n in -nFE


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Jielena_Bragasin2G
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n in -nFE

Postby Jielena_Bragasin2G » Sat Feb 22, 2020 9:52 am

I know that he went over this in class, but can someone explain to me where the n comes from ? Thanks!

Jordan Young 2J
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Re: n in -nFE

Postby Jordan Young 2J » Sat Feb 22, 2020 10:18 am

n is the number of moles and it comes from the fact that the value of Faraday's constant is the charge per mole so that's why you have to multiply it by the number of moles.

Jasmine Kim 1L
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Re: n in -nFE

Postby Jasmine Kim 1L » Sat Feb 22, 2020 10:21 am

In the 7th edition of the textbook, the equation is explained on pg. 547. Basically, the change in Gibbs free energy equals the max nonexpansion work of an isobaric and isothermal reaction (). Work is done when electrons move through a potential difference, which it can be calculated by multiplying their total charge with the potential difference. So you need n, the number of electrons in moles, to calculate

Betania Hernandez 2E
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Re: n in -nFE

Postby Betania Hernandez 2E » Sat Feb 22, 2020 12:46 pm

n is the number of electrons in moles. This can be found by writing the two half-reactions and seeing how many electrons are transferred.

Megan Cao 1I
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Re: n in -nFE

Postby Megan Cao 1I » Sat Feb 22, 2020 1:49 pm

n is the moles of electrons, just be sure not to confuse it with the moles of the agents. you can figure out the number of moles of electrons by writing both the half reactions and seeing the number of electrons transferred.

Zoe Gleason 4F
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Re: n in -nFE

Postby Zoe Gleason 4F » Sat Feb 22, 2020 1:50 pm

n represents the moles of electrons. In each balanced half reaction, the number of moles of electrons should be the same, but on different sides of the reaction.

Shivam Rana 1D
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Re: n in -nFE

Postby Shivam Rana 1D » Sat Feb 22, 2020 2:03 pm

The n in -nFE represents the mole of electrons transferred in a redox reaction.

Anish Natarajan 4G
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Re: n in -nFE

Postby Anish Natarajan 4G » Sat Feb 22, 2020 2:10 pm

In this particular case, since we're trying to talk about charge, n is the number of moles of electrons

Norman Dis4C
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Re: n in -nFE

Postby Norman Dis4C » Sat Feb 22, 2020 3:47 pm

Be aware that since an equation is composed of two half reaction. You only need to know the moles of electron transferred in one group but not adding up that of both reactions.

ramiro_romero
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Re: n in -nFE

Postby ramiro_romero » Sat Feb 22, 2020 7:50 pm

You need the moles of electrons transferred in a cell in order to cancel out the moles in Faraday's constant (mol x C/mol = C). To find the moles of electrons transferred, you write out both the oxidation and reduction half reactions, balance the reactions so the electrons cancel on both sides, and the coefficient of the electron in the balanced half reactions is your value for n. Review example 6L.1 and 6L.1A on page 685 in the 7th edition textbook for practice.

Michelle Song 1I
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Re: n in -nFE

Postby Michelle Song 1I » Sun Feb 23, 2020 1:36 am

n is the number of moles of electrons, which you can find after balancing the half reactions.

Emily Vainberg 1D
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Re: n in -nFE

Postby Emily Vainberg 1D » Sun Feb 23, 2020 2:26 pm

You need the moles of electrons transferred in a cell in order to make sure the units cancel out and you get your final answer in the correct units.To find n write out the half reactions to determine the amount of electrons are being transferred. Make sure not to add the electrons from both sides. If one one side you have -2e and the other you have -3e you find the lowest number that would cancel out electrons on both sides. For this example the number would be 6.

Hannah Pham
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Re: n in -nFE

Postby Hannah Pham » Sun Feb 23, 2020 2:40 pm

n is the number of moles of electrons in a balanced half reaction.


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