What to use for n in -nFE?


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HilaryPhan_3N
Posts: 6
Joined: Fri Sep 26, 2014 2:02 pm

What to use for n in -nFE?

Postby HilaryPhan_3N » Wed Feb 04, 2015 8:58 pm

How would I know what to use as n for Gibbs free energy=-nFE? Which amount of moles would I use if there are different coefficients for each compound in a reaction?

Niharika Reddy 1D
Posts: 127
Joined: Fri Sep 26, 2014 2:02 pm

Re: What to use for n in -nFE?

Postby Niharika Reddy 1D » Wed Feb 04, 2015 9:12 pm

n is the moles of electrons transferred. If you write the balanced redox reaction correctly, both half reactions will have the same number of moles of electrons gained or given off, so it shouldn't matter which one you use.
Last edited by Niharika Reddy 1D on Wed Feb 04, 2015 9:13 pm, edited 1 time in total.

martha-1I
Posts: 76
Joined: Fri Sep 26, 2014 2:02 pm

Re: What to use for n in -nFE?

Postby martha-1I » Wed Feb 04, 2015 9:13 pm

I had this same question and it was answered in a forum I started
Chem_Mod wrote:The easiest way to find out n is to look how many electrons are transferred in your balanced HALF-reactions. For example, in the Copper/Zinc cell, n=2
In the Silver/Zinc cell, silver transfers 1 electron and zinc transfers 2. Since these don't match, the silver reaction must be multiplied by 2 before we could add up the half-reactions. In this case, n=2.
In an Aluminum/Zinc cell, aluminum transfers 3 electrons and zinc transfers 2. To make these match, aluminum reaction is multiplied by 2 and zinc by 3. Then, n=6.

General rule: Find the number of electrons in each balanced HALF-reaction. If they match, that is n (First example). If they don't match, take the lowest common multiple, and that is n (Second/third examples).


Basically the n is the amount of electrons that are being added to the reduction 1/2 reaction and the oxidization 1/2 reaction. Since we balance them, they should end up being the same.


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