Nernst Equation 2.303RT/F = .059V


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Kyra LeRoy 1E
Posts: 52
Joined: Sat Jul 22, 2017 3:00 am

Nernst Equation 2.303RT/F = .059V

Postby Kyra LeRoy 1E » Thu Feb 15, 2018 9:23 pm

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!

John Huang 1G
Posts: 46
Joined: Thu Jul 13, 2017 3:00 am
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Re: Nernst Equation 2.303RT/F = .059V

Postby John Huang 1G » Thu Feb 15, 2018 10:23 pm

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.

Kyra LeRoy 1E
Posts: 52
Joined: Sat Jul 22, 2017 3:00 am

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

Postby Kyra LeRoy 1E » Thu Feb 15, 2018 11:38 pm

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.

Matthew Lee 3L
Posts: 51
Joined: Fri Sep 29, 2017 7:07 am

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

Postby Matthew Lee 3L » Fri Feb 16, 2018 12:06 pm

I would also like to know where the 2.303 came from. Thanks!

William Satyadi 2A
Posts: 31
Joined: Sat Jul 22, 2017 3:00 am

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

Postby William Satyadi 2A » Fri Feb 16, 2018 4:25 pm

The 2.303 comes from the conversion from using logQ instead of lnQ. Either equation works and they're equivalent.


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