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2013 Final Practice Exam, Question 4a; n=?

Posted: Thu Mar 16, 2017 11:55 am
by 104733309
In the 2013 Final Practice Exam, for question 4a how do you know that n=4?

Re: 2013 Final Practice Exam, Question 4a; n=?

Posted: Thu Mar 16, 2017 12:09 pm
by Shushanna S 3F
The iron goes from an oxidation state of 0 to +2 (and the oxygen goes from 0 to -2). And since there are two moles of iron and oxygen, 2*2electrons=4 electrons. Hope that helps!

Re: 2013 Final Practice Exam, Question 4a; n=?

Posted: Fri Mar 17, 2017 12:20 pm
by Franklin Kong 3D
I believe each oxygen in the O2 in the reactants initially has an oxidation number of 0, while in the product all 4 oxygens have an oxidation number of -2. This would make each Fe in the product have an oxidation number of +2, while in the reactants each Fe had an oxidation number of 0. Thus, each Fe was oxidized from 0 to 2+ and each oxygen atom in the O2 reactant was reduced from 0 to 2-. Therefore, since there were 2 Fe that were oxidized from 0 to 2+ and 2 oxygens that were reduced from 0 to 2-, the number of electrons transferred would be 2 atoms x 2 electrons = n = 4.

Re: 2013 Final Practice Exam, Question 4a; n=?

Posted: Fri Mar 17, 2017 9:14 pm
by 104781135
Why does 2Fe(s) in the reactants have an oxidation number of zero again? Is it because it is in it's most stable form or something like that?

Re: 2013 Final Practice Exam, Question 4a; n=?

Posted: Fri Mar 17, 2017 9:35 pm
by Kelly Tran 3N
This was very helpful! I had the same question too

Re: 2013 Final Practice Exam, Question 4a; n=?

Posted: Fri Mar 17, 2017 10:24 pm
by 104781135
I have one more question! Can anyone go through the math portion with me? I don't really understand what goes on from 0.044V=0.0257/4lnQ. How do we get that Q=942.23? I get a much smaller answer doing this math.

Re: 2013 Final Practice Exam, Question 4a; n=?

Posted: Fri Mar 17, 2017 10:38 pm
by lilyjustine_1G
104781135 wrote:I have one more question! Can anyone go through the math portion with me? I don't really understand what goes on from 0.044V=0.0257/4lnQ. How do we get that Q=942.23? I get a much smaller answer doing this math.


You should divide 0.0257 by 4 first before doing anything else because it is technically one entity that is being multiplied to lnQ. You should get 0.006425 which you would then divide 0.044 by. Now you have 6.848 = lnQ so you take the e of both sides to get rid of the ln and badda bing badda boom --> 942.23