(aq) - Pb2t(aq) Iz(s)Which is the half-reaction occurring at the anode?2 I-(aa) Izls) + 2 ePbOzls) + 4 Ht(aq) + 2 ePb2tlaq) + 2 HzO()Pb2tlaq) + 2 HzO() PbOz(s) + 4 Htlaq) + 2e"Iz(s) + 2 e 72 |-(aq)

Question 1 pts An electrochemical cell has the following cell reaction: PbOz(s) + |-(aq) - Pb2t(aq) Iz(s) Which is the half-reaction occurring at the anode? 2 I-(aa) Izls) + 2 e PbOzls) + 4 Ht(aq) + 2 e Pb2tlaq) + 2 HzO() Pb2tlaq) + 2 HzO() PbOz(s) + 4 Htlaq) + 2e" Iz(s) + 2 e 72 |-(aq)



Answers

Use the following information to answer questions 25-28. A voltaic cell is created using the following half-cells: $\begin{array}{ll}{\mathrm{Cr}^{3+}+3 e \rightarrow \mathrm{Cr}(s)} & {E^{\circ}=-0.41 \mathrm{V}} \\ {\mathrm{Pb}^{2+}+2 e \rightarrow \mathrm{Pb}(s)} & {E^{\circ}=-0.12 \mathrm{V}}\end{array}$ The concentrations of the solutions in each half-cell are 1.0 M. Which of the following occurs at the cathode? (A) $\mathrm{Cr}^{3+}$ is reduced to $\mathrm{Cr}(\mathrm{s})$ (B) $\mathrm{Pb}^{2+}$ is reduced to $\mathrm{Pb}(\mathrm{s})$ (C) $\mathrm{Cr}(s)$ is oxidized to $\mathrm{Cr}^{3+}$ (D) $\quad \mathrm{Pb}(s)$ is oxidized to $\mathrm{Pb}^{2+}$

Good morning. Today we're gonna be looking at a voltaic cell. Ah, specifically, a voltaic cell built on this equation right here to solid chromium atoms plus three iron cat ions becomes two chromium cat ions and three iron atoms. So, um, you know, this is a standard will take. So a Z, you can see we've got the end of the Catholic electrons traveled to the external circuit, and you have the salt bridge here for, um and ions and kate ions to travel. So let's talk about this research. Uh, which way do electrons traveling the circuit? Well, electrons are negative, and they're attracted positive. So should be pretty clear. Electrons travel from the negative electrode to the positive electrode, or they travel from the an ode to the cathode. And you know, Anote is negative because an ode and I on Catherine is positive because Catholic Cata cronies. Now, our second question is what direction? Um, how do the ions How do negative ions travel over the salt bridge? Well, negative ions are going to travel from the cathode through the anote over the salt bridge, so they're really gonna go. This'll sway. And so what? You have to know what this is. The an on the and ions that are traveling here. Aren't any of, you know, the chromium or the iron, right? The an eye owns that air traveling here are generally speaking, uh, the and I honest from the solution that these two electrodes, Aaron and, uh, that's what travels over the salt bridge. Okay. And then our last question is, um, the two different half reactions. So that's also pretty simple. Um, so at the cattle were always gonna have reduction. Take place. Right. Because if we gain electrons, then there's going to be a net positive charge. Here is those electrons are soaked up, which is why the cattle this post at the an odd oxidation takes place. Because when we lose electrons, this entire thing, it's a net negative charge. Right? When you know the a. Quist, um, substance starts attaching to here. So let's talk about our 2/2 reactions. 1/2 reaction is with iron. The F three f e two plus a curious comes three f e solid plus six U minus. All right, six because three times two plus charge that has to be canceled. Also, my mistake. Six was correct, but it should be fix on this side. And then our half reaction for the chromium is to chromium solid. I was sorry. Accomplice era two chromium five plus a nucleus. Close 10 e of this half reaction, it's just not appear to be balanced, but okay, We're just gonna work with it as it is. Okay, so, um, so this is reduction, right? You know that because the electrons are on the left side of the equation. And also, the oxidation number here is going from plus two, 20 All right, when the oxidation number decreases, that means it's gain electrons, the oxidation number for chromium. And this equation is going from zero to a whopping plus five, which means that this must be oxidation. You can confirm that, basing that the electrons are on the right side of the and now we just talked about you know what goes where. So we know that that at our catalogue, we're gonna have our reduction equation 60 minus Equus plus three f e two plus equally. This the gives us three. Effie saw it and then are an old we're gonna have our oxidation equation. just talked about that to see our solid equals to see our five plus incurious plus 10 u minus equals, and that is how you take care of a voltaic.

