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In a Hittorf cell, a solution of cadmium iodide,CdI2, having a molality, m, of7.545x10-3 mol kg-1, was electrolyzed.In a coulometer that is series with the Hittorf ...

Question

In a Hittorf cell, a solution of cadmium iodide,CdI2, having a molality, m, of7.545x10-3 mol kg-1, was electrolyzed.In a coulometer that is series with the Hittorf cell the mass of Cddeposited at the cathode was 0.03462g. Solution with a mass of152.64g was withdrawn from the anode compartment and was found tocontain 0.3718g of CdI2. Calculate the transport numbersof Cd2+ and I-

In a Hittorf cell, a solution of cadmium iodide, CdI2, having a molality, m, of 7.545x10-3 mol kg-1, was electrolyzed. In a coulometer that is series with the Hittorf cell the mass of Cd deposited at the cathode was 0.03462g. Solution with a mass of 152.64g was withdrawn from the anode compartment and was found to contain 0.3718g of CdI2. Calculate the transport numbers of Cd2+ and I-



Answers

A cell was set up having the following reaction. $$ \begin{aligned} \mathrm{Mg}(s)+\mathrm{Cd}^{2+}(a q) \longrightarrow \mathrm{Mg}^{2+}(a q)+\mathrm{Cd}(s) & \\ E_{\mathrm{cdl}}^{\circ} &=1.97 \mathrm{~V} \end{aligned} $$ The magnesium electrode was dipped into a $1.00 \mathrm{M}$ solution of $\mathrm{MgSO}_{4}$ and the cadmium electrode was dipped into a solution of unknown $\mathrm{Cd}^{2+}$ concentration. The potential of the cell was measured to be $1.54 \mathrm{~V}$. What was the unknown $\mathrm{Cd}^{2+}$ concentration?

So here we are given information essentially about specific reactions and essentially are honored in this case will be essentially the platinum hydrogen electrode. So we have an oxidation reaction occurring here. So the process for oxidation BH two yields to age plus plus to minus. And we're already given information that the pressure essentially of hydrogen gas is one atmosphere. Well, basically the concentration hydro sodium hydrogen is one moller. So these are both understand your conditions. Yeah. However, in this case for a catheter or reduction occurs, were given information that we have essentially a silver electrode with solution. However, we're going to saturate that solution with a gcl. So here we sort of have different conditions. So the first thing we can do is essentially break up the A G C. L reaction to constituents, steps, possible constituent steps. So this would be a possible reduction reaction. While a possible essentially basically the other possible reaction that could occur here would be the oxidation of silver. So we can see that essentially this cancels out to form precipitation of essentially silver cord. So essentially that you not sell for this reaction for the first, basically part of that reaction which were interested in isn't as well known. So basically we're going to be interested in that part and we can determine essentially the case, basically the chaos P. We can determine from the K sp the not sell for this reaction. And this basically value is readily known for essentially the oxidation of silver to the one plus ion. So essentially we can write N F. E not sell equals R. T. Natural log in this case K S. P. So why not sell we can find is R. T over N F. Natural log of K sp. And here we can just substitute and RKSP is 1.8 times 10 to the negative 10th. So, plugging all our values and where N equals one mole of electrons based on the information given here, Mhm And taking the natural log of r k S p, we can find that in the end are not sell Is equivalent to zero 58 And also remember that essentially ari don't sell for this reaction here Is zero because it's arbitrarily set to 0V. So that's not particularly of our interest right now, but well, that will be important later. So essentially, since we're given information basically for some type of a cell and we have not given information that whether it's galvanic and will take, We just are going to have to assume that we have to construct the spontaneous cell, which means that in itself is greater than 0V. So we have to choose basically reactions that satisfy this. So in this case, if we had essentially this type of a system, essentially this condition would not be satisfied. And so let's try to find first of all the known values that we can easily find. So one of the known values we can easily find is essentially that the E not for essentially the oxidation of silver Is equivalent to negative 0.80V. And basically by summing the reactions together, we can find that the not of the first unknown reduction reaction is equivalent to 0.80 -18, Which is equivalent to 0.22V. So that's one more in part of basically our question. And we are still given information That the concentration of essentially chloride is about one moller. And essentially we can assume that the concentration of silver ion is relatively minimal, since the concentration of chloride is so high, most of essentially the silver is essentially precipitated and gcl form. So essentially a reaction here would basically be the initial catholic reaction. So a gcl, yeah plus electron healed A G plus C on minus while essentially are an ode in this case would be hydrogen gas yields to H plus Plus two E-. So this essentially becomes our reaction interest in this case. And since the concentration of chloride and the concentration of hydrogen H plus is one mould as well as the pressure of H two equals one atmosphere, we're essentially under standard conditions. So we don't even have to use the Narcs equation And we already know that China of oxidation is 0V. So understand your conditions, we can just use the not sell so that you not sell in this case Would just be the sum of the values maybe 0.22V. And that's it.

