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The standard cell potential (E"cell) of the reaction below is -0.55 V: The value of AG? for the reaction isJlmol:I2 (s) 2Br" (aq)21- (aq) Brz ()1.1* 105Bs...

Question

The standard cell potential (E"cell) of the reaction below is -0.55 V: The value of AG? for the reaction isJlmol:I2 (s) 2Br" (aq)21- (aq) Brz ()1.1* 105Bs 0.555.5 x 10-6D. the temperature is needed for the calculationE. 1.3

The standard cell potential (E"cell) of the reaction below is -0.55 V: The value of AG? for the reaction is Jlmol: I2 (s) 2Br" (aq) 21- (aq) Brz () 1.1* 105 Bs 0.55 5.5 x 10-6 D. the temperature is needed for the calculation E. 1.3



Answers

Determine the overall reaction and its standard cell potential at 25 °C for this reaction. Is the reaction
spontaneous at standard conditions?
$\mathrm{Cu}(s)\left|\mathrm{Cu}^{2+}(a q) \| \mathrm{Au}^{3+}(a q)\right| \mathrm{Au}(s)$

In this problem were given with the reaction between aluminium metal and chromium tree on and were given the standards. Sell part in shell for this reaction is at 25 degrees C as their 1 92 fold, and we're has to find out the standard free energy change for the station. At 25 the receipt salutes the first right down the reaction equation, which is aluminium made till react to it. Chromium three plus I own and that will give us aluminium. Three players bless chromium. And to find out the standard free energy change, we can utilize the formal off dull G goingto minus in if e not hi in is the number off most of electrons transferred between the reactions, if is the further constant and in order the standard cell port and shield. So we know the value of Billy, which is given us 0.9 to unfold. And we know the fairly constant which is 4 96,084 Coolum fire more Elektrim And and then what we have to do is we have to just black in this values in this formula, and we need develop in, and we can find the well off in from the reintegration. So from the reaction in question again, see that on the oxidation number for aluminium in the reactive side is zero and in the product scientists plus three. So for aluminium, we have a changing oxidation number off plus three. And for chromium, it was plus three in the reactor inside and Syrian, the product side. That means there is a D equals in the oxidation stating gets of chromium. And from here you can see that the number of electrons transferred in between L. William and chromium is three more. So three moles of electrons are transferred in between that within the value for in his three mole. So now we know the value for all of the's increase years. We can just plug in the values to find out the value of Dell C, and that will give us minus three Mall electron and with multiplied by the value off averages 96 for 84 Colome Barmal let it run times 0.92 fold one vertical one Jewell per column if we can. Right down. It says zero pon meant to Jewell per column. Now, if we just multiply them. It could be to 66 to 95 point 84 and that unit will be chill so the electrons electrons will become turn. And there Coolum per column in the country. So on the jewelry windier. And if we want to convert it to kill a jewel, the value will be to 66 point, um, tune in six killer chill. So that's the standard free energy change for the reaction given.

So now we'll work on problem one fourteen 14 from chapter 20 and this problem, we're told that all takes that was based on silver plus silver and iron three iron to have cells in party. They asked us what is the standard EMF of the cell so we can get thes standard values from appendix E for silver plus going to silver and first iron three goingto iron too. And we see that their new 0.8 and 0.77 So we put the lower one second and we get a value of 0.3 faults. No, in part B, they ask us wish reaction occurs at the cathode and which at the anode of the cell. So the cathode reaction will be the reduction reaction. It will be this value here, which is for silver. So we have silver plus being reduced to silver neutral and then at the anode we have the reduction reaction, which is the value for 0.77 And we have iron too, being oxidized to iron three plus in part, see of this problem we're told to use the standard entropy values and appendix e in the relationship between cell potential and the free energy change to predict whether the standard cell potential increases or decreases when the temperature is raised above 25 degrees Celsius. So first, let's write down some equations. Everything. We know that Delta G eyes equal to negative and f e. And we also know that Delta G from the chapter on, uh, thermodynamics. We know that Delta Ji is equal to Delta H minus T Delta. No. Yes, So we can go ahead. And now look at the values that were asked for. You know that zero and jewels Permal Per kelvin. We have silver plus we have iron too, producing silver and iron three and the values that we have from appendix C A 73.93 for silver, plus 113.4 for iron too. 42.55 for silver and 2 93.3 for iron three. So we can see that these entropy values here the products are more waited, then the reactions. So this means that if temperature is raised for the change in entropy here is favorable. So that means if he is raised, then Delta G will increase. So if t increases, that means that Delta G they're well and will actually decrease in that it will become more negative. And that means that e will increase based on these two reactions here.

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.

This question, we need to use the equation DELTA G equals negative N F E. Here we have our balanced chemical reaction. And from this balanced chemical reaction we should be able to figure out what N is the number of moles of electrons that are transferred. If DELTA G standard equals negative N F E standard and the standard will be equal to our delta G in units of jewels. So will multiply that by 1000 divided by N what is in. We have silver. Going to silver plus, there are two of them. So that's one electron for each. That's two total. And we have iron three plus going to iron two plus. So that's one electron for each and two total. So and is two and then F is Faraday's constant, giving us an E standard of 0.3 to four volts sl which is what we have determined up here is equal to eat cathode minus the anodes cathodes where reduction is occurring, we see that iron three plus is going to iron two plus minus E an ode. Any anodes is the couple that we are interested in. So rearrangement gives us E an ode is equal to e. Cathode, which we will look up. The F E three plus deputy plus couple is +771 volts minus. When we move this over here and move this over here, well minus our 0.3 to four and we get 0.447 volts for the couple of But for the couple in question, the Silver chrome eight, going to the Silver Medal and chrome eight, An ion.


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