Calculate standard potential, Gibbs free

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Grace Han 2K
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Joined: Tue Nov 15, 2016 3:00 am

Calculate standard potential, Gibbs free

Postby Grace Han 2K » Thu Mar 15, 2018 4:58 pm

On test 2, #7, we have to use Gibbs free energy to calculate the standard potential because standard potentials cannot be added since they are not a state function.
However, when it asks to calculate standard cell potential in #6D, we just add -1.24V + 2.04V to get the answer. I dont understand when to use the gibbs free= -nFE and when to just add.

Anh Nguyen 2A
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Joined: Fri Sep 29, 2017 7:05 am
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Re: Calculate standard potential, Gibbs free

Postby Anh Nguyen 2A » Fri Mar 16, 2018 11:39 am

Because in 6d, its a galvanic cell so you use E=E cathode - E anode to calculate the overall standard potential of the whole cell. This has nothing to do with E not being state function. While in question 7, the reaction is made up of two steps (A turns into B and then B turns into C). For state function, you can just add the two steps together but since E is not a state function you have to calculate it through G=-nFE and since G is a state function, you can add G together.

Karen Ung 2H
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Joined: Fri Sep 29, 2017 7:04 am
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Re: Calculate standard potential, Gibbs free

Postby Karen Ung 2H » Fri Mar 16, 2018 4:34 pm

I may have had a different test from you nurmerically, but I'm pretty sure in 6D you're able to add the standard potentials because they're from the half reactions. So, based of the equation E(anode)+E(cathode) = E cell, you get the overall standard cell potential. On #7, you can't do the same thing because the half reactions are not derived from the overall reaction. You have to find the delta G of each half reaction and add the together to find the delta G of the overall reaction. (You can add the delta G values because it is a state function). With the delta G of the overall reaction, you can plug that value into ∆G° = - n F E°to find the standard cell potential fo the overall reaction.


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