Um, the reaction they have some reaction are at the not, and this is a better. And the older reaction is this. The first thing the relation between charge, current and times is charged is a girl toe current multiplied by time. And we know that one embryo is a grilled one column up one Chris. One second may go getting room so can store a judge equivalent to a current off 300 mega and fear falling for one hour. So we have guarantees 300 make a Imperia converted toe. Yeah, column per second. That is 300 mega and fear multiplied by one m pair. It tend to the power three. Meghan Pier then converted to call the person that is 300 into 10 to the power minus trickle lumber. Second and the time we have this one hour that is text A 600. They can send the judge therefore judges current into time guarantees 310 to the power minus three and diamonds 3600 on point you right into a tentative or triple and therefore the capacity of this ls one point. See right into 10 to the power three room. On the second is, um, the charge used to deliver 300 peer guarantees. Yeah, is 12 the 08 into 10 to the part three column first, um, calculate the number off. Also, Fullerton's used by dividing the charge with Faraday. That is 0.8 in 10 to depart. Trickle them. I have one 9.65 to 10 to the par four that is 1.12 into 10 to defer minus to more soft electron. In over a reaction. One more off cadmium releases toe malls off electron. So let us calculate the malls off. Cadmium needed for 1.12 into 10 to the power minus two moles off electrons. There's 1.12 into 10 to the minus two most off electron into one more lof cadmium. A point two more off electron hits. 5.6 into 10, 24 minus three months off. Catch me and the molar Mass off cadmium ISS. 112.4 g per no, no. Calculate the mass off cadmium by multiplying with it. Well, the market is 5.6 into 10 to depart, minus three moles off cadmium multiplied by 112.4 grand. Far most that is their 0.6 to 9 times off. Get me, Therefore must required, please. Is there a 0.62? Nice. The overall cell reaction to most of electron required for right have trances required for to my old self night. Oh, which the address calculating also needed from also off. And I always needed for 11 0.12 uh, 1.12 and dio 1.12 into 10 to the power minus two malls off electron into malls off, uh, into two moles off and I owe or age upon to malls off electrons. That is 1.1 toe into tend to diplomat in this tomb also, and night old which Moeller mass off and I 00 h is 91.68 g per more. Therefore, myself and I always required ISS calculated by multiplying it with the Moola mass that is 1.1 doing to tend to the par minus to yeah, multiplied when 91.68 upon Well, that is 1.3 g off. And I always uh huh two moles off electrons required for two mole soft edge to off. So the uh huh has, uh, moved malls offish to needed for 1.12 into 10 to the four to multi 1.12 into 10 to the four minus two into two months off H two upon two months off, Electron is 1.1 going to turn to the par minus two moles off H 20 and Mama Savidge to Who is 18 g far more the fourth day. Uh, myself. It's, too, by multiplying it small with Murayama's every 1.1 toe in to turn to the par minus two. Um, also h 20 when deployed by 18 upon. When more that is there a point toe 02 g off H 20 Therefore the mass off XTO required. Is there a 0.202? Now calculate the total mass off the rear. 10 days Moss off. Jack Dance is myself good, ma'am. Plus Moss off And I all boy George bless mus off h 24 is 1.86 g. No ah, the weight of battery is 18.3 g and the mass off the reject entries 1.86 g now percentage myself. Resistance is myself. The reactant upon mass off the battery in 200% putting the Valley is it is 10.2 percent. 20 states they had 10 consists 10.2% off.

