finding "n"


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Kaylee Nezwek 1D
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finding "n"

Postby Kaylee Nezwek 1D » Fri Mar 12, 2021 10:17 am

hey! I'm getting a bit confused on how to find "n" when using the Nernst equation or calculating K. does it refer to the number of the moles, or the number of electrons transferred?

for example, in solving for K for the reaction Mn (s) + Ti^2+ (aq) --> Mn^2+ (aq) + Ti (s), in the solutions manual n=2. I wasn't sure if this represented moles or the two e- transferred in each half-rxn. For the reaction In^3+ (aq) + U^3+ (aq) --> In^2+ (aq) + U^4+ (aq), n was also = 2, even though for the half-rxns, only one e- was transferred. can anyone explain why n=2 for these reactions/how to determine n?

Alexis Sanft 1E
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Re: finding "n"

Postby Alexis Sanft 1E » Fri Mar 12, 2021 10:20 am

n is found by the number of moles of electrons being transferred. You have to make sure the number of electron transfer is balanced though by balancing both electron count for the half reactions.

Keerthana Sundar 1K
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Re: finding "n"

Postby Keerthana Sundar 1K » Fri Mar 12, 2021 3:26 pm

Kaylee Nezwek 1D wrote:hey! I'm getting a bit confused on how to find "n" when using the Nernst equation or calculating K. does it refer to the number of the moles, or the number of electrons transferred?

for example, in solving for K for the reaction Mn (s) + Ti^2+ (aq) --> Mn^2+ (aq) + Ti (s), in the solutions manual n=2. I wasn't sure if this represented moles or the two e- transferred in each half-rxn. For the reaction In^3+ (aq) + U^3+ (aq) --> In^2+ (aq) + U^4+ (aq), n was also = 2, even though for the half-rxns, only one e- was transferred. can anyone explain why n=2 for these reactions/how to determine n?


In the first example you stated, the n=2 represents the two e- transferred in each half-rxn, as you mentioned. For the second reaction you mentioned, are you sure the textbook did not make a typo or the reaction was typed wrong? The electrons transferred should be the value of n, so in the example you provided, it should be n = 1.

Ken_Cheng_2A
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Re: finding "n"

Postby Ken_Cheng_2A » Fri Mar 12, 2021 3:30 pm

Other people asked questions about this particular textbook problem, and it was revealed that the solutions manual has a typo. For the reaction with In^3, it should be n=1, as Keerthana explained.

shevanti_kumar_1E
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Re: finding "n"

Postby shevanti_kumar_1E » Fri Mar 12, 2021 3:30 pm

n refers to the moles of electrons transferred in the balanced redox reaction.

Aditya Sundaram 3D
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Re: finding "n"

Postby Aditya Sundaram 3D » Fri Mar 12, 2021 5:26 pm

First, you can use a balanced redox equation. Once you have that you can count the electrons being transferred and that should give you n.

Will Skinner
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Re: finding "n"

Postby Will Skinner » Fri Mar 12, 2021 5:37 pm

N represents the number of moles of electrons transferred. You are correct about the n in your first example, but for the second equation if the textbook uses n=2 it must be a typo. It should be 1.

Also, always remember to balance the half reactions before determining n.

Lillian
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Re: finding "n"

Postby Lillian » Sun Mar 14, 2021 6:37 am

n refers to the number of moles of electrons transferred. So to find it, balance the redox reactions :)

Rachael Cohen 3G
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Re: finding "n"

Postby Rachael Cohen 3G » Sun Mar 14, 2021 9:21 am

n is the number of moles of electrons being transferred in the reaction.

Sean Wang 1F
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Re: finding "n"

Postby Sean Wang 1F » Sun Mar 14, 2021 11:57 am

N is the amount of electrons transferred in the redox reaction. Make sure to scale the equations so that the electrons transferred in both half reactions are the same.

Evelyn Silva 3J
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Re: finding "n"

Postby Evelyn Silva 3J » Sun Mar 14, 2021 11:59 am

N is the number of electrons being transferred after balancing the reduction and oxidation reactions

Ziyan Peng 3A
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Re: finding "n"

Postby Ziyan Peng 3A » Sun Mar 14, 2021 12:42 pm

n is the number of electrons in the balanced equation.

Eve Gross-Sable 1B
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Re: finding "n"

Postby Eve Gross-Sable 1B » Sun Mar 14, 2021 12:57 pm

Yes, so the "number" and moles are going to be the same thing, it's the coefficient next to the electron, but it has to be taken from the balanced reaction instead of the half equations.


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