For this problem. We're giving a Redox reaction and we have Teoh fill in a few blanks to tackle this problem. What I'm gonna do, I'm gonna use this diagram and fill it in a labelling And, um, just placing things in the cried or the correct orientation, and then you can use the drawing in order to figure out the answer to how to fill in the blanks. Okay, so the first thing that we need to identify is the direction in which the electrons are going to flow in the extent in the external circuit. Excuse me. So that's gonna be through our whole Tom Attar right from 1/2 cell to the other. We know that electrons are always going to flow from the an ode to the cathode because in the an ode, we're going to have an oxidation half reaction and the cathode is going to be the reduction half reaction. Um, and if you're unsure about that, just look up the definitions for an an ode in a cathode half. So and maybe review what an oxidation or reduction half reaction is if you guys are a little confused with that going back to the problem. We know that again electrons we're gonna flow from are an ode. You are a cathode. I'm just gonna label that in yellow and we've answered the first blank. Now we need to identify how the and ions move in the salt bridge, which I'm going to write in read. So if we know that our electrons are moving from the ano too that cathode, right then the species that exists in solution in the an ode is getting more positive in charge while the charges getting increasingly negative in the cathode. Right, Because electrons are moving to the cathode and are leaving from the an ode. So in order to replenish the charge, we know that an ions must move from the cathode. Do the an ode. And I think for this guy you can say that it just moves from the cathode to the an ode. Um, you could also say Well, actually all get to that in a bit. But yes, it moves from the cathode to the an out. So now what we need to do is look at our Redox equation. Actually, first write it off to the side and determine which species is getting reduced in which species is getting oxidized. So we got to chromium solid plus three iron two plus Equus reacts to create two chromium. The charge will not is three plus Equis and three f e s. Okay, so we should start by writing out. We're identifying the oxidation states of each individual molecule first. Oh, you to go a little bit farther. There we go. OK, so we know that solid chromium has an oxidation state of zero. Because chromium is the Onley Adam here and it has a neutral charge. And for the same reason, solid iron also has a zero oxidation state. Because iron is the Onley Adam that exists in this molecule and eggs and it exists as a cat eye on. We can say that it's oxidation State is plus two in the oxidation state of chromium three plus is plus three god. So if we look at chromium, I'm gonna start in red again. We start with an oxidation state of zero and we move to an oxidation state of plus three. So because we are increasing our oxidation state chromium is becoming oxidized in this reaction. If we look at iron It goes from a plus two oxidation state to a zero oxidation state. And because the oxidation state is decreasing, we know that that is a reduction that's happening. So we know that I'll just write it out below here again. C. R s is being reduced. It is to see our three plus bakery a solution and f e two plus speak with this. It's getting oxidized. Two solid iron. Okay, The next thing that we need to do is write out the rest of the half reactions for these guys. And we do that by neutralizing the charge. All we have to do is add electrons to either decide because there are no oxygen atoms. President, there are no hydrogen atoms, President. And we've already balanced out the atoms that are not oxygen or hydrogen in these half reactions already. Oh, sorry. That should not be three minus. So we have two chromium chromium, three plus ions, so we need actually a total of six electrons. My mistake, because we have three f e two plus, we're gonna add our six electrons over here. We have one last thing that we need to do. We've identified which species is getting reduced. We've identified which species is getting oxidized, and now we need to assign them to either the Anote or the cathode. Earlier in the video, I said that the Anote is where the oxidation reaction is occurring and the cathode is where the reduction reaction or half reaction is occurring. Excuse me. So now that we know all that information, you should be able to fill in the blanks and answer this question.

This question is like that is similar to the previous one. So let me write the 2/2ves Positive plus two electrons, you and the potentially is 1.68 votes. And for the second one we have So for two negative needs to sulphur plus Loss of two electrons. And the potential is 0.14 vote adding these two. You're positive. Sell for two negative leads you to you plus cellphone. Yeah. And the oral cell potential by wearing these are 1.82 bowls. We see that the self potential is positive. That means the reaction will be spontaneous.

So in this question, were given a Redox reaction in a voltaic cell. A few things we have to remember about Baltics cell is that it has two compartments where oxidation and reduction half reactions takes. Please. Oxidation takes place in a compartment. Would negative electrode, which is unknown and reduction take place, takes please in the compartment with positive electrode, which is Katherine and the electrons produced from these Redox reactions flow from Anil to Catholic through through an external circuit. For this reaction were asked to find the equations for oxidation, introduction, half reactions where they occur and how the electrons and I owns and cat ions move through these self. So first of all, we're going to break this reaction into two, um, to reaction, figuring out which one is oxidation and which one is a reduction half reaction. So are magnesium, which is in solid has zero oxidation number, while mine museum two plus has plus two. Since the oxidation number increased, we have. This is our oxidation reaction, and to balance it by charge, we can add two electrons on the right hand side and our reduction reaction. We have two protons and acquis, which goes to hydrogen, and we can violence it by charged by adding two electrons on the left hand side. Our second question is, where do these react oxidation and reduction? Helfrich reaction occurs, like I said before oxidation occurs and Adam and reduction occurs in cathode. So to complete the sentence, um, in part c, we know that the electrons in the external circuit flows from an hour to Catholic, so it flows from magnesium electrode, two hydrogen electrode. Similarly, since magnesium is losing electron, it's getting more and more positive. So are negative. Ions have to move from cattle to Anote. So our negative ions will be moving from hydrogen to magnesium. While our positive, um, I owns cat ions will be moving from, um ano too cathode.


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