1. EQUILIBRIUM AND SPONTANEITY
Combining Multiple Topics

2. Equilibrium (K) and Spontaneity (G)
A + B C
What does it mean if the Keq is large? Small? One?
Keq value
Free Energy Change (∆G)
Explanation
Very Large
Keq >> 1 —
Spontaneous
Lots of Products form and little of the Reactants remain
(Forward is favored)
Very Small
Keq << 1 +
Non- spontaneous
Little Products have formed and lots of Reactants remain (Reverse is favored)

3. Mathematical Relationship
Why would a large K value produce a negative ∆Go ?
∆Go = - RT ln K
This equation (given on Reference Sheet) allows you to calculate the ∆Go once you know the K.

4. 2 NaF (aq) + Mg(OH)2 (s) 2 NaOH (aq) + MgF2 (s)
Practice Problem
2 NaF (aq) + Mg(OH)2 (s) NaOH (aq) + MgF2 (s)
At 298K, the reaction is at equilibrium. The concentration of NaF is M and the concentration of NaOH is 0.047M.
Calculate the Keq.
Determine the ∆Go of this reaction at 298K.
Is this reaction spontaneous or non-spontaneous? 4

5. Not a Equilibrium (Q not K)
∆G = ∆Go + RT ln Q
∆G of forward reaction is negative
∆G of forward reaction is positive

6. REVIEW of VOLTAIC CELL What is getting reduced?
What conditions are we assuming?
What is getting oxidized?
The solutions are 1.0 M and the temperature is 25°C.
What is the overall voltage?

7. Electrochemistry and Equilibrium
The zinc electrode becomes smaller as zinc atoms are oxidized to Zn2+ ions, which go into solution.
The copper electrode becomes larger as Cu2+ ions in the solution are reduced to copper metal.
The concentration of Zn2+ ions at the anode increases and the concentration of the Cu2+ ions at the cathode decreases.
Cu2+ + Zn  Zn2+ + Cu
What would happen if the [Zn+2] increased?
What would happen if the [Cu+2] increased?

8. Nernst Equation (not on AP)
E is the cell potential at some moment in time
Eo is the cell potential when the reaction is at standard-state conditions
R is the ideal gas constant in units of joules per mole
T is the temperature in Kelvin
n is the number of moles of electrons transferred in the balanced equation for the reaction
F is the charge on a mole of electrons (96500 C per mole)
Q is the reaction quotient at that moment in time.

9. Sample Problem Cu|Cu2+ (0.024 M)||Ag+(0.0048 M)|Ag
Calculate the potential at 25oC for the following cell. Cu in a M solution of Cu2+ ions and Ag in a M solution of Ag+ ions. (NOT 1.0 M solutions)
Cu|Cu2+ (0.024 M)||Ag+( M)|Ag
2 Ag+ + Cu  Cu Ag
Would the voltage (.46 V if 1.0 M solutions were used) become greater or smaller?

10. Cu(s) + 2 Ag+(aq) Cu2+(aq) + 2 Ag(s)
SOLUTION
Cu(s) + 2 Ag+(aq) Cu2+(aq) + 2 Ag(s)
Oxidation: Cu  Cu e-  Eo ox = - ( V)
Reduction:  Ag+ + e-  Ag  Eo red = V
    _________________________________________    
 Eo = Eored + Eoox = V
We now set up the Nernst equation for this cell, noting that n is 2 because two electrons are transferred in the balanced equation for the reaction.
Substituting the concentrations of the Cu2+ and Ag+ ions into this equation gives the value for the cell potential.