suppose-that-you-make-an-electrochemical-cell-one-beaker-has-a-calcium-metal-electrode-and-a-solut-2

Question

Suppose that you make an electrochemical cell. One beaker has a calcium metal electrode and a solution of aqueous calcium ions, $\mathrm{Ca}^{2+}(a q) .$ The other beaker has an iron metal electrode and a solution of aqueous iron ions, $\mathrm{Fe}^{3+}(a q) .$
a. Write the half-reactions for each beaker.
b. Use the activity series table on page 570 to decide which metal is oxidized and which metal cation is reduced.
c. Make a sketch of the electrochemical cell. Show the reaction that occurs in each beaker.
d. Write a net ionic equation for the reaction that takes place to generate electricity.

David Collins · Accepted Answer

question Number three in this section is associate ID with the formation of an electrochemical cell from calcium metal, and it&#x27;s cat ion and iron metal, and it&#x27;s cat ion. A cat eye on that they have is Iron three plus, which is not found in this section. So the only way to answer this question using the information provided in this section is to assume that it&#x27;s Iron two. Plus. They ask you to write half reactions that are occurring at each of the half cells. To do this, we need to look at the activity Siri&#x27;s and find out which is more active. Calcium or iron. Turns out that calcium is more active, so electrons are going to leave the calcium and calcium is going to become calcium two. Plus. As electrons leave, they&#x27;re going to enter the iron so the iron is going to take on the electrons. Right now I&#x27;m working with F E two plus three. Plus, we&#x27;ll talk about two plus in a minute. So anyway, I&#x27;m just going with three plus for now, taken on three electrons becoming iron. So the calcium is oxidized and the iron is reduced. The entire reaction is going to be. The calcium is a reactant, and the Iron three plus is a reactant forming calcium ion is a product and iron as a product. Then we need to balance it by putting threes in front of the calcium and twos in front of the iron so that we have the same net charge on both sides two times three is six plus and three times to a six plus. To answer a follow up question later on, where we actually have to calculate the electrochemical potential or the voltage, we&#x27;re gonna have to assume that this is two plus because it&#x27;s, uh three plus is not found in the activity Siri&#x27;s.

Morgan Sizemore · Answer

All right. So here we have thes overall reaction for an electrochemical cell, and we are going to find the half reactions for both the and out in the cathode. So using our to react, INTs will be able to find our 2/2 reactions. So if we start with our and I two plus, then we look up our half reaction, as in I tube less aqua. He is plus two electrons forms in I solid. Um, and then if we do the same for, um, the CD, we will just have to flip it for what is found in the appendix. And we will get that CD solid forms CD two plus AQ we ISS plus two electrons. And then in order to figure out which ones, the cathode and the anote. If the electrons are on the left side of the equation, then it would be the cathode. And if the electrons are on the right side, then it would be the an ode. So that is all of heart a, um and then going on to part B, it asked us to write a balanced net ionic equation. So we will take those two and it combined them into one reaction. So we&#x27;re combining them together. Then our ju electrons on the left will cancel out with our two electron center On the right, So are two reactions that are left are the in I two plus UPA qui ists plus cd solid forms and I solid plus cd two plus awkward. So we just take those two parts of the reactions and combine them in tow one keeping our reactant on the left and our products on the right. And then the last part, which was part C, says just to draw the cell diagram. So we take that, uh, reactant, and we put all of our an nodes on the left and are cathodes on the rights to do this in a new color. Um, so our an ode waas from our green equation above. So we have, see solid. And then we have a line representing the electrode solution boundaries. Then we also have the CD two plus, which was Aquarius Then we have a double line which represents our bridge. And then we have our cathode on the right. So the first part of our cathode equation was the inside two plus which is obvious. Another line separating the two parts of the reaction. And then we had our and I solid. So this waas you three news? This was part A. This is part B and that is part C.

