## E of a Redox [ENDORSED]

$\Delta G^{\circ} = -nFE_{cell}^{\circ}$

Lydiaelson1C
Posts: 15
Joined: Fri Sep 25, 2015 3:00 am

### E of a Redox

Could someone please explain why when you multiply a half reaction, the E does not change? I'm just confused why entropy and enthalpy change when the reaction is flipped or multiplied bye E does not. I know that there is a difference because S and H are thermo and E is electric, but I do not get why the value is not affected by the coefficients changing.

Bryan Lau 3H
Posts: 53
Joined: Fri Sep 25, 2015 3:00 am

### Re: E of a Redox

It's because entropy and enthalpy are EXtensive properties. The more wood you burn, the more heat it will give off (enthalpy). The more of something you have, the more disordered it can be (entropy). On the other hand, E is an INtensive property. By definition, it doesn't matter how much of something you have; E will remain the same.

804594277
Posts: 18
Joined: Fri Sep 25, 2015 3:00 am

### Re: E of a Redox

It doesn't matter if you double the size of the cell or half it, the E value will always be the same. Because, as stated before, it is an intensive property.

Michael Lee 2I
Posts: 51
Joined: Fri Sep 29, 2017 7:05 am

### Re: E of a Redox

The E value will always remain the same because it is an intensive property.

Phi Phi Do 2E
Posts: 27
Joined: Fri Sep 29, 2017 7:05 am

### Re: E of a Redox  [ENDORSED]

E of a redox is an intensive property meaning that it does not matter the size or amount of material so there is no need to multiply it to another number.

Shreya Ramineni 2L
Posts: 50
Joined: Fri Sep 29, 2017 7:07 am

### Re: E of a Redox

E is intensive, so the amount does not affect its potential.