# Electrochemistry question: n value in ΔG° =-nFE°cell

1. Jan 13, 2016

### rockinrack

If I want to find the standard ΔG for a reaction in an electrochemical cell, I can use the equation ΔG° =-nFE° where n is the number of moles transferred, F is Faraday's constant, and E° is the standard emf of the cell. If I know the standard reduction potentials of the redox reactions happening in each half-cell, I can calculate the standard emf with the equation E°cell=E°reduction, anode-E°reduction, cathode. Standard reduction potentials are calculated under standard conditions, so all concentrations are at 1 molar. Now, consider the redox reaction:
2Fe3+ + 2Cl- ---> 2Fe2+ + Cl2
The two half reactions are:
2Fe3+ + 2e- ---> 2Fe2+
2Cl----> Cl2 + 2e-

So if I calculate E°cell and want to use it to find ΔG°, I just have to determine the value of n, the number of moles of electrons transferred, and plug the values into ΔG° =-nFE°cell. Two electrons are transferred in the balanced equation above, so I could understand using n=2. However, this is the number of moles of electrons transferred for every 2 moles of Fe3+ and Cl- that react, as you can see from the equation. I am trying to calculate the standard ΔG, in which all concentrations are 1 molar, so n=1, correct? 1 mole of electrons is transferred for every one mole of Fe3+ and Cl- that react, and since we are dealing with concentrations of 1 molar, the "concentration" of electrons being transferred is 1 molar (I put concentration in quotes because the electrons themselves are dissolved in solution, but I refer to the number of moles in terms of concentration just to make my point clear.

So does anyone know if n should be 1 or 2? Thanks!

2. Jan 14, 2016

### James Pelezo

n = number moles electrons involved in 'balanced' oxidation-reduction process. Example ...
Given
Cr3+ + 3e- => Cro; Eo = - 0.74v (Oxidation Rxn)
3Ag+ + 3e- => 3Ago; Eo = + 0.80v (Reduction Rxn)
Cro => Cr3+ + 3e-
3Ag+ + 3e- => 3Ago
Cro + 3Ag+ => Cr3+ + 3Ago (Net Oxidation-Reduction Rxn)

∆Go = -nFEo
n = 3 moles e- for balanced redox equation
F = 96,500 Coulombs
Eo = EoRedn - EoOxdn = (+0.80v) - (-0.74v) = +1.54v
∆Go = - (3)(96,500 C)(1.54v) = -445,830 volt-Coulombs = -445,430 joules = -446 Kj (3 sig. figs)
For your problem, n = 2.