Krish Desai · Answer

All right. Hello, guys. Welcome. Today, we&#x27;re gonna be looking at this. Um, we&#x27;ll take so talking about, you know, the diagram parts of it. So let&#x27;s get started. First things first. What is the What do we know? We know that there is magnesium two. Plus, being oxidized, that means that we&#x27;re going to have the electrons on the right side of the equation. So if we work backwards from there, that means that the M G two plus has to be on the same side of that equation because these to balance each other out, which means that we&#x27;re actually starting with solid magnesium on this side. Now, let&#x27;s think about the A. G. Plus, this is reduced. So which means that we&#x27;re gonna have three electrons on the left side of the equation, which means that again, a G plus is on the same side. And so this is our half reaction there again, um, right side, oxidized left side reduced. Pretty simple. And then, of course, we have our n n 03 sold bridge. So now let&#x27;s start diagramming out. Ah, General will take so right. Voltaic cells. Pretty simple. I mean, it is if you know, you can draw straight lines on a time blip device. All right, I think we got this. Okay, then you have your electrodes. All right. Now, here is our external circuit wire. Actually. Here, let me just let&#x27;s say that this is connected to a light bulb, right? Just for effect and technical difficulties. Okay, this is connected to a lightbulb. So this is our external circuit. And then the last piece of this is our salt bridge connecting these two sides. And so this is our salt bridge made off, and a n 03 We&#x27;ll call. This are an ode. Remember, An notes are negatively charged. Well called. This are cathode, and this is our external circuit. And those were really the major parts of all takes up. No animals are negatively charged. Cattles are positively charged, which means that an odes the half reaction taking place in an ode is oxidation Because an oxidation, you lose electrons. And so those electrons air floating freely around here, providing that negative charge. Let me just carry to those and for the same for the inverse logic, cathodes or reduction, because it&#x27;s taking up any free electrons. Okay, so because this oxidation would go back up, we specified RMG is oxidized. Ages reduced. So with those half reactions, we can say that this oxidation is solid. Magnesium becoming Equus. Magnesium and electrons, meanwhile, are reduction. Half reaction is Ah, silver is a quiz silver ions plus an electron. Sorry, I forgot the eighties becoming solid silver. So that&#x27;s our half reaction is done. Now, if we want to look at the direction of electron flow, will see that because the Anote is negatively charged and the cathode is positively charged, electrons are gonna travel from the negative side. True, the positive side and in the middle they&#x27;ll go up here and we&#x27;ll get light or whatever else happens in the external circuit. Drawing is really not conducive with this set up. Anyways, that is the flow of electrons. And now we also want to see the right we need. We still need an inclusion for the net reaction, so we&#x27;ll notice there&#x27;s only one electron needed here, and two electrons last year will actually have to multiply this entire reaction by two. Um, and then when we add them together, we&#x27;ll get solid magnesium plus two silver ions gives us Ah, magnesium two Plus I ARN plus two silver solid. And that is our net equation for this full takes up. And now finally, what we need to do is we need to talk about the movement of the N A and the n 03 and the Salt Bridge. So first of all, let&#x27;s talk about what happens when you get this current when you get this current, because the electrons are now going here, right? You&#x27;re getting an imbalance of charge. Getting too much negative on this side. Too much positive on this site. And when that happens, the reaction won&#x27;t occur anymore. Right? Sorry, sorry. That&#x27;s incorrect. What? What happens is you have your half reaction here, right? You have solid magnesium. Then you have magnesium two plus here. So, yes, there is too much excess positive. Too much negative here. But it&#x27;s not directly because of the electrons. It&#x27;s because as this half reaction progresses, you&#x27;re going to get too much magnesium two plus in the solution here. And so when that happens, right, there&#x27;s a solution. When that happens, you need something to neutralize this so you can get more magnesium to pulls And that&#x27;s where the n a and the N 03 minus come in. Because N a n 03 is really and a plus ions and n +03 minus ions. By the way, same thing is happening on this site. You&#x27;re getting too much a g here. Um, sorry. You&#x27;re getting too little A g plus here and too much silver here. All right, so you need to balance it out. And so what happens is in the salt bridge, you&#x27;re an A plus is going to travel to the cathode. Meanwhile, your N 03 minus is gonna travel to the ANA and so that Enel three minus will come and neutralize the magnesium. It&#x27;ll become mg and on 32 and the n a plus will essentially create the positive charge in the solution here that&#x27;s required to keep the reaction go. Uh, so, yeah, so that&#x27;s you know, everything about why salt bridges are required in the direction they travel. And basically, in the salt bridge, the an ion will always travel to the an ode. And the cat ion will always travel to the cap, and it&#x27;s required to make sure that the reaction keeps going and doesn&#x27;t stall because the solution is to statured. Okay. Thank you. Have a good day.

