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adding/subtracting half-redox rxns

Posted: Sun Feb 23, 2020 11:22 am
by dtolentino1E
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?

Re: adding/subtracting half-redox rxns

Posted: Sun Feb 23, 2020 11:29 am
by Kimberly Koo 2I
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.

Re: adding/subtracting half-redox rxns

Posted: Sun Feb 23, 2020 1:29 pm
by VPatankar_2L
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.

Re: adding/subtracting half-redox rxns

Posted: Sun Feb 23, 2020 1:45 pm
by Lauren Tanaka 1A
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.

Re: adding/subtracting half-redox rxns

Posted: Sun Feb 23, 2020 1:55 pm
by rabiasumar2E
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.