What is the Faraday Constant and Why is it Important?

[Jhenrique]

The faraday constant is prayed:
$$\\ 1\; \text{F} = \frac{96 500\;\text{C}}{1\;\text{mol}}$$
(approximately...)


I'm wrong if I say that the faraday constant is:
$$\\ 1\; \text{F} = \frac{96 500\;\text{C}}{1\;\text{mol e}}$$
?

 

[adjacent]

You are right, mole and mol is the same.

 

[Jhenrique]

You are right, mole and mol is the same.

I'm not talking about mol and mole, I'm talking about mol and mol of electron!

 

[Born2bwire]

Then that is incorrect then. It is a constant and the units are specifically defined. The fact that it is representative of the charge of a mole of electrons does not change its units to reflect that it is of electrons or the fact that it is just a number.

 

[Jhenrique]

Hummm, but 1 mol e ≠ 1 mol p, so the electrical charge associated to 1 mol of eletrons is very different to charge associated to 1 mol of protons.

looks this article: http://en.wikipedia.org/wiki/Stoichiometry#Determining_amount_of_product

Make very much sense say x mol A, y mol B and z mol C than simply say x mol, y mol or z mol.

Until today I don't understand why what is being counted is omited of front of mol...

 

[Born2bwire]

The Faraday constant for a mole of electrons is the same as a mole of protons since it does not carry a sign as far as I can see. The Faraday constant could be talking about mole of sodium ions or many other things. We do not include such things in the units. It is just a bookkeeping factor you have to carry forward on your own. It is done all the time in physics. Coulomb's Law is for unit charges but we easily apply a constant prefactor to adjust it for bodies with multiple electrons worth of charge. Yet we do not show this in the units we choose for e the equation.

 

[dauto]

Hummm, but 1 mol e ≠ 1 mol p, so the electrical charge associated to 1 mol of eletrons is very different to charge associated to 1 mol of protons.

looks this article: http://en.wikipedia.org/wiki/Stoichiometry#Determining_amount_of_product

Make very much sense say x mol A, y mol B and z mol C than simply say x mol, y mol or z mol.

Until today I don't understand why what is being counted is omited of front of mol...

Saying mol A, mol B, etc.. is like saying the dimensions of an object are 3 meters high, 2 meters wide and 5 meters long. high, wide and long are added to describe what is being measured but they are not part of the unit.

 

[Jhenrique]

Saying mol A, mol B, etc.. is like saying the dimensions of an object are 3 meters high, 2 meters wide and 5 meters long. high, wide and long are added to describe what is being measured but they are not part of the unit.

*I think* that when about particles, not should be omited, IMO. Cause you have an infinity of particles, elements and molecules all with different properties...

 

[nasu]

You are missing the whole point of introducing this constant.

What Faraday has observed is that the SAME amount of charge (or multiples of it) is necessary to go through the solution in order to separate or deposit one mole of ANY substance.
This "ANY" is reflected in the unit of C/mole. Any mole.

The same charge is carried by a mole of either electrons, protons, sodium ions, chlorine ions, etc.
Or twice as much charge is carried by one mole of calcium ions, copper II ions, etc.

 

[D H]

Or even a mole of moles, with each mole being stripped of exactly one electron.

 

[Jhenrique]

You are missing the whole point of introducing this constant.

What Faraday has observed is that the SAME amount of charge (or multiples of it) is necessary to go through the solution in order to separate or deposit one mole of ANY substance.
This "ANY" is reflected in the unit of C/mole. Any mole.

The same charge is carried by a mole of either electrons, protons, sodium ions, chlorine ions, etc.
Or twice as much charge is carried by one mole of calcium ions, copper II ions, etc.

I don't believe!

 

[nasu]

OK. No problem.

 

[UltrafastPED]

I don't believe!

See section 2, on Michael Faraday:
http://iopscience.iop.org/1468-6996/14/4/043502/pdf/1468-6996_14_4_043502.pdf