Andrew Engley · Answer

thistles. A bigger problem. Um, consists of three different parts. A through C. Um, just asking it Just asked various questions about a voltaic cell. Were given some information about how it set up and from that were asked to solve some questions. So let&#x27;s take a look at part A were asked to label each part of a cell. Uh, and we&#x27;re told that we have magnesium metal being oxidized and silver cat ions being reduced. Okay, so we know that the magnesium and the silver half cells must be separated. We know that. Looking at the notation, we&#x27;re told that they&#x27;re two different half cells, and then we&#x27;re told that they are in two different phases. That&#x27;s what that line in between species represents. From that we know that magnesium metal man pens kind of a little wacky. There, you magnesium metal is gonna be right here in magnesium. Cat ion is going to be floating around in solution, and it&#x27;s gonna be the same case for our other half. So we&#x27;re gonna have solid silver metal right here and then silver and a careerist solution for this half cell. Okay, We&#x27;re also told that magnesium metal is oxidized. So that means that this must be are an ode. That&#x27;s just the definition of an anote is where the oxidation half reaction occurs. And because reduction it&#x27;s occurring in this other half cell, we know that this must be the cathode. Okay? And I can&#x27;t kind of labelled the other parts already. The Volt here stands for volt ometer. And then I also labeled the Salt Bridge, which we&#x27;re going to get to in part C and a little bit, but everything else should be labeled and good to go. Um, I actually can&#x27;t see the image that you guys are gonna have access to, So this is the best that I can do with information that I was given. But this should touch on most, if not all, of the fill in the blanks or whatever you have for that problem. Okay, So now, for part B were asked to write half reactions, uh, occurring at the end note and the cathode. And then we also need to, um right the overall net reaction of the cell. So I&#x27;m gonna do that underneath. We&#x27;ll start with the an ode for the An ode. We know that we start with magnesium metal and that the magnesium metal cot I on that we&#x27;re generating as a two plus charge. So therefore, in order to balance the charge, we know that we have to generate two electrons as well. Okay for the cathode. We&#x27;re told that we have silver cat eye on and is being reduced, so we know that it is going to turn into its solid state. We have our silver cat I own here. And then in order to make it neutral, we add one electron that&#x27;s gonna produce solid silver metal in order to write the overall war net reaction. All we have to do is that is, um, make sure that the number of electrons and these half reactions are equal to each other. So if we take this half reaction multiplied by two, notice how we have the same number of electrons in both half reactions, and then we can just write him across. Cancel out are electrons because we&#x27;re gonna have the same amount of electrons on each side that&#x27;s gonna look something like this. We&#x27;ve got solid magnesium plus two silver ions in a solution. I would like you horribly there. You go is going to react to produce the magnesium cat eye on an area solution and solid silver medal or two moles of solid silver Medal. And that&#x27;s gonna be our net overall reaction for the voltaic cell. So now we&#x27;re asked to trace how the electrons will flow in the external circuit. Um, so that&#x27;s gonna be well, actually, we&#x27;ll start here looking at our net reaction and are half reactions. We know that the magnesium is getting oxidized and that the silver is becoming reduced. So that means that magnesium is losing electrons and silver is gaining electrons. You also might know that just from the general definition oven anote in a cathode. But electrons always flow from the an ode to the cathode. So I&#x27;m gonna indicate that yellow my e minus for electron to indicate that electrons air flowing from the an ode to the cathode. Okay. And we&#x27;re told that we have, um, sodium nitrate in our salt bridge were asked to draw the movement of those ions. Okay, so we know that, um, our nitrate Aquarius must move to the an ode, and our sodium must move to the cathode. And the reason for That is because we have, um, an excess buildup of positive charge in the an ode and an excess buildup of negative charge in the cathode. Um, because of the loss of electrons. So when we&#x27;re oxidizing solid metal, we are creating a metal cat I own are our magnesium metal Qaeda cat ion. Excuse me. So we need to neutralize that charge with nitrate. And the exact opposite is true. The Catherine, we&#x27;re generating a more negatively charged species, so we need to introduce a positive cat ion in the salt bridge. And the reason the salt bridge is so important is for two reasons, um, one, it completes the circuit, right? You can&#x27;t have a circuit if it&#x27;s not a loop, you can&#x27;t really get in electrical reading from that. Um so it connects the 2/2 cells together to complete a circuit that&#x27;s very important. And the other thing that is important for the main reason why is required is if we didn&#x27;t have the Salt bridge here, the reaction would occurred so quickly that we would not really be able to observe a reading on the Volt ometer. So it really does two things it prevents a rapid reaction from occurring so we can actually get a voltage output from Marvel Tom Attar. And it also maintains electric neutrality, which is important so that we don&#x27;t have that access charge buildup. And that should be all the parts answered for this. If you see yes, so that&#x27;s how you tackle this problem.

