Reaction E  [ENDORSED]

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Angela 1K
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Joined: Fri Sep 29, 2017 7:05 am

Reaction E

Postby Angela 1K » Thu Feb 22, 2018 8:05 pm

How come when figuring out the reaction E, we don’t manipulate the potentials?

For example, for homework problem 14.11, the question involves the half reactions oh Ni(s) + 2e- —> Ni with E = -0.23V and Ag+ + e- —> Ag with E = +0.80V.

In order to make the reaction, we must reverse the first reaction and multiply the second reaction by 2. However, we don’t manipulate their reaction potentials. Why?

Thanks.

Sammy Thatipelli 1B
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Re: Reaction E

Postby Sammy Thatipelli 1B » Thu Feb 22, 2018 8:20 pm

This is because the E potential does not change. Its sign never changes, because potential doesn't change even if the reaction is reversed.

Kelly Seto 2J
Posts: 30
Joined: Thu Jul 27, 2017 3:00 am

Re: Reaction E

Postby Kelly Seto 2J » Fri Feb 23, 2018 2:02 pm

E values are intensive properties, meaning that their values don't scale with the other components of the reaction

Christina Bedrosian 1B
Posts: 33
Joined: Fri Sep 29, 2017 7:05 am

Re: Reaction E

Postby Christina Bedrosian 1B » Fri Feb 23, 2018 2:05 pm

E values never change, even if the chemical reaction is being multiplied by a factor

Mika Sonnleitner 1A
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Re: Reaction E

Postby Mika Sonnleitner 1A » Fri Feb 23, 2018 2:15 pm

Like density, cell potential is an intensive property. This means that cell potential does not depend on the amount of substance present, so even if you multiply your chemical reaction by a coefficient in order to balance the half reactions, you do not need to do anything to change the cell potential of each half reactions.

William Satyadi 2A
Posts: 31
Joined: Sat Jul 22, 2017 3:00 am

Re: Reaction E  [ENDORSED]

Postby William Satyadi 2A » Fri Feb 23, 2018 2:19 pm

E is an intensive property, meaning we do not need to manipulate it or multiply it by stoichiometric coefficients.


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