n in ∆G = -nFE
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n in ∆G = -nFE
I know that n is the number of moles in the reaction. How are you supposed to find this out when given a chemical reaction? Thank you!
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Re: n in ∆G = -nFE
n refers to the number of moles of electrons being transferred. When you are presented with a redox reaction, you need to balance the reaction and find how many e- are being gained/lost. n should be the coefficient in front of e-.
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Re: n in ∆G = -nFE
Like Will said above, n is the number of moles of electrons being transferred (and not moles of reaction like in previous units). To find n, split your overall chemical reaction into an oxidation half reaction and a reduction half reaction. N should be the number in front of the e- after your half reactions are balanced.
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Re: n in ∆G = -nFE
The n value we are accounting for is the number of moles of electrons after balancing the redox reaction. Making sure the reaction is balanced is necessary when determining the n value for the equation.
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Re: n in ∆G = -nFE
Its the number of moles of electrons moving around. if 2e- are being taken from Cu and given to say, Fe2+, then n=2
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Re: n in ∆G = -nFE
It really depends on the type of question you're being given. For instance, if they gave you all the other values, you could easily solve for n.
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Re: n in ∆G = -nFE
From what Prof Lavelle said in lecture, "n" is equal to the number of electrons being transferred in a redox reaction. He was mentioning that some students get confused with if its positive or negative (because some compounds lose e- and some gain e-), but you have to look at the broad prospective that a certain amount of e-, "n", are being moved in the reaction - you use that number for n in the equation.
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Re: n in ∆G = -nFE
When balancing half reactions you will see how many electron are being transferred from one species to another. So this number of electrons being transferred will be the n value you use for the gibbs free energy equation or the Nernst equation.
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Re: n in ∆G = -nFE
n is the number of electrons transferred in the reaction. You can find this by balancing the half-reactions.
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Re: n in ∆G = -nFE
Hi! Just to add on, n should be the same in both oxidation and reduction half-reactions. If they are not, that means you will not get the proper balanced overall reaction.
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Re: n in ∆G = -nFE
This refers to the moles of electrons that are being transferred in the reaction. To find this, you can write out the half reactions for each molecule, and then make sure you take into account coefficients in the equation if necessary.
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Re: n in ∆G = -nFE
n in electrochem is the number of moles of electrons that are transferred during the reaction.
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Re: n in ∆G = -nFE
I think n refers to the number of electrons transferred in the reaction. You could find this through balancing half reactions.
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Re: n in ∆G = -nFE
n is the moles of electrons being transferred and can be found from the balanced half reactions.
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Re: n in ∆G = -nFE
n= moles of electrons so in the chemical equations however many electrons are transferred equals the number of moles of electrons transferred.
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Re: n in ∆G = -nFE
n in the equation ∆G = -nFE is the number of electrons that are being transferred, which you can find by balancing the half-reactions
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Re: n in ∆G = -nFE
after balancing the half reactions, you can find the value of n which corresponds to the number of electrons being transferred
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Re: n in ∆G = -nFE
n refers to the number of e- being transferred. You can find this once you balance the equation (make sure the charges are balanced as well).
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Re: n in ∆G = -nFE
n is the number of electrons transferred, which you can find once you create the half reactions. Hope this helps!
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Re: n in ∆G = -nFE
I think that in this formula, n actually represents the number of electrons transferred, but I am not entirely sure about that
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Re: n in ∆G = -nFE
In the equation ∆G° = -nFE, the n refers to the number of electrons transferred, rather than moles which is how we would normally use n in other equations. To find the number of electrons transferred you need to look at the half reactions and see the number of electrons when both half reactions have the same amount to electrons transferred, which is found after multiplication of the equations (if necessary).
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Re: n in ∆G = -nFE
N refers to the number of moles of electrons transferred, which is determined between the two half reactions.
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Re: n in ∆G = -nFE
I solve for n by subtracting the moles of gas on the left from the moles of gas on the right.
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Re: n in ∆G = -nFE
The n is referring the moles of electrons being transferred so when you look at a redox reaction first look at the charges from the left and the right. If the right is stable and the left is 3- then there was a transference of 3 electrons.
Re: n in ∆G = -nFE
n is the moles being cancelled out as people have mentioned above. It is super simple with practice and is just the number of moles we scaled and cancelled out when we add the half reactions together.
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Re: n in ∆G = -nFE
In this equation, n refers to the number of moles of electrons that are transferred in a given chemical reaction. This value can be figured out when balancing your half reactions.