Morgan Sizemore · Answer

Okay, So for this problem, they give us the oxidation of CD solid to Studio H two and then the reduction of M N 04 minus two him and 02 solid. On Ben, we are to write the half reactions for the an out and cathode. So for the, um, oxidation, that one is going to be our an ode because it&#x27;s being oxidized. So the equation for that, based on the appendix will just flip. It is going to be CD solid Bless 20 H minus oddly, is gives a CD Oh h two solid plus two electrons and then for the reduction because it is being reduced that is going to be our cathode. And then looking that up in the appendix, the half reaction would be in in oh, for minus plus two h +20 plus three electrons gives us in a to less for oh, age minds in that spa quis. So that is part a. Then for part B, we are to write a balanced net ionic equation. So we&#x27;re gonna take those 2/2 reactions and combine them. But because of the number of electrons and are red equation there are two electrons on the right in our green equation There three electrons on the left. But we want those to cancel out. So the least common multiple would be six. Which means that we need to multiply everything in our red equation by three so that there will be six electrons. So I&#x27;m gonna put a three, multiply that by three, which gives us six and then a three there and then for our green equation. We also want our number of electrons to be sick so that they&#x27;ll cancel out, which means we have to multiply. Everything by two will be a two there two times two will become four. There&#x27;s a to there and then four times two will become eight. Uh, another thing. So now when we look at the equation Sorry, Uh, those six electrons canceled cause they&#x27;re on opposite sides. There&#x27;s also 60 h on the left side of a red and 80 H on the right side with are green. So those six will cancel out with six of these, which will make that eight become a to So now our final equation. Um, for part B, we&#x27;ll take our three reactions we have so three CD solid plus two I m in 04 minus plus four age, +20 liquid. I forgot that. That was awkward. Assists in there. And then we will keep our products, Which was three CD. Oh, h two. I just realized that. Okay, um solid Plus two. I m in o Teoh, which is solid. Plus only +20 h minus, which is? Ah, quests. And then the last for part C. We&#x27;re just drawing a cell diagram. Eso We&#x27;re going, Teoh, put it our Sorry, we&#x27;re gonna put part on the left side, which is the Anote, and we&#x27;re going to put our cathode on the right separate, separated by just one line. So we have our CD solid. And then we have our, um cd Oh h two. And then we do a double line showing that barrier and then on the right, we have m in 04 minus in an 02 solid. And our platinum from our electrode may add in those states that I keep skipping over and there

