## n in NFE

$\Delta G^{\circ} = -nFE_{cell}^{\circ}$

Karina Kong 2H
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### n in NFE

For n in the equation Delta G=-nFE, how do you know how many moles you need?

Sidharth D 1E
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### Re: n in NFE

N should be the number of electrons transferred in that scenario.

lilymayek_1E
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### Re: n in NFE

When you are solving for a redox reaction, and you split the original reaction into its respective half-reactions, you have to balance both the molecules involved and the electrons transferred. That final balanced value of electrons transferred is your n value. The top answer in this forum helps!
https://chemistry.stackexchange.com/questions/5122/how-is-it-determined-how-many-electrons-are-transferred-in-redox-reactions

J Medina 2I
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### Re: n in NFE

n is the number of electrons transferred which cannot be determined until your redox half-reactions are balanced. These half-reactions should have the same number of electrons transferred on opposite sides.

SarahCoufal_1k
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### Re: n in NFE

n is the number of electrons transferred in the redox equation. When you do your reduction half reaction there is a certain amount of electrons transferred and same case for the oxidation half reaction. You then have to multiply the equations to balance the electrons. when they match and cancel that number is n

Jesse H 2L
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### Re: n in NFE

the moles of electrons transferred

Kaitlyn Ang 1J
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### Re: n in NFE

After you balance the redox reactions so that the "number" of electrons from the half reactions cancel, that common "number" of electrons is your n

Posts: 104
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### Re: n in NFE

n is the number of moles of electrons that are transferred.

Nawal Dandachi 1G
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### Re: n in NFE

n is the number of moles of electrons transferred

Sanjana K - 2F
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### Re: n in NFE

Remember that it's the number of electrons after you balance your equations (so the number of electrons transferred in the cathode half reaction should equal the number of electrons transferred in the anode half reaction, which should equal your value of n).

Charlene Datu 2E
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### Re: n in NFE

Since it's n is the number of moles of electrons in the reaction after both half-reactions are balanced, it's good to note that it doesn't matter which half-reaction you take this value from. This value should be the same in both half-reactions.

Hope Hyland 2D
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### Re: n in NFE

n is the number of electrons being transferred (after balancing the half-reactions)

Ryan Yoon 1L
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### Re: n in NFE

n is the whole number mole value of electrons that were shared in the half reactions.

Ryan Yoon 1L
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### Re: n in NFE

n is the whole number mole value of electrons that were shared in the half reactions.

ramiro_romero
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### Re: n in NFE

n is how many moles of electrons are transferred. First you need to determine the half rxns, then balance the oxidation and reducing reactions so that the electrons cancel out (and don't appear in overall rxn). Finally, once both half rxns are balanced, the coeffecient of both electrons is your value for n.

Kaylee Clarke 1G
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### Re: n in NFE

n is referring the electrons transferred after balancing the half-reaction!

Altamash Mahsud 1I
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### Re: n in NFE

The n in that equation is not the number of moles, but it is instead the number of electrons being transferred in the redox reaction. You can get this number by balancing the redox half reactions and combining these half reactions to get the completed redox reaction, which will tell you the number of transferred electrons.

ThomasNguyen_Dis1H
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### Re: n in NFE

n is the number of electrons transferred in the final equation

205154661_Dis2J
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### Re: n in NFE

N stands for the number of electrons transferred. So you would have to separate your redox rxns into the reduction and oxidation rxn in order to see how many electrons are being transferred after you balanced both reactions.

chimerila
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### Re: n in NFE

It can be kind of confusing, especially since I'm so used to seeing "n" and thinking "MOLES". But to help myself, I like to remember that we're using this equation when solving problems related to electrochemistry, so "n" actually signifies electrons.

"E"lectrochemistry ------> "e"lectrons :)

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