# Electrochemisty Concept- Help Plz

1. May 1, 2005

### DieCommie

If E(cell) = E(cathode)-E(anode)

and

E(cell) = (RT/nF)ln K

Then shouldnt E(cathode)-E(anode)=(RT/nF)ln k ?

I have a question asking to find the voltage of the cell and I am given a equilibrium constant.... which means i should use the second equation with "k" in it, and i get the right answer. But why does this value differ from the first equation's E(cell)?

I hope this explains my question, Thx!

2. May 1, 2005

### r3dxP

i thought E(cell) = E(anode) + E(cathode)

3. May 1, 2005

### DrMark

Welcome to the wonderful world of electrochemistry, perhaps the most confusing area of analytical chemistry. First off the cell potential can be expressed as:

Ecell =E0red + E0ox

or

Ecell =E0red - E0red

As to Ecell=RT/nF ln(K) This is incorrect. Think about it where are the standard redox potentials ? This is actually part of the nernst equation and used to calculate half cell effective potentials (i.e. concentration dependence).

for:

ox + ne- -> red

E=E0+RT/nF ln(K) ... K=[ox]/[red]

So then:

Ecell=Ered - Eox

Last edited: May 1, 2005
4. May 1, 2005

### DieCommie

Hmm... my text explicitly says Ecell0=(RT/nF)(ln k). The description is "Relating the standard emf of the cell to the equilibrium constant". This is why i thought it should equal the emf calculated from Ered+Eox. Where are the standard redox potentials? I dont know...seems like they should be there. I suppose i understand less than i thought Thanks for trying to help though.....

5. May 2, 2005

### DrMark

Sorry, some confusion here on my part. What your talking about is the standard emf, that is when all activities for products and reactants are 1. This then leads to E0=RT/nF ln(K) for a give redox reaction. The value of E0red - E0red should give the same result. Edit: Edit: Are both reduction potentials using the same reference electrode.

Last edited: May 2, 2005