## Reaction E [ENDORSED]

Angela 1K
Posts: 80
Joined: Fri Sep 29, 2017 7:05 am

### Reaction E

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
Posts: 63
Joined: Thu Jul 27, 2017 3:01 am
Been upvoted: 1 time

### Re: Reaction E

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

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

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

Mika Sonnleitner 1A
Posts: 50
Joined: Fri Sep 29, 2017 7:04 am
Been upvoted: 2 times

### Re: Reaction E

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.