14.27


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Liam Giffin 2B
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14.27

Postby Liam Giffin 2B » Mon Feb 06, 2017 12:20 am

For question 14.27 in the textbook I don't understand why we can't just add the potentials of reactions A and B (shown in solution manual answer) together. The problem and solution manual solution are shown below.
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1427A2.jpg
1427A1.jpg
1427Q.jpg

Manpreet Singh 1N
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Re: 14.27

Postby Manpreet Singh 1N » Mon Feb 06, 2017 8:27 am

Hello,

For this problem, we want the standard potential for U^4+ +4e- ----> U

since we don't have that exact standard potential, we can use the information we have so we have:
U^4+ +e- ----> U^3+ E^o=-0.61
U^3+ +3e- -----> U E^o=-1.78 When we add, the U^3+ cancels out because they are on opposite sided of the reaction.

We know the formula is for E is E=-nE^oF so we add them and make the equal to the one we want. There are 4 moles in the one we want, there is 1 mol is the first half reaction and 3 in the second and since F is a constantsI am gonna leave it like it for simplicity sake.

-4E*F=(-1(-0.61)*F)+(-3(-1.78)*F) and you solve for E

We cant just add the values because there are different values of moles, and this way it is more precise.


hope that helps :)


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