Lyndon's Review Question 7 clarification

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Michael Novelo 4G
Posts: 64
Joined: Fri Sep 28, 2018 12:28 am

Lyndon's Review Question 7 clarification

Postby Michael Novelo 4G » Sat Mar 16, 2019 6:19 pm

Did anyone who attended Lyndon's review session know how he got E (knot) = 1.26 I know he said to use Delta G = -nFE since it is a state function but I'm confused on the math aspect of it. He said n=3 for the first line and , n=1 for the second line and the total should equal to n=2. are we subtracting delta G when n= 3 from delta G when n=1 or some other way? I got a very large number trying to do that but if anyone can clarify that'd be great thank you.

405112316
Posts: 62
Joined: Fri Sep 28, 2018 12:17 am

Re: Lyndon's Review Question 7 clarification

Postby 405112316 » Sat Mar 16, 2019 8:45 pm

delta G = delta G (eq 1) - delta G (eq 2)

You are subtracting the delta Gs from each other, but also be very aware of the negative signs from -nFE.
In total it should be:

-nFE(total) = -nFE(eq1) + nFE(eq2)

You will get 1.26 V

Michael Novelo 4G
Posts: 64
Joined: Fri Sep 28, 2018 12:28 am

Re: Lyndon's Review Question 7 clarification

Postby Michael Novelo 4G » Sat Mar 16, 2019 9:44 pm

alright I understand we subtract them from each other but what values are we plugging in to get the 1.26V? delta G = -(3 x 96,485 x 1.40) - (-1 x 96,485 x -1.69) is what I'm doing and that's wrong.


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