Nernst K vs Q


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Siddiq 1E
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Joined: Fri Aug 09, 2019 12:15 am

Nernst K vs Q

Postby Siddiq 1E » Sat Mar 14, 2020 1:53 pm

How does the Nernst equation change based on if you use K or Q?

Chem_Mod
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Re: Nernst K vs Q

Postby Chem_Mod » Sat Mar 14, 2020 1:59 pm

If you use K, that means you are using the ratio of products to reactants at equilibrium. This would give you E of 0, as a battery or cell is essentially dead at equilibrium. Voltage is generated by electron flow, and electrons flow until the system reaches equilibrium. Thus once equilibrium has been reached, electron flow essentially stops and the voltage will go to 0.

Hui Qiao Wu 1I
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Joined: Fri Aug 30, 2019 12:16 am

Re: Nernst K vs Q

Postby Hui Qiao Wu 1I » Sun Mar 15, 2020 1:56 pm

The major difference between the two is if the rxn is at equilibrium or not. If you use K, then you are taking into consideration that the rxn is at equilibrium. If a rxn is at equilibrium, certain factors would be affected like E.

Hannah Pham
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Re: Nernst K vs Q

Postby Hannah Pham » Sun Mar 15, 2020 1:58 pm

You use K when the reaction is at equilibrium and Q if it is not.

Philomena 4F
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Re: Nernst K vs Q

Postby Philomena 4F » Mon Mar 16, 2020 5:43 pm

Chem_Mod wrote:If you use K, that means you are using the ratio of products to reactants at equilibrium. This would give you E of 0, as a battery or cell is essentially dead at equilibrium. Voltage is generated by electron flow, and electrons flow until the system reaches equilibrium. Thus once equilibrium has been reached, electron flow essentially stops and the voltage will go to 0.

When using Q in the eq, does it imply that the Ecell potential will gradually decrease until it reaches equilibrium (E=0)?

Ayushi2011
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Joined: Wed Feb 27, 2019 12:17 am

Re: Nernst K vs Q

Postby Ayushi2011 » Mon Mar 16, 2020 5:58 pm

You use K when the reaction is at equilibrium, and Q when the reaction is not.


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