## adding/subtracting half-redox rxns

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

dtolentino1E
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Joined: Thu Jul 11, 2019 12:17 am

### adding/subtracting half-redox rxns

when we combine half-redox rxn like in Hess's law, why can't we also add together the E as if it was delta H?

Kimberly Koo 2I
Posts: 99
Joined: Sat Aug 17, 2019 12:17 am

### Re: adding/subtracting half-redox rxns

This is because delta H is a state function while E is not. So delta H doesn't depend on anything but the initial and final state, which is why we can use Hess's law.

VPatankar_2L
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### Re: adding/subtracting half-redox rxns

Also, unlike in Hess' law where you could multiply the enthalpy values by the same coefficient that you multiplied the entire reaction by, you can't use the same strategy with cell potential. Standard reduction potential gives the voltage difference between two standard electrodes which is always the same. E* is an intensive property so it does not depend on the number of times the reaction occurs.

Lauren Tanaka 1A
Posts: 109
Joined: Sat Aug 17, 2019 12:18 am

### Re: adding/subtracting half-redox rxns

Since E is not a state function you can't apply the same concepts that we used in Hess's Law. E is an intensive property too so no matter how many times the reaction occurs the E will not change.

rabiasumar2E
Posts: 108
Joined: Thu Jul 11, 2019 12:15 am

### Re: adding/subtracting half-redox rxns

Since E is an intensive property, it doesn't depend on how many times the reaction occurs and therefore, will never change! H is an extensive property, which is why you can use Hess's Law.

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