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Nernst Equation 2.303RT/F = .059V

Posted: Thu Feb 15, 2018 9:23 pm
by Kyra LeRoy 1E
Can someone explain where the 2.303 comes from in the Nernst Equation when
(2.303RT)/F = .059 V
so
E°= - (.0592)/n log(Q)

& why does this use log instead of ln?

Thanks!

Re: Nernst Equation 2.303RT/F = .059V

Posted: Thu Feb 15, 2018 10:23 pm
by John Huang 1G
Hi!
Sorry, but I'm not sure in what context your first question is in, so I'm not sure I follow...


But for your second question, they could have used ln as well. The only difference between using log and using ln is the difference in constants. So it does not matter if they used log or ln, they will still get the same answer.

Re: Nernst Equation 2.303RT/F = .059V

Posted: Thu Feb 15, 2018 11:38 pm
by Kyra LeRoy 1E
A follow up: I'm not sure if we went over the 2.303RT/F in class but I was watching videos on the Nernst equation & it talked about this.

I think it may just be a conversion factor if you want to use log instead of ln.

Re: Nernst Equation 2.303RT/F = .059V

Posted: Fri Feb 16, 2018 12:06 pm
by Matthew Lee 3L
I would also like to know where the 2.303 came from. Thanks!

Re: Nernst Equation 2.303RT/F = .059V

Posted: Fri Feb 16, 2018 4:25 pm
by William Satyadi 2A
The 2.303 comes from the conversion from using logQ instead of lnQ. Either equation works and they're equivalent.