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Re: n in ∆G = -nFE
I believe the n refers to the number of moles of electrons transferred in the half redox reactions.
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Re: n in ∆G = -nFE
n stands for the number of electrons (in moles) that are being transferred. You find this by balancing the half reactions.
Last edited by Violet Kwan 3H on Sun Feb 28, 2021 10:43 pm, edited 1 time in total.
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Re: n in ∆G = -nFE
N is the number of electrons that are transferred in the reaction so it changes from reaction to reaction. You can find it by balancing the half-reactions.
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Re: n in ∆G = -nFE
n is the number of moles of electrons that are being transferred. You find n after balancing the reaction and splitting it into half-reactions.
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Re: n in ∆G = -nFE
In this case, n represents the number of e- being transferred. You can find this by balancing the half-reactions.
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Re: n in ∆G = -nFE
In previous concepts, n signified number of moles of reactant/product. In this chapter, n would actually signify the number of moles of electrons transferred between.
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Re: n in ∆G = -nFE
In this case, n is the number of electrons transferred in the equation. It'll be the coefficient for e- after you've completely balanced the equation.
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Re: n in ∆G = -nFE
I would just look at the number of moles of electrons that are transferred in the reaction that you are looking at, that is what n refers to here
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Re: n in ∆G = -nFE
n is the number of moles of electrons that are being transferred after balancing the redox reaction.
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Re: n in ∆G = -nFE
n refers to the number of moles of electrons transferred in the chemical reaction. In order to find the value of n, balance the half-reactions and determine how many electrons are needed when putting the two half-reactions together into one balanced chemical reaction.
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Re: n in ∆G = -nFE
n is the moles of electrons being transferred. You can figure it out thru balancing your half-reactions and then the complete reaction
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Re: n in ∆G = -nFE
N is referring to the number of electrons that are being transferred in a given chemical reaction. We can find this when we write out the half reactions and multiply so that we have even numbers of electrons on each side being transferred, and then you can just plug that number into the equation and solve.
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Re: n in ∆G = -nFE
You balance your half reactions to see how many electrons are transferred in the whole reaction! :))
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Re: n in ∆G = -nFE
n is actually number of moles of electrons TRANSFERRED in the reaction.
Example: X 2+ + Y --> Y 2+ + X
By looking at the charges we can see the transfer of two e-, therefore n = 2 here.
Example: X 2+ + Y --> Y 2+ + X
By looking at the charges we can see the transfer of two e-, therefore n = 2 here.
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Re: n in ∆G = -nFE
Hi, n actually refers to the number of mol of electrons being transferred, which you can find through a balanced reaction.
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Re: n in ∆G = -nFE
Hi,
n refers to the moles of ELECTRONS transferred in a redox reaction. In order to find this amount, you would need to split the overall reaction into two half reactions (oxidation and reduction) and then balance them accordingly. However, when you're balancing you need to ensure the number of electrons on the reactant side and product side cancel out when you combine the two equations. That number of electrons you get here will be n in the equation you referenced.
Hope this helps! :)
n refers to the moles of ELECTRONS transferred in a redox reaction. In order to find this amount, you would need to split the overall reaction into two half reactions (oxidation and reduction) and then balance them accordingly. However, when you're balancing you need to ensure the number of electrons on the reactant side and product side cancel out when you combine the two equations. That number of electrons you get here will be n in the equation you referenced.
Hope this helps! :)
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Re: n in ∆G = -nFE
In the equation ∆G = -nFE, the n stands for the moles of electrons. If you have your overall balanced redox reaction, then however many electrons you needed to balance the reaction will be your n value. If you have two skeletal equations (not balanced) then you must balance them each accordingly and combine them to get your overall redox reaction. When you combine them the coefficient that you will have on both sides of your reaction for the number of electrons will be your n value.
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Re: n in ∆G = -nFE
to find n you have to balance the redox reaction that is given to you, so that you can plug that number (which is the moles of electrons) into the equation!
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Re: n in ∆G = -nFE
After you balance the redox reaction, you can find n (the moles of electrons transferred).
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Re: n in ∆G = -nFE
N is the moles of electrons. You have to balance the redox reaction using the half-reactions to calculate how many electrons are being transferred
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Re: n in ∆G = -nFE
n= moles of electrons, so when given a chemical equation you need to find the two half reactions from a balanced redox reaction. Then you can find how many electrons are needed for the balanced equation and that is your n value.
