Chemistry problem

[pink84]

Can anyone tell me where the number, 2.303, comes from?

Q. Given that the E(standard) for Au+3/Au is 1.50V and the E for Li+/Li is -3.05V, what is Ecell?

Sol.

Ecell= Ecell(standard)-RT/nF logQ
Ecell= Ecell(Standard)-(0.0591/n)logQ

0.0591 is a constant equal to 2.303RT/F at 298K


I don't know how to get the 2.303.

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[airbornemedic]

em you do know you calculating a non-standard EMF for a cell right? why would you use 298K for the second part? Estandard uses that. you need a new temperature besides 298k for the -RT/nF logQ, also go find Q using a balanced redox reaction, where the exponent for each concentration is spectrometric coefficient like any equalibrium equation. once you find that get the temperaure of your non-standard cell, you should be able to solve

is that your own solution or did the book give you those two steps and an answer of 2.303?

 

[Short Round]

Can anyone tell me where the number, 2.303, comes from?

Q. Given that the E(standard) for Au+3/Au is 1.50V and the E for Li+/Li is -3.05V, what is Ecell?

Sol.

Ecell= Ecell(standard)-RT/nF logQ
Ecell= Ecell(Standard)-(0.0591/n)logQ

0.0591 is a constant equal to 2.303RT/F at 298K


I don't know how to get the 2.303.

You don't calculate the 2.303...

They are saying that at 298*K:

(.0591/n) is equal to the value (2.303RT/nF), where R = 8.314 J/mol-K , T = temp in K(so 298*), n = # electrons transferred, F=Faraday's constant(96,485 Coulombs/mol).

This is part of the Nernst Equation:

E=E(standard) - (RT/nF)lnQ
or
E=E(std) - (2.303RT/nF)logQ

So if you're given a problem at 298*K you can substitute (.0591/n) for (2.303RT/nF) when you're doing the calculations:

E=E(std) - (.0591/n)logQ

I think they just want you to be able to recognize the Nernst Equation, since I don't think you can use calculators on the PCAT and thus figuring out actual values by hand would be