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Use the followlng cell to answer the questions:Cr / CrJt (1,00 M) Ni?+ (1.00 M)0) How many milligrams of Ni would be plated on the cathodecurrent of 0.30drawn for 1...

Question

Use the followlng cell to answer the questions:Cr / CrJt (1,00 M) Ni?+ (1.00 M)0) How many milligrams of Ni would be plated on the cathodecurrent of 0.30drawn for 1.5 hrs?426 7mq Nib) What Is [Cr J+] when [Ni?+] has dropped to 10 4 M?What the cell potential the condltlons descrlbed I Part 67[CrJ

Use the followlng cell to answer the questions: Cr / CrJt (1,00 M) Ni?+ (1.00 M) 0) How many milligrams of Ni would be plated on the cathode current of 0.30 drawn for 1.5 hrs? 426 7 mq Ni b) What Is [Cr J+] when [Ni?+] has dropped to 10 4 M? What the cell potential the condltlons descrlbed I Part 67 [CrJ



Answers

Consider a concentration cell similar to the one shown in Exercise $67,$ except that both electrodes are made of Ni and in the left-hand compartment $\left[\mathrm{Ni}^{2+}\right]=1.0 M .$ Calculate the cell potential at $25^{\circ} \mathrm{C}$ when the concentration of $\mathrm{Ni}^{2+}$ in the compartment on the right has each of the following values.
a. 1.0$M$
b. 2.0$M$
c. 0.10$M$
d. $4.0 \times 10^{-5} M$
e. Calculate the potential when both solutions are 2.5$M$ in $\mathrm{Ni}^{2+}$ .
For each case, also identify the cathode, anode, and the direction in which electrons flow.

This question is similar to the previous question. Number 69. Again, it's helpful to recognize what the nurse equation is for a concentration cell. It will always take on this form. He cell is gonna be equal to East L Standard, which will be zero minus 00.5916 over the number of moles transferred. In this case, it's too. And then it will be the concentration for the yeah node, which will be the smaller concentration over the concentration for the Catholic, which with a larger concentration, it needs to be set up in such a way that we get a positive cell. So with this information, then for the 1st 1 you sell a zero for the 2nd 1 E cell is going to be equal to this where we've got one Moeller in one compartment and we've got to Mueller and the other so long as being a smaller one up top theano it up top. When 0089 go to the next one where one concentration is 10.1 Moeller and one concentration is one Mohler an ode up top. Smaller concentration 10.30 volts. De we've got a really small concentration that wouldn't be a top will be the Anote one. Moeller will then be the cathode 0.13 volts and then the last one when they're both 2.5 Moeller, the ratio here ends up being zero. Following the cell potential is zero.

Here. In this question we will collect the standard self potential for the given cell. So first we will right. They have reactions for the given selling notation and 25°C. So here our reaction and and not And reaching eight Qatar. So we have your reaction eight an artist chromium by losing electrons. Can work into Keita in three positive charge. And having self potential Not is equal to 0.74 world. Okay. And Next one is this is observation. And at the thought your reduction take place. I'm not curry. Yeah Gain two electrons and convert into mercury in solid form. So here in or out of the same is equal to 0.80 world. So now we will find the sale potential cell potential by using the given uh cell potential values here here is the self potential limitation is not officer is equal to he nod. Get thought minus in owned and not By putting their news behave 0.80 world -0.74 world is equal to one 1.4 1.54 Fold

