## Nernst Equation 2.303RT/F = .059V

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

### Nernst Equation 2.303RT/F = .059V

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)

Thanks!

John Huang 1G
Posts: 46
Joined: Thu Jul 13, 2017 3:00 am
Been upvoted: 1 time

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

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

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

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