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Julia Ward 2C
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Joined: Wed Nov 16, 2016 3:03 am


Postby Julia Ward 2C » Sun Feb 25, 2018 8:11 pm

Just to clarify--the value 0.0592 V can only be used at 25 degrees C and with log Q? Can it be used with ln as well?

Ozhen Atoyan 1F
Posts: 50
Joined: Thu Jul 27, 2017 3:01 am
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Postby Ozhen Atoyan 1F » Sun Feb 25, 2018 8:14 pm

Yes because to get that number we use 298K, which is 25 degrees Celsius.

Ridhi Ravichandran 1E
Posts: 35
Joined: Sat Jul 22, 2017 3:01 am


Postby Ridhi Ravichandran 1E » Sun Feb 25, 2018 8:17 pm

It cannot be used for ln as well. For ln, there is a different constant (it is listed in the book).

Sean Monji 2B
Posts: 66
Joined: Fri Sep 29, 2017 7:06 am


Postby Sean Monji 2B » Sun Feb 25, 2018 8:18 pm

Sadly you cant use ln, as the constant ratio from ln to log was used to find the constant. It would be a different constant if you are using ln instead of log.
EDIT: ^^^

Shane Simon 2K
Posts: 30
Joined: Fri Sep 29, 2017 7:05 am


Postby Shane Simon 2K » Sun Feb 25, 2018 9:05 pm

And just to clarify because I know that this confused me initially as well, the lnQ is converted into a logQ using the conversion factor lnx = 2.303 logx. So the 0.0592 value is calculated by (2.303*R*T/F) which would be 2.303*8.314*298/96485.3.

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