Question 145 simply wants you to estimate what the K A value is for the buffer solution. Mops, which is a weak acid often used in RNA analysis. To determine the K A value. We recognize that the pH range that is best for a buffer is typically somewhere around, plus or minus one pH unit of P K A. Because the range is 6.5 time, or 6.5 to 7.9 than the K value is likely to be right in the middle of this range. So the best estimate of the K A would be the 6.5 plus the 7.9, which gives us 7.2 and then pH is going to be equal to 10 to the negative 7.2 Oops, negative 7.2, which is going to be six whips, which is going to be six point 31 times 10 to the negative eight as our okay, a value

Okay. This is a process that's happening in an electrochemical. Sorry. An electrochemical cell. Yes, but it's an electrolysis cell. So a trawl Assis, um is the process that you are forcing electrochemical energy, electrical energy, um, into a compounds and forcing a chemical change that is generally not spontaneous. That is made spontaneous by the input of that energy. Um, so you'll see why these these answers were gonna be different from answers for spontaneous processes. Um, so the question just says the electrolysis of a quiz caveat chloride into its elements. So that's all I wrote here. So here's cadmium chloride as broken into element. Cadmium So elemental cadmium has an auction stations the oxidation of zero and elemental chlorine has an oxidation state of zero. But they have oxidation states in calcium chloride, right? What are they so Korean always has minus one? Um, not always. But generally the halogen is have minus wanted group 17 and there's two of those, which means the balance. The charge cadmium needs to have a positive to charge. That means the half reactions are gonna be us followed. So cadmium two plus going to cab bm zero cadmium is going to be reduced in this process, and then Ah, si l two is not You're not gonna write it as seal to because it's really to chloride, right? It's cadmium chloride. So you're going to do, um, to see l minus is going to be oxidized because it's going from nine. It's 1 to 0. So that's gonna be one electron each it's giving up to get Teoh seal to, You know, that's that. I didn't balance my initial equation, right? Because, um, I didn't balance it in terms of, um, electrons. I just wrote the, um, the electrolysis. So, um oh, it actually is balanced much. Um, sorry, I'm not what I was thinking anyway. So, um, what you're going to do here is get the e cell potentials for those values, right? So, um, for the process of cadmium going to cab bm zero um, these are gonna be the reduction potential values. So cadmium. So I should be required. This is going to the cathode, right? This is gonna be the an ode. So you're cathode is going to be, um e not is equal to negative 0.403 volts and your, um, e cell for the n owed is going to be, um, positive 1.358 volts now, depending on which one you learned it. I always give this speech in all the videos. That I do, um is because depending on which way you learned that you might have been taught to flip the half reaction reduction potential at the Anote. Because reduction these of the reduction potentials. And if you want the potential of oxidation, you would flip the value that's given to you in the book. So if it's a positive value here, it's gonna be a negative value if the reverse is happening. Right, So And when you do that, you're not gonna follow the math that's in the book. You're gonna add the two values together. So generally you would do reduction potential of the cathode minus the reduction potential of the material at the an ode off the metal or non metal that's reacting at the an ode. Um, but if you're gonna account for the sign flipping of the anote half reaction, then you could add the two together. So you might notice that your answer is going to be negative because we would do. 0.403 negative. 0.403 volts. Um, plus negative 1.358 volts. And you're going to get ah, super negative answer, which is negative 1.761 volts. So that shows you that an electrolytic cells, um, the react the overall half reaction is actually negative. It's non spontaneous. And the electric vehicle, the electrical energy you're forcing through your compound is actually causing the electron transfer. Um, conversely, if you decide not to flip the annual value so you just do Catholic minus an ode, you're gonna get the exact same thing I just wrote wishes Cathode. That is negative 0.403 volts. Um, minus the potential of the an ode. So if Corinne was being reduced, the reduction potential would be 1.358 um, votes. And you're going to get the same thing here. You're gonna get negative 1.761


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