Posted by chem help please on .
QUESTION: Calculate the standard freeenergy change (deltaG°) for the following reaction at 25 °C in kJ.
2Ag^3+(aq) + 3Zn(s)<==> 2Au(s) + 3Zn^2+(aq)
here's what i have so far:
first i wrote out the half reactions:
Au^3+(aq) + 3e^ > Au(s) = +1.52*2
Zn(s) > Zn^2+(aq) + 2e^ = +0.76*3
so E°cell= (1.52 * 2) + (0.76 * 3)=5.32V
then i plugged it into the equation..
deltaG°= nFE°cell, where F=96,485 C/mol
am i on the right track? how do i find n so i can solve for deltaG°?

chemistry 
Anonymous,
There is one error, you have to have each equation have the same number of electrons. You did, however, get the E values right.
The correct equations should be like this:
2Au^3+(aq) + 6e^ > 2Au(s) = +1.52*2
3Zn(s) > 3Zn^2+(aq) + 6e^ = +0.76*3
"n" refers to the amount of electrons in the reaction. Since now they are equal in both reactions, n = 6. 
chemistry 
DrBob222,
There is a second error. The Ecell is not 5.32 v.
Ecell = 1.52 + 0.76 = ?
You do NOT multiply by the coefficients 2 and 3. Yes, n will be 6.