The net reaction in a voltaic cell with Eºcell= +.726V

Question

2Fe^3+(aq) + 3Zn(s) --> 2Fe(s) + 3Zn^2+(aq)

What is the delta Gº for this reaction at 25ºC

I tries using this equation:
delta G = -nFEºcell
= -n(90485)(.726)

However, I don't know how to calculate n in this. I understand that n= # of electrons tranfered but how do you calculate that?

Answer 1

One way is to write the two half reactions.
Fe^+3 + 3e ==> Fe(s)
Zn(s) ==> Zn^+ + 2e
======================
The first reaction was multiplied by 2 and the second by 3 for a total of 6 electrons.

Answer 2

So you just multiply by the coefficients of the 2 half reactions?

then let's say if you had:
Cu^2+ + Zn ==> Cu + Zn^2+

this would give you 1 electron?

or if you had
Mg + 2AgNO3 ==> 2Ag + Mg(NO3)2

this would give you 2 electrons? because one of the reactions is multplied by 2 and the other by 1?

this would give you

Answer 3

Yes and no. It's ther total number of electrons in the reactions. For the reaction in question, you had a total of 6e gained for Fe (after multiplying by 2) and a total of 6e lost by Zn after multiplying by 3. To complete this perhaps I should have just added the two equations together to show you. Here they are again.
2*(Fe^+3 + 3e ==> Fe(s))
3*(Zn(s) ==> Zn^+2 + 2e)
============================
Now multiply and add to get this.
2Fe^+3 + 6e + 3Zn(s) ==> 2Fe(s) + 3Zn^+2 + 6e
Then the electrons cancel and we never see them again but that is the total number of electrons. You have 6e on one side and 6e on the other.
For the Cu and Zn example you gave, your conclusion is wrong because it isn't the coefficient that you use. It's the total number of electrons. The Cu(II) gains 2e and the Zn loses 2e so the total number of electrons is 2.
Cu^2+ + Zn ==> Cu + Zn^2+
The Mg/AgNO3 example you gave is correct. The total is 2 electrons.

Answer 4

Thank you so much! I understand now!