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Re: n in ∆G = -nFE
n is the number of moles of e- transferred. you can either look at the change in oxidation number (for a balanced final equation) or write out the two half reactions and balance them to find the change in e-
Re: n in ∆G = -nFE
You need to balance the chemical reaction. Typically this happens with the same process you would use when splitting the reaction into half reactions. n is the number of electrons transferred once the reaction is balanced.
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Re: n in ∆G = -nFE
n is the moles of electrons transferred and you can figure this out by balancing your half reaction equations!
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Re: n in ∆G = -nFE
The n is the moles of electrons transferred. This can be found by figuring out the oxidation numbers of the reactants and products, making two half reactions, and determining the amount of electrons introduced or released in each half reaction.
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Re: n in ∆G = -nFE
While n is moles of electrons, the usage of it in this formula, when given a redox reaction, is simply the number of electrons transferred.
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Re: n in ∆G = -nFE
n refers to the mols of electrons that are transferred and we can see this when the number of electrons of both half reactions are the same once everything is balanced.
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Re: n in ∆G = -nFE
n is the number of moles of electrons being transferred. You can find this by balancing the half-equations.
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Re: n in ∆G = -nFE
n would be the coefficient in front of the e-. For instance, if a reaction transfers 2 electrons from reactants to products then n would be 2.
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Re: n in ∆G = -nFE
Balance the half reactions and see how many electrons are transferred. This value will be your n.
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Re: n in ∆G = -nFE
To find n, you balance the two half reactions to see how many electrons were transferred! :))
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Re: n in ∆G = -nFE
You would have to write the two half reactions and multiply both by any coefficients necessary to get the same number of electrons to cancel out. This is because the number of electrons lost has to be the number of electrons gained. This number of electrons is the number you should use.
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Re: n in ∆G = -nFE
Hi,
Whenever we are balancing half reactions into an overall reaction, you must balance the electrons on both sides to ensure they cancel out. That number of electrons will be considered "n."
For example, assume you have a redox reaction where there are 3 e- on the left and 2 e- on the right of the reaction in their half reactions. You would need to multiply each number by a factor to get 6 e- so they cancel. The number 6, in this case, would be your "n."
Hope this helps! :)
Whenever we are balancing half reactions into an overall reaction, you must balance the electrons on both sides to ensure they cancel out. That number of electrons will be considered "n."
For example, assume you have a redox reaction where there are 3 e- on the left and 2 e- on the right of the reaction in their half reactions. You would need to multiply each number by a factor to get 6 e- so they cancel. The number 6, in this case, would be your "n."
Hope this helps! :)
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Re: n in ∆G = -nFE
Hi!
You first need to balance the two half-reactions so that the electrons cancel out when the half-reactions are added. The coefficient on the electrons in the half-reactions after balancing is the n.
You first need to balance the two half-reactions so that the electrons cancel out when the half-reactions are added. The coefficient on the electrons in the half-reactions after balancing is the n.
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Re: n in ∆G = -nFE
n refers to the number of mols of electrons being transferred and to do find this look at the difference in oxidation numbers of the species being reduced and oxidized
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Re: n in ∆G = -nFE
The number of electrons that cancel out in the half-reactions when making the overall reaction is the value of n.
Re: n in ∆G = -nFE
n is the number of electrons transferred or cancelled out when balancing the half reactions.
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Re: n in ∆G = -nFE
the n is the number of electrons that are transferred in the reaction. You can find this by observing the balanced half reactions for a redox reaction.
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Re: n in ∆G = -nFE
n is the number of electrons transferred so you balance out the electrons for both of the half-reactions. The electrons in the balanced half-reactions (which the electrons should cancel out in the overall reaction) is the value of n.
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Re: n in ∆G = -nFE
Agreed with the above. n refers to number of moles of electrons transferred, so to find out from the equation it's best to split it into the half equations and using the charges to determine how many were moved.
Re: n in ∆G = -nFE
n refers to the transfer of electrons in the process. To find it, balance the redox reactions.
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Re: n in ∆G = -nFE
you would want to balance the half equations to determine the value of n which would be in front of the electrons
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Re: n in ∆G = -nFE
n is the number of moles of electrons transferred, and you can find this by finding and balancing the half-reactions of oxidation and reduction.
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Re: n in ∆G = -nFE
Like in another question posted, n is just moles transferred but this can be found by balancing the reaction given to you.
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