n in NFE
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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
https://chemistry.stackexchange.com/questions/5122/how-is-it-determined-how-many-electrons-are-transferred-in-redox-reactions
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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.
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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
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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
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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).
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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.
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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.
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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.
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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.
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 :)
"E"lectrochemistry ------> "e"lectrons :)
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Re: n in NFE
n is the number of moles of electrons lost by what is oxidized and gained by what is reduced.
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Re: n in NFE
You can find n by looking at the redox reaction and the number of electrons that cancel out on each side of the reaction
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Re: n in NFE
n will be the moles of electrons transferred. Remember to balance your redox reaction to get the correct number of moles of e- transferred.
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Re: n in NFE
Once you balance the redox reaction or even balance the half equations, you should be able to see the coefficient in front of the electrons or see how many electrons are transferred. This number would be the n in the NFE equation.
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Re: n in NFE
Once you balance your oxidation and reduction half reactions, n is going to the number of electrons that transferred in the equation. Since its balanced, it should be the same in both reactions.
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Re: n in NFE
N is the number of electrons transferred in the redox reaction. Be sure to balance the reaction to get the correct number.
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Re: n in NFE
The value of n is going to be the amount of electrons transferred:) you can find this out by doing the redox half reactions!
Hope this helps!
Hope this helps!
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Re: n in NFE
You would plug in the number of moles of electrons being transferred in the redox reaction for the variable n.
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Re: n in NFE
To find n, write the two half reactions with equal charges. The number of electrons transferred in the overall balanced reaction is the value of n.
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Re: n in NFE
n is the number of electrons that are involved in our redox reaction. Make sure to balance out the charges though.
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Re: n in NFE
How do I calculate the n in NFE? Do I have to balance the half-reactions? What if the electrons is 0? Would that ever happen?
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Re: n in NFE
Susanna Givan 2B wrote:How do I calculate the n in NFE? Do I have to balance the half-reactions? What if the electrons is 0? Would that ever happen?
You can calculate n (number of electrons transferred) by balancing the redox reactions. I'm pretty sure the number of electrons wouldn't be 0 since it wouldn't be a redox reaction if no electrons were transferred.
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Re: n in NFE
N is the number of the electrons being transferred for that given reaction! When you balance the redox reactions and break down how many e- are being transferred between molecules/sides, you can just plug that number in.
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Re: n in NFE
N is the number of electrons transferred, but it cannot be determined until the reaction is balanced.
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Re: n in NFE
you would want to balance the half reactions and the number of electrons would be the value of n.
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Re: n in NFE
n represents the moles of electrons transferred. Therefore, you can look at the balanced redox reaction and see how many moles of electrons were transferred and use that value for n. Hope this helps!
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Re: n in NFE
"n" in this equation is referring to the number of moles of electrons transferred in the redox reaction. To see how many moles of electrons you have, you must first balance the two half reactions to see where the electrons are going, and how many of them there are. Once you have balanced them, n should be easy to find and plug in.
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Re: n in NFE
Recall that dG is energy used to perform work. In studies of electricity and circuits, work is often expressed as the (dot product) between Charge and Voltage . In dG = -nFE, F represents the ratio between charge and one mole of electron, and E represents voltage. As a logical continuation of this, n would thus need to represent the mole of ELECTRONS.
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Re: n in NFE
Hello!
The n in -nFE stands for the number of electrons being transferred, hope it helps!
The n in -nFE stands for the number of electrons being transferred, hope it helps!
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Re: n in NFE
n is the number of electrons transferred in the redox reaction. Just remember to make sure that the equations for the oxidation and reduction reaction need to have equal moles of electrons transferred.
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Re: n in NFE
n is equal to the number of electrons transferred in the redox reaction! I find that this is easiest to calculate when you balance each half reaction, as it will explicitly tell you how many e- cancels out when the entire redox reaction is put together.
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Re: n in NFE
n is the number of electrons transferred which can be determined from the half-balanced redox reaction.
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Re: n in NFE
n is the number of electrons transferred, which you would determine after balancing the half reactions.
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Re: n in NFE
N does not refer to moles of the substances I believe, but instead the moles of electrons transferred.
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Re: n in NFE
n would be the number of electrons transferred, which you can obtain by balanced half reactions/a redox reaction.
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