Jennifer Landry · Answer

okay, in this problem, we have a vote take. So that&#x27;s gonna be similar to a figure in your book. 20.5. And it has an electrode half sell. One of the electric have cells consists of a silver strip placed in a solution of silver nitrate. So we have one container with our silver strip electrode in there. Yes, this is silver here, and it&#x27;s in our solution of a g n 03 And then we have another container, other half solution here, which has a strip of metal in it. Um, in the other medal is iron. We have strip of iron here, and it it is in a solution of iron to chloride. So we&#x27;re gonna have our solutions connected by salt bridges here and then our electrodes connected by some sort of wire. Um, and the overall reaction for this system is that f e solid plus to a G plus. And that&#x27;s a qu&#x27;est CE is going to yield Effie two plus plus two eiji solids. So the first question is what is being oxidized in what is being reduced. So what is being oxidized? Our little pneumonic? Here&#x27;s oil rig oxidation is losing electrons and reduction is gaining electrons. So the substance that is losing electrons is iron. It is going from zero to plus two oxidation number, so it is losing electrons. Therefore, my iron metal is being oxidized. And then my silver ions air going from plus 1 to 0 oxidation numbers. So they&#x27;re gaining electrons. So my silver ions that are part of the silver nitrate they are being reduced heart be asks, right the half reaction that occurs in the 2/2 cells. So the half reaction with iron is we have iron metal and it is becoming iron two plus ions acquis plus two two electrons to balance that charge out. And that is my oxidation half reaction. And then my reduction, half reaction. I have silver ions. They&#x27;re curious plus an electron to form silver medal solid. So that is my reduction reaction. Laugh reaction questions See? Asks us which electrode is the anode in which electrode is the cathode? So the little rule here is vowel, vowel, constant, constant. So the note is where oxidation occurs and the cathode, the continent constant struggle, and the cathode is where reduction occurs. So the anode is where my oxidation reactions occurring. So that is where the, uh the iron half reaction is where my iron metal is. And the cathode is where the reduction half reactions occurring. And that is where my silver metal strip is. My silver ions, that air being reduced. But in terms of the actual cathode, it&#x27;s the sorry. It&#x27;s the silver solid silver strip that is acting as the cathode and the solid iron strip here. That&#x27;s acting as the anode question D asks us indicate the signs of the electrodes. Um, and that&#x27;s so for electrodes. We have our again. We have our iron, which is acting as our anode, and we have our silver, which is acting as our cathode. And because electrons always flow from the anode, um, the anode is going to be an electrons are gonna want to flow towards positive, right? So they&#x27;re flowing from the an ode to the cathode Elections are negatively charged, so they&#x27;re gonna want to go towards the positive electrode. So by definition, the cathode is the positive electrode, and the anode is the negative electrode. So they&#x27;re flying from the negative to the positive. Um, at least they are in a spontaneous voltaic cell, Part E Do electrons flow from the silver electron to the iron electron or from the iron to the silver. So again, electrons are always going to flow the way they&#x27;re gonna flow. Is there going to flow from the anode? So from the oxidation which is losing electrons to the cathode, which is gaining the electrons, And since the anode is the iron, they&#x27;re going to flow from the iron to the silver on. And then finally, part F Heart F says in which directions do the cat I ends and Annan&#x27;s migrate through the solution. All right, so that&#x27;s talking about up here. So what&#x27;s gonna happen is at the cathode start with what&#x27;s happening at the cathode. So that would be on the left here on the cathode. Um, we have silver ions that are changing from silver ions into silver medal. Right, So we&#x27;re losing a positive charge. So are positive ions are cat ions are going to flow from the in. Our Selfridge are positive. What I might say are, positive ions are cat lions are gonna flow toward the cathode because we&#x27;re going to need to make up for that positive charge, so they&#x27;re going to flow within this into the silver solution, the cathode at the anode. Let&#x27;s look at what&#x27;s happening. We have f e changing into Effie two plus, so we&#x27;re actually getting a buildup of positive charge at the anode. Therefore, the negative ions the anti EMS in our salt bridge are going to need to flow towards the anode in order to balance out that extra positive charge.

Problem

Repeat Exercise 3 with the pairs of metals here. …

Problem 3 Easy Difficulty

Answer

$3 \mathrm{Ca}+2 \mathrm{Fe}^{3 t} \rightarrow 3 \mathrm{Ca}^{2+}+2 \mathrm{~F}_{2}$

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$3 \mathrm{Ca}+2 \mathrm{Fe}^{3 t} \rightarrow 3 \mathrm{Ca}^{2+}+2 \mathrm{~F}_{2}$

Living by Chemistry

Lizabeth T.

Allea C.

Morgan S.

Karli S.

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Angelica M. StacyLiving by Chemistry

Lizabeth T.

Allea C.

Morgan S.

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