## Winter Final Exam 2013 4A

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

Kyle_Diep_4E
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Joined: Fri Sep 25, 2015 3:00 am

### Winter Final Exam 2013 4A

For the first part of the question, how do we know that n=4? Do we find it through the given equation or do we have to delve deeper?

Jacob Afable 3J
Posts: 41
Joined: Fri Sep 25, 2015 3:00 am

### Re: Winter Final Exam 2013 4A

n represents the number of electrons transferred in the balanced reaction. The reaction shows that Fe(s) is getting oxidized since it's oxidation state changes from 0 as Fe(s) to 2+ in 2 Fe(OH)2(s). So for every Fe(OH)2 formed, 2 electrons are being transferred. Since the reaction states that there are 2 molecules of Fe(OH)2, a total of 4 electrons are being transferred in the reaction.

Samiksha Chopra 2I
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### Re: Winter Final Exam 2013 4A

For that I understand that two electrons are being transferred, but how come it needs to be multiplied by 2? There are 2 molecules of Fe(OH)2 by there are also 2 of Fe, so shouldn't it be a one to one ratio?

Juliana Logan 1A
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Joined: Fri Sep 25, 2015 3:00 am

### Re: Winter Final Exam 2013 4A

The oxidation state of 2Fe(s) is 0. The oxidation state of Fe in 2Fe(OH)2 is +4, so 4 electrons are being transferred.

Zane Mills 1E
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### Re: Winter Final Exam 2013 4A

On this question why is the oxidation # for O2 zero but oxygen in other atoms is -2? can you explain in terms of e- pairs?

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