Problem 6L.1


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Theodore_Herring_1A
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Problem 6L.1

Postby Theodore_Herring_1A » Mon Feb 25, 2019 4:47 pm

In the problem 6L.1 (7th Edition),

Calculate the standard reaction Gibbs free energy for the following cell reactions:

(a) 2 Ce4+(aq) + 3 I−(aq) →2 Ce3+(aq) + I3−(aq), Ecell°=+1.08V,

What value n do I use for Ecell = -nFE?

marisaimbroane1J
Posts: 69
Joined: Fri Sep 28, 2018 12:23 am

Re: Problem 6L.1

Postby marisaimbroane1J » Mon Feb 25, 2019 4:50 pm

Your value for n will be the number of moles of e- in the balanced half reactions. So write out the two half reactions and then make sure the number of e- is the same for both of them. This number will be n in the equation.

David Effio 1H
Posts: 38
Joined: Wed Feb 14, 2018 3:01 am

Re: Problem 6L.1

Postby David Effio 1H » Mon Feb 25, 2019 4:51 pm

To find the value of n, the number of electrons, you should try to find what the half reactions will look like. Make sure the charges, reactants and products are all balanced first. You will then be able to plug in n for the formula.

Kirsty Star 2H
Posts: 48
Joined: Fri Sep 28, 2018 12:24 am

Re: Problem 6L.1

Postby Kirsty Star 2H » Mon Feb 25, 2019 5:26 pm

Once you do what those people mentioned, I believe the balanced reaction comes out to 2 e-. So n = 2

Angela Grant 1D
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Joined: Fri Sep 28, 2018 12:25 am

Re: Problem 6L.1

Postby Angela Grant 1D » Mon Feb 25, 2019 7:23 pm

You'd use n = 2 electrons because when you split up the half-reactions you get:

Oxid.: Ce4+ + e- --> Ce3+
Red.: 3I- --> I(3)- + 2e-

Multiply the oxidation half-reaction by 2 to balance out the electrons and cancel them out, leaving you with two electrons getting cancelled.


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