1. Electrochemistry Part IV: Spontaneity, Nernst Equation & Electrolysis
Jespersen Chap. 20 Sec 4 to 7
Skipping Sec 6 & 8
Dr. C. Yau
Spring 2013 1 1

2. Spontaneity of Reaction
We know to have a spontaneous rxn…
E > 0
ΔG < 0
How are these two related?
ΔG = - n FEcell where n = moles of e-
F = Faraday's constant
9.65x104 C/mol e-
(remember 1 V = 1J/C)
ΔGo = - n FEocell
(under standard conditions) 2

3. NiO2 (s) + 2Cl(aq) + 4H+ (aq)  Cl2 (g) + Ni2+ (aq) + 2H2O (l)
Example 20.7 p. 937
Calculate ΔGo for the reaction, given that its standard cell potential is V at 25oC.
NiO2 (s) + 2Cl(aq) + 4H+ (aq)
 Cl2 (g) + Ni2+ (aq) + 2H2O (l)
ΔGo = - n FEocell F = 9.65x104 C/mol e-
(1 V = 1J/C, so Eo = J/C)
How do we figure out what n is?
Ans kJ
Do Pract Exer 13 & 14 p. 938

4. Calculating K from Cell Potential
We know ΔGo = - n FEocell
We also know ΔGo = - RT ln K (Chap. 19)
so n FEo = - RT ln K
Example 20.8 p. 789
Calculate K for the reaction in Example 20.8.
NiO2 (s) + 2Cl(aq) + 4H+ (aq)
 2Cl2 (g) + Ni2+ (aq) + 2H2O (l)
Collect all the constants we need.
Do Pract Exer 15 & 16 p. 939

5. Derivation of the Nernst Eqn
What happens when it is not under standard conditions? Divide both sides of eqn by (-nF) we get...
ΔGo = - n FEocell
ΔG = - n FEcell
Nernst Equation

6. Common simplified version of the Nernst Equation for 25.0 oC:
This version of Nernst Eqn will be given also, but remember it’s only for 25.0oC

7. Nernst Equation
Eo is the cell potential under standard conditions (for aqueous soln, 1M)
What if it is not 1 M?
Example 20.9 p. 940
Suppose a galvanic cell employs the following:
Ni2+ + 2e-  Ni Eo = V
Cr3+ + 3e-  Cr Eo = V
Calculate the cell potential when [Ni2+] = 4.87x10-4M
and [Cr3+] = 2.48x10-3 M
This type of quest will be on your final exam.
Ans V

8. The rxn of tin metal with acid can be written as
Example p. 941
The rxn of tin metal with acid can be written as
Sn (s) + 2H+ (aq)  Sn2+ (aq) + H2 (g)
Calculate the cell potential
(a) when the system is at standard state.
(b) when the pH = 2.00
(c) when the pH is 5.00.
Assume that [Sn2+] = 1.00 M and the partial pressure of H2 is also 1.00 atm.
Do Pract Exer 17, 18, 20 p. 942
Ans V
Ans V

9. What we are skipping in Chap. 20:
Concentration from E Measurements
Sec 20.6 Electricity
Batteries: Lead Storage Batteries
Zinc-Manganese Dioxide Cells (LeClanche cell)
Nickel-Cadmium Rechargeable Batteries
Nickel-Metal Hydride Batteries
Lithium Batteries
Lithium Ion Cells
Fuel Cells
Photovoltaic Cells

10. Electrolytic Cell (Sec. 20.7)
We had been considering 2 half-reactions, one of which is the "drive" for the forward reaction.
In the electrolytic cell (or electrolysis cell), the driving force is the electric current.
"Lysis" means to break apart.
"Electrolysis" means to break with an electric current.
We know 2Na + Cl2 2NaCl is spontaneous.
We can do the reverse reaction by applying an electric current.
2NaCl  2Na + Cl2
What are the physical states?
Usually the electrolytic cell has only one chamber.

12. Electrolysis in Aqueous Solutions
Can we "break apart" potassium sulfate?
It is not easy to melt an ionic compound.
What if we did this in an aqueous solution?
Water can be oxidized as well as reduced.
Why? How?
H2O  H reduction at the …
H2O  O oxidation at the…
Overall: 2 H2O (l)  2H2 (g) + O2 (g)
This cannot take place without an electrolyte present (acting as a salt bridge).
Do Pract Exer 23, 24 p. 958

13. Example p. 957
Electrolysis is planned for an aq soln that contains a mixture of 0.50 M ZnSO4 and 0.50 M NiSO4. On the basis of standard reduction potentials, what products are expected to be observed at the electrodes? What is the expected net cell reaction?
First determine which species are present.
Next determine which might be reduced, which might be oxidized.
(Remember those to be reduced are on the left side of Table 19.1, those to be oxidized are on the right side. Which is reduced the easiest on the table?
Which is oxidized the easiest?)

14. Which is most easily reduced?
Example p. 957 (continued)
Ni2+ + 2e-  Ni Eo = V
Zn2+ + 2e-  Zn Eo = V
2H2O + 2e-  H2 + 2OH- Eo = V
Which is most easily reduced?
Now examine what is most easily oxidized.
S2O e-  2SO Eo = V
O2 + 4H+ + 4e-  2H2O Eo =+1.23 V
What is the net cell rxn?
What is the Eocell?
Ans. Ni2+
Ans. H2O

15. Ni2+ + 2e-  Ni Eo = - 0.25 V O2 + 4H+ + 4e-  2H2O Eo =+1.23 V
Example p. 957 (continued)
Ni2+ + 2e-  Ni Eo = V
O2 + 4H+ + 4e-  2H2O Eo =+1.23 V
2H2O O2 + 4H+ + 4e Eo =-1.23 V
What is the net cell rxn?
2Ni2+ + 4e-  2Ni Eo = V
2H2O O2 + 4H+ + 4e Eo = V
2Ni2+ + 2H2O  2Ni + O2 + 4H+ Eo = V
Do Pract Exer 23, 24 p. 958
How can it be negative?

16. Easiest to be reduced
Easiest to be oxidized

17. Periodic Table That’s what nonmetals do in redox rxns:
X X- (reduction)
Easiest to be reduced
Periodic Table
Easiest to be oxidized
That’s what metals do (oxidation): M M+