E of a Redox [ENDORSED]
Moderators: Chem_Mod, Chem_Admin
-
- 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.
-
- 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.
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.
-
- Posts: 51
- Joined: Fri Sep 29, 2017 7:05 am
-
- 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.
-
- Posts: 50
- Joined: Fri Sep 29, 2017 7:07 am
Return to “Work, Gibbs Free Energy, Cell (Redox) Potentials”
Who is online
Users browsing this forum: No registered users and 8 guests