## 14.27

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

Michelle_Tan_1G
Posts: 35
Joined: Fri Jun 17, 2016 11:28 am

### 14.27

According to the back of the book, the answer is -1.5 V, but I got -2.4 V. My work is below. What am I doing wrong?
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Marie_Bae_3M
Posts: 25
Joined: Wed Sep 21, 2016 2:57 pm

### Re: 14.27

I think that, in this case, we cannot add the E*'s together because the number of moles of electrons are not the same in the half-reactions (1e- vs. 3e-). According to the solutions manual, we have to use the deltaG*=-nFE* equation, to find the deltaG* of each individual half-reaction. Then we add these together to find the overall reaction's delta G*, which can be replaced by -4 (4 as the number of moles of e- for the reaction given in the problem) times Faraday's constant (F) times E*total (what we are trying to find). We divide the added half-reaction deltaG*'s by -4 and F to get the final answer of E*total.

Michelle_Tan_1G
Posts: 35
Joined: Fri Jun 17, 2016 11:28 am

### Re: 14.27

That answer makes sense, but I'm still getting an answer of -.905 V while following those instructions. Is there something else I'm doing wrong?
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Marie_Bae_3M
Posts: 25
Joined: Wed Sep 21, 2016 2:57 pm

### Re: 14.27

I think you just have the E potentials mixed up. U+3+3e-->U is E=-1.79, U+4+e-->U+3 is E=-.61