## 2013 Final Q4 Part A

$E_{cell} = E_{cell}^{\circ}-\frac{RT}{nF}\ln Q$

Raymond_Guan_1H
Posts: 21
Joined: Fri Jul 15, 2016 3:00 am

### 2013 Final Q4 Part A

How do I get n=4 for part A in question 3? I thought it was the amount of electrons, but I can't seem to get 4.

Simone Seliger 1C
Posts: 30
Joined: Fri Sep 25, 2015 3:00 am

### Re: 2013 Final Q4 Part A

I am having the same problem :(

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

### Re: 2013 Final Q4 Part A

You look at the chemical equation and see that Fe (neutral charge) changes to 2Fe(OH)2 in the products, indicating Fe has a positive +2 charge. 2 e- are transferred, but there are 2 moles of Fe, so a total of 4 e- transferred, n=4.

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