Midterm 2012 Q: 7A


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Antonio Melgoza 2K
Posts: 33
Joined: Fri Sep 25, 2015 3:00 am

Midterm 2012 Q: 7A

Postby Antonio Melgoza 2K » Sun Feb 07, 2016 6:43 pm

How do you get -0.35? I got -2.02 for my answer
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Cesar Martinez 2A
Posts: 10
Joined: Fri Sep 25, 2015 3:00 am

Re: Midterm 2012 Q: 7A

Postby Cesar Martinez 2A » Sun Feb 07, 2016 7:23 pm

I'm not quite understanding how you're getting -2.02 with those values. You're given Ecell=-0.689 and Eanode=0.34

The equation you will use is Ecell=Ecathode-Eanode, where you want to find the value of Ecathode.

Rearrange the equation to solve for Ecathode:

Ecathode=Ecell+Eanode

So, plugging in the numbers for Ecell and Eanode:

(Ecathode)=(-0.689v)+(0.34v)

Ecathode=-0.35

Antonio Melgoza 2K
Posts: 33
Joined: Fri Sep 25, 2015 3:00 am

Re: Midterm 2012 Q: 7A

Postby Antonio Melgoza 2K » Sun Feb 07, 2016 7:55 pm

I was dividing 0.689 by 0.34, I get it now. Thanks!


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