## 14.27

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

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### 14.27

Using data in Appendix 2B, calculate the standard potential for the half-reaction U4+ + 4e ---> U

I understand that we need to use the standard potential from appendix 2B, but why do we first need to calculate delta G? Is it because the standard potential is an intensive property?

Ramya Lakkaraju 1B
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### Re: 14.27

I think it might have to do with the fact that delta G is a state function so you can add each step's delta G to get the total.

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### Re: 14.27

I also do not understand why we would need deltaG values to calculate the potential of a half-reaction

Phillip Winters 2F
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### Re: 14.27

If you add the deltaG values together, then you can use the formula deltaG=-nFE to calculate the cell potential

Jenny Cheng 2K
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### Re: 14.27

Gibbs Free Energy is a state function, so you can add intermediate components to reach your desired final value. Cell potential is NOT a state function, so you can't use the same addition method. However, you can relate Gibbs Free Energy and cell potential through ΔG=-nFE.

Alexandria Weinberger
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### Re: 14.27

Gibb's Free Energy is a state function, so the values of individual reactions can be added together to find the free energy of the overall reaction. Standard potential is not a state function, so you cannot add potentials together or take away to find the potentials of intermediate reactions.

Clarisse Wikstrom 1H
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### Re: 14.27

It would seem easier to just add up the E values for each half reaction, correct? However, you can only do this when solving for the potential of an entire reaction, not a half reaction. To solve for the potential of a HALF-REACTION, you must find the delta G values of each one, add them together, then convert to E. This is because deltaG is a state function, as well as the fact that E values correspond to a specific half reaction based on experimentation and cannot just be added together unless solving for the E value for the total rxn.