[garijniv]

Hi

My level is around high school level, but I tend to go quite deeply.

Currently i'm looking into the Nernst equation, in the context of the electrolysis of water.

Given this half equation

Anode (oxidation) 2 H2O(l) → O2(g) + 4 H+(aq) + 4e−  E(ox)=-1.23

and

F = faradays constant = 96,485 C/mol
T = 298K (25C)
n= number of electrons transferred per cell reaction
R = gas constant = 8.314 J/mol K


n=4 ('cos oxygen get oxidised, going from oxidation state of -2 in H2O, to an oxidation state of 0 in O2, and there's two oxygen atoms).

Nerst equation:   E_nernst = E_SEP - (RT/nF)lnQ

Q = reaction quotient.  Let's suppose hypothetically, Q=2  (which might not be a realistic value for Q but let's say 2 for now)


-1.23 - ((8.314*298)/(4*96.485)) * ln(2) = -5.6797

But if I put those values into

http://calistry.org/calculate/nernstEquation

I get -1.2388



Why the difference?

[Borek]

Beware of the comma in the Faraday's constant.

[garijniv]

Beware of the comma in the Faraday's constant.

-1.23 - ((8.314*298)/(4*96485)) * ln(2)
= -1.23444971251084174955

Amazing, thanks, that solves that

[garijniv]

Actually funnily enough even after fixing F, I still get different results from a calculator and from that website.

-1.23 - ((8.314*298)/(2*96485)) * ln(2)
= -1.23889942502168349909

whereas on that site it's = -1.230000184482319   https://i.imgur.com/aIlvXzP.png

And even bigger difference if I use a low value for Q

So


-1.23 - ((8.314*298)/(2*96485)) * ln(0.000001)
= -1.05262049995275606301


Whereas on that site  https://i.imgur.com/af86hYM.png  = -1.22999632297800

[garijniv]

ah sorry is fine.. I think I had entered the wrong number in the wrong box for n.. Besides that I should have done n=4!