Before we calculate the electrochemical potential of these concentration cells, let's review some basic principles associated with concentration cells. In a concentration cell. We have the exact same species that could be oxidized or reduced in both compartments, the an ode and the cathode. For this particular concentration cell, one compartment contains the silver ion at a one molar concentration, and the other one contains the silver ion at some specified concentration. Both electrodes themselves are pure silver in a concentration sell. The electrons move in such a fashion to make the concentrations in both of the half sales equal. So if we have a high concentration of silver here, then that means electrons need to come towards this half cell so that they can combine with silver and decrease this silver concentration so that it equals the lower concentration in the other half cell. If, however, the other half cell has a higher concentration, then electrons they're going to go from left to right to decrease this higher concentration until it equals this lower concentration, so to review in a concentration cell that contains cat ions such as this one, electrons will always flow from the low concentration half cell, which will be the an ode to the high concentration half cell, which will be the cathode, remembering that electrons always flow from the an ode to the cathode and with the concentration cell containing cat ions, the lower concentration half cell is going to be the an ode. Let's identify these half cells as a and be. This will help us. We answer the rest of the questions. Here is the reduction half reaction of silver. I am going to silver solid, so we recognize that the Silver Ion is a reactant. This will be important when we employ the Nerdist equation for part A. The second half cell has the same concentration as the first half cell. If the two concentrations were equal, namely, they're both one Mueller, then the cell potential is going to be zero based on what we just talked about. If there's no difference in concentration, we've reached a state of equilibrium. Electrons cannot flow. There is no cathode. Nor is there an an ode, and the self potential is zero in Part B. We have this half cell with a higher concentration to Mueller instead of one Mueller. So based on what we just talked about up here Electrons air, then going to flow from a which is the lower concentration which is theano to be, which is the higher concentration which will be the cathode. We can calculate the cell potential by taking the Catholic potential minus theano potential. Each of these potentials could be determined using the nursed equation where e cathode is going to be equal to the standard reduction potential of silver minus 0.5912 divided by N and is going to be one multiplied by the log of the concentration of anything that has a changing concentration. That's a product. Our product is a solid. So we just put a one there divided by the concentration of silver in the cathode compartment. And we identified the cathode compartment to be be with the higher concentration of to Mueller. So we have a two down here minus the concentration of the an ode. I'm sorry. The potential of theano reduction potential of the Anote, which will be the standard production potential of silver minus again 0.5912 divided by an which is one the log of one over one, Mueller and this gives us a self potential of 0.18 volts. For part C, we have a lower concentration 0.1 Moeller in Compartment B than we do in Compartment A. So electrons air going to flow in the opposite direction. Based on what we mentioned appear, electrons will flow from B, which is the lower concentrate concentrated half cell to A, which is the higher concentrated half cell. A. Now being the cathode, be being the an ode will calculate. Self potential is the same way that we did before, where we've got our cathode, which were defining as a, which has one Mueller concentration. So it's standard reduction potential of silver minus 0.5912 over in, which is one log of one for the product. Silver divided by one for the molar ity of Silver Ion and then in compartment be. It'll be the same thing, but the concentration is going to be 0.1 Mueller for the Silver Ion, and we'll get a cell potential 0.59 Vault's. And for part D, we haven't even lower concentration in compartment B four times, 10 to the negative. Five electrons air going to flow in the same direction as we described in C. Because Compartment B has a lower concentration than compartment A. So electrons will flow from Compartment B, which will be the an ode to Compartment A, which will be the cathode. The set up is the same as we did before to calculate the self potential. However, now in Compartment de, we're going to have a concentration of four times 10 to the negative five Mueller. We can then calculate ourself potential to be 0.260 volts. How for the last one e it asks. What if both of these were 0.1 Mueller Well, if both of them are 0.1 Moeller instead of one Moeller, the same result will occur. The two concentrations air equal, so there's no cathode. There's no an ode. There's no flow of electrons, and the cell potential is also zero. This will be the result when these two concentrations are equal, regardless of what those concentrations are

I ordered an Through all parts of this particular question, it's helpful to write the nursed equation for a concentration cell. That Ernst Equation for a concentration cell is always a this form where we hav e cell is equal t cell standard, which will always be zero minus 00.5916 divided by an log of we always have the an old concentration on in the numerator and the Catholic concentration in the denominator so that we get a positive E cell where the an old compartment concentration is always the lower one and the Catholic compartment concentration is always the higher one. That way this ratio will be less than one, giving us a lot of value that is negative. And then, with this negative right here, we're going to end up with a positively cell, which is what we need for the concentration. So so then it becomes relatively easy. He self, for the 1st 1 will be zero because thes two concentrations will be one and the log of 10 So everything ends up being zero, and then for B, we've got one side with one Moeller that will be the an ode, and that's a lower concentration divided by the cathode gives us in any celeb 0.18 the 2nd 1 We now have 0.1 Moeller and one Moeller. So the 10.1 Mueller goes on top. It's smaller, divided by the one Moeller, we get a positive 0.5916 Then we have a really small one. That one's gonna go on top and be the antidote. And then one Moeller for the other compartment will be the cathode 10.26 volts. And then, if both of them are 0.1 Moeller, same thing potential will be zero. And electrons always slow from whatever we define the Catholic. I'm sorry the animal to be to whatever we define the Catholic to be.


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