Page 1 of 1

### Winter 2013 final Q4

Posted: Mon Mar 12, 2018 10:26 am
For this question, I know how to calculate everything except n. Could someone please explain how they got n because it does not explain that part? Is it the half reactions? because those are not given either...

### Re: Winter 2013 final Q4

Posted: Mon Mar 12, 2018 2:21 pm
Sorry without the question unable to give an answer.

### Re: Winter 2013 final Q4

Posted: Tue Mar 13, 2018 11:07 am
Here is the question

### Re: Winter 2013 final Q4

Posted: Tue Mar 13, 2018 1:32 pm
n is the movement of electrons so if you do the oxidation and reduction half reactions you should find that n = 4. Also I think you can just look at the charge of iron going from 0 in the reactants to +4 in the products and determine that the iron is being oxidized and 4 electrons are leaving, but if someone would confirm that would be great.

### Re: Winter 2013 final Q4

Posted: Wed Mar 14, 2018 4:51 pm
Since it is basic reaction, you have to get two half equations to find out the electrons, which is n.

### Re: Winter 2013 final Q4

Posted: Fri Mar 16, 2018 6:45 pm
Kinda stumped too but I thinkkkkk it's because Fe has an oxidation state of Fe0 -> Fe+2 + 2e-. And because there are 2 of Fe and 2 Fe(OH)2, 2x2e- = 4e- = n?

### Re: Winter 2013 final Q4

Posted: Sat Mar 17, 2018 7:20 pm
Yeah, n=4 and that refers to electrons moved because this type of n is a pure number and not moles.

### Re: Winter 2013 final Q4

Posted: Sat Mar 17, 2018 10:59 pm
n refers to the number of electrons moved, or electrons on the product side, therefore since 4 electrons were transferred n=4

### Re: Winter 2013 final Q4

Posted: Sun Mar 18, 2018 2:41 pm
How do you find these old finals?

### Re: Winter 2013 final Q4

Posted: Sun Mar 18, 2018 5:04 pm