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Practice Exercise 6 Alvoltaic cellis based on Co? /Co half-cell (Ercd =-0.28V and an AgCVAg half-cell (Ered 7+0.22V(a)What half-reaction occurs at the anode? Co GER...

Question

Practice Exercise 6 Alvoltaic cellis based on Co? /Co half-cell (Ercd =-0.28V and an AgCVAg half-cell (Ered 7+0.22V(a)What half-reaction occurs at the anode? Co GER cathod ) Q2As' ~2LEO (anedc)(b) What is the standard cell potential? E t41 E Ena cethau) Erea 5 Wlucte) Cli 2 + 0= 22 V

Practice Exercise 6 Alvoltaic cellis based on Co? /Co half-cell (Ercd =-0.28V and an AgCVAg half-cell (Ered 7+0.22V (a)What half-reaction occurs at the anode? Co GER cathod ) Q2 As' ~2 LEO (anedc) (b) What is the standard cell potential? E t41 E Ena cethau) Erea 5 Wlucte) Cli 2 + 0= 22 V



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.

Hello everyone. So in this question we have to calculate the standard self potential. So for the first part we have been given the reaction nickel to positive Equus list. M. G magnesium solid gives an A. So let's plus MG two positive and request. So the half reaction will be with the standard electrode potential values will be like this for oxidation. How you can see a node. It will be MD solid gives mg two positive plus two electron and the reduction or we can see cattle or cattle or reduction. The equation will be like this to positive Equus Plus two Electrons. Yes and I solid. So now we have to calculate the cell potential and It comes out to be 3.74. We -0.23 which is nearly Positive 2.14. What? This is the answer for the first part. Now coming on to the next part we have been given the reaction to H positive request. Splice FBI agent solid. He has H two cassius plus every two positive request. So half reaction will be the standard electoral potential values will be like this. So for oxidation or we can say for an order direction will be if he solid is if you do positive last two electrons this is a close and for the reduction or cattle the reaction will be to age positive Equus less to e negative H two cases. So now the we have to calculate the standard potential as follows. So it can be calculated like this. We have been given the values so the value for this And would it will be 0.45 world And for cattle. 80 world. It's No coming out to be your .45 years. And that is the answer for be part no see part the reaction given her is two and no three negative equals plus A positive acquis splice three See You phillips gives to know gases plus for H two liquid lists less today. See you to positive experts. So now we'll have to calculate the standard cell potential. And we have been given for the oxidation. The E note oxidation will be equal to minus zero point 34. We and the for the reduction, it's Euro 96 weeks. It comes out to be positive 0.623. And that is the answer for 3rd party and thanks.

Here we are continuing with our redox reactions. And so we have oxidation of MG two MG 2 plus. There's an acquis medium, add two electrons. My reduction is Nickel two plus. What I need to add is two electrons. And what I get is nickel in the solid form. And so the you know Of oxidation is equal to positive 2.37V. The not for reduction Is equal to negative, not .23 bolts. And so in order for the south is equal to the north of oxidation, add E note of reduction. And so in this case, what we have is 2.37V -123 Bolts. And this gives me positive 2.14 bolts For the first example. And now we've got to further examples where we get not .45 for part B Units of Vaults again and we get 9.62 in part c units of vaults again.

Were given these 2/2 reactions that occur within a galvanic cell in a party. We want to determine what the overall cell potential is for this galvanic cell. We know that four galvanic cells, the cell potential, has to come out to a positive value. And so, in order to do that we need we know that we need to reverse one of these reactions so that we cancel out the electrons on either side. So both of these half reactions represent reduction, but one of them will have to undergo oxidation In order for the cell potential to be positive, we have to keep the standard reduction potential that has a higher value, and so that will be the one that is reduced since it has a higher potential to be reduced. So the one with the lower standard reduction potential will be the one that is oxidized, since it has a lower potential to be reduced in there for a greater potential to be oxidized. So we see that since 1.50 is greater than negative 2.37 that means we have to flip the second reaction. So we're oxidizing solid magnesium and when we combine those 2/2 reactions in that way, we see that this is what that results in. And now we have to cancel out the electrons on each side. So in order to do that, we need to multiply the top half reaction by two in the bottom one by three so that we're cancelling out a total of 60 electrons on each side. And now the overall cell reaction becomes to a you three plus acquiesce was three mg solid goes to three OMG two plus a qui ists plus two solid gold. And now, since we re arrange these half reactions in this way, we see that the so potential comes out to a positive value, which is what we want for a galvanic cell. So he had those two potentials together 1.50 plus 2.37 We see that comes out to 3.87 volts, which is the answer for part A. And now, in part B, we need to use the inner city equation because we're at non standard conditions. We know that that means that they sell potential is equal to the standard cell potential minus 0.0 591 over end times The log of Q. We conform it an expression for Q. In terms of species concentrations, the quickest product is MG two plus. So the new Maria is a concentration of M G two plus and we Cuba to due to its tricky metric coefficient of three and the acquiesced reacting is a U three plus, and we square it so concentration of a U three plus squared. We're told what the value is of the nonstandard self potential, and that is 4.1 volts. This is equal to the standard cell potential, which we calculated important to be 3.87 volts and then subtract 0.0 591 over end. And we saw for the overall reaction, we had to cancel a total of six electrons on each side. So that is why N is equal to six times the log of Q. Where Q. Is this expression that we just form? We have a value for the concentration of M G two plus ions and acquia solution. You're told that that is 1.0 times 10 to the power of negative five Moeller and we cubit based on that Q expression. Then we divided by what we're solving for the concentration of a U three plus ions, and we square that. And now we rearrange this equation to isolate and solve for a U three. Plus, we subtract over these potentials and then divide by this negative ratio and we raise tend to the power of whatever that is. That result comes out to be so that we get rid of this log, and then we can easily rearrange by dividing and taking the square root to solve for the concentration of a U three plus I owns. And when we do that, we should get a final answer for the concentration of a U three plus to be about zero 0.40 Moller.


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