2013 Midterm 8B

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Kira Conde 2O
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Joined: Fri Jul 15, 2016 3:00 am

2013 Midterm 8B

Postby Kira Conde 2O » Tue Feb 14, 2017 4:37 pm


I'm having a really hard time getting the value they specified for [Fe2+] for this problem. I get how we got from the Nerst Equation to

0.05916/2 log [Fe2+](0.015)^2/(0.015)^2 =2.05-2.20

but I'm not sure why you cant just simplify 0.05916/2 to 0.02958 instead of multiplying by two and then dividing by 0.05916. I thought they would end up with the same answer, but I only get the right answer when I multiply by two and divide by 0.05916.

Thank you!

Posts: 15
Joined: Fri Sep 25, 2015 3:00 am

Re: 2013 Midterm 8B

Postby 404639305 » Tue Feb 14, 2017 8:20 pm

you can simplify it first. When you simplify it you get 0.02958xlog [Fe]0.015^2/0.015^2=-0.15. So now you have to get log alone so you divide it over and get
log[Fe]0.015^2/0.015^2=-5.07, and then you just solve it normally!
Hope that helps.

Angela To 2B
Posts: 20
Joined: Wed Sep 21, 2016 2:55 pm

Re: 2013 Midterm 8B

Postby Angela To 2B » Tue Feb 14, 2017 8:45 pm

Is there a difference between using ln vs log in the equation? Because the solutions manual and the course reader has written it both ways.

204751840 2E
Posts: 19
Joined: Tue Oct 04, 2016 3:00 am

Re: 2013 Midterm 8B

Postby 204751840 2E » Wed Feb 15, 2017 12:08 am

Angela, both equations work. For whatever reason, they sometimes come out to slightly different numbers, but so long as you show your work and and your calculations are correct you should receive credit. One of the past midterms (either 2016 or 2013, I can't remember) has a situation like this where it says to give credit for two different answers depending on which method was used to get there.

Emily Wang 1H
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Re: 2013 Midterm 8B

Postby Emily Wang 1H » Wed Feb 15, 2017 8:22 am

You can use either log or ln and the answers will turn out to be the same. Just remember that ln is log base e. If you want to convert between the two, lnx = 2.303logx.

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