1. Gibb's Free energy and Electric cells
Supa Chemistry
Mrs. Paparella
Spring 2016

2. DG = -n F Eo (need to memorize)
DG = Gibb’s Free Energy
n = number of moles of electrons exchanged in reaction
F = Faraday’s Constant = 96,500 Coulombs/mole of e- Given on Test
Eo = Cell Potential measured in Volts; Also referred to as emf or Electromotive Force.
Voltage = Work/Charge
1 V= 1Joule/Coulomb of Charge
A Volt describes how much energy it takes to move 1 coulomb of charge
These problems will use KJ for DG that need to be changed back to Joules ( x by 1000) It’s never easy!

3. How does E Cell relate to Gibb's Free Energy?
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Spontaneous reactions have
a + E value ( + Volts) and
a – DG value.
Nonspontaneous Reactions have a + DG value and a – E value (- Volts)

4. Calculate the Gibb's Free Energy in a zinc /Copper +2 electrochemical cell which has a voltage of 1.1 V
DG = -n F E How many e- are transferred? 2 mol e-
1.1 V = 1.1 J/C
Plug and Chug!
DG = -2 mol e- x 96,500 C/mol e-x 1.1J/C
DG = - 212,300 J or KJ

5. You can also find the cell potential if you are given the equation and the Free Energy information.
2Al(s) + 3Cu2+(aq)-->2Al3+(aq) + 3Cu(s)
D G = KJ
Question: How many moles of e- are involved?
Answer: 6 moles of e-
Now you can proceed!

6. Remember to change KJ to Joules for delta G
DG = -n F E
-1,152,200 J = - 6moles of e- x 96,500 C/ mole of e- x E
Solve for E (Cell Potential)
-1,152,200 J/ -579,000 C= 1.99 J/C or V

7. Alternative Method
1 Faraday is equal to 96.5KJ / V mol e- This is also given on the test
Compare with 96,500 C /mol e-
So, if you know the D G and the # of mol of e- you can find the cell voltage by DG = -n F E
Using the previous problem data:
1,152.2KJ = -6 mol e- x 96.5KJ/Vmol e- x E and solve for E.
1,152.2KJ / 579 KJ/V = E
E = 1.99 V

8. Book Practice: p
You will first have to calculate the Cell Potential using the Eo Red half-reaction information given. Remember to change the sign if the substance is oxidized!
For this set of problems, remember if something is oxidized its charge goes up! So Fe2+ is going to be oxidized to Fe 3+.
You will be doing 3 separate problems.