1. Chapter 17 Electrochemistry
Electrochemistry deals with the relationships between electricity and chemical reactions.
Oxidation-reduction (redox) reactions were introduced in Chapter 4
Can be simple displacement reactions:
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
Cu(s) + 2Ag+(aq)  Cu2+(aq) + 2Ag(s)

2. Zn + Sn2+  Zn2+ + Sn

3. Zn + Cu2+  Zn2+ + Cu

4. Oxidation-Reduction Reactions
Redox reactions can also be more complex, with structural and composition changes as well as an exchange of electrons.
5VO2+(aq) + MnO4-(aq) + H2O(l) 
5VO2+(aq) + Mn2+(aq) + 2H+(aq)
5Fe2+(aq) + MnO4-(aq) + 8H+(aq) 
5Fe3+(aq) + Mn2+(aq) + 4H2O(l)
21m03vd1

5. Oxidation and Reduction You can’t have one without the other!
Decrease in number of electrons
(loss of electrons)
Increase in oxidation number
Oxidation number:
Increase in number of electrons
(gain of electrons)
Decrease in oxidation number
Reduction

6. Half-Reactions
Oxidation-reduction reactions can be written as separate oxidation and reduction reactions, called half-reactions.
Oxidation:
VO2+(aq) + H2O(l)  VO2+(aq) + 2H+(aq) + e-
VO2+ is called the reducing agent (or reductant), because it causes the reduction of another substance; the reducing agent is oxidized in the process

7. Half-Reactions Reduction:
MnO4-(aq) + 8H+(aq) + 5e-  Mn2+(aq) + 4H2O(l)
MnO4- is called the oxidizing agent (or oxidant), because it causes the oxidation of another substance; the oxidizing agent is reduced in the process
Oxidation-reduction can be considered to be the competition between two substances for electrons. The one with the greater attraction for additional electrons becomes the oxidizing agent; the one with the least is the reducing agent.

8. Spontaneous Oxidation-Reduction Reactions
If we place a metal in a solution of another metal ion, sometimes we get metal deposition, sometimes not. How do we decide?
Zn + Sn2+  Zn2+ + Sn
Sn + Zn2+  no rxn

9. Spontaneous Redox
How do we know which substances will act as oxidizing agents or as reducing agents?
An activity series gives the relative activity of substances as oxidizing or reducing agents. Review from Chapter 4.
Determine an activity series in several ways:
activity in displacing H2 from water
activity in displacing metals from metal ion solutions (more active metal displaces a less active metal from solution)
generation of an electrochemical potential or voltage

10. Spontaneous Redox Na displaces H2 from H2O Zn displaces Pb from a
solution of Pb2+

11. Spontaneous Redox
Pb + 2Ag+  Pb2+ + 2Ag
Ag + Pb2+  no rxn

12. Spontaneous Redox
The reverse reactions can be made to occur with electric current, but they are not spontaneous.
General rule, using an activity series:
stronger stronger weaker weaker
oxidizing + reducing  reducing + oxidizing
agent agent agent agent
Depending on the relative strengths, the reaction can go to completion, or reach a state of equilibrium.

13. Activity Series

14. Group Work
Predict which of the following combinations will undergo an oxidation-reduction reaction:
Mg + K+
Mg + Zn2+
H2 + Ni2+
H2 + Cu2+

15. Group Quiz
Predict which of the following combinations will undergo an oxidation-reduction reaction:
Sn + Pb2+
Na + H2O
Pb + H2O
Co + H+(aq)
Ag + H+(aq)

16. 20.2 Balancing Oxidation-Reduction Equations
Some redox equations can be balanced by inspection, just like other types of reactions.
Zn + CuCl2  ZnCl2 + Cu
Net ionic equation:
Zn + Cu2+  Zn2+ + Cu
Need to balance atoms and charge
What is wrong with the following?
Al + Cu2+  Al3+ + Cu
(Al-can demo)

17. Balancing Equations 5Cr3+ + 3MnO4- + 8H2O  5CrO42- + 3Mn2+ + 16H+
How do we balance an equation as complex as this?
Two methods:
half-reaction method
oxidation number change method
Will focus on the half-reaction method since it is useful in more circumstances.

18. Half-Reaction Method
Write an unbalanced half-reaction for either oxidation or reduction.
Balance the half-reaction:
a. Balance all atoms other than H and O.
b. Balance O by adding H2O to the equation.
c. Balance H by adding H+ to the equation.
d. Balance ionic charges by adding electrons to the equation.
e. If in basic solution, add OH- to each side of the equation to neutralize all the H+.
f. Cancel any H2O occurring on both sides.

19. Half-Reaction Method continued
Write an unbalanced half-reaction for the other process.
Balance by the same procedure.
Equalize the number of electrons lost and gained by multiplying each coefficient in each half-reaction by the appropriate constant.
Add the two half-reactions and cancel equal amounts of anything occurring on both sides.
Make a final check of atom and charge balances.

20. Balancing Equations Cr3+ + MnO4-  CrO42- + Mn2+
How do we balance this equation in acidic solution?
1. Cr3+  CrO42-
2a. Cr already balanced
2b. Cr3+ + 4H2O  CrO42-
2c. Cr3+ + 4H2O  CrO H+
2d. Cr3+ + 4H2O  CrO H+ + 3e-

21. Balancing Equations 3. MnO4-  Mn2+ 4a. Mn already balanced
4b. MnO4-  Mn2+ + 4H2O
4c. MnO4- + 8H+  Mn2+ + 4H2O
4d. MnO4- + 8H+ + 5e-  Mn2+ + 4H2O
5. 3 e- lost, but 5 e- gained. Equalize by multiplying half-reactions:
5(Cr3+ + 4H2O  CrO H+ + 3e-)
3(MnO4- + 8H+ + 5e-  Mn2+ + 4H2O)

22. Balancing Equations 5Cr3+ + 20H2O  5CrO42- + 40H+ + 15e-
3MnO H+ + 15e-  3Mn H2O
6. Add the equations:
5Cr3+ + 3MnO H2O + 24H+ + 15e-  5CrO Mn H2O + 40H+ + 15e-
Cancel any substances on both sides of the equation:
5Cr3+ + 3MnO4- + 8H2O  5CrO Mn H+
7. Check atom and charge balance:
5 Cr, 3 Mn, 20 O, 16 H, 12 + charges on each side

23. Balancing Equations in Basic Solution
Neutralize any H+ with OH- to form H2O and add the same number of OH- to the other side of the equation.
Cr3+ + 4H2O  CrO H+ + 3e-
Add 8OH- to each side of the equation:
Cr3+ + 4H2O + 8OH-  CrO H+ + 8OH- + 3e-
Form water by neutralization:
Cr3+ + 4H2O + 8OH-  CrO H2O + 3e-
Cancel any water occurring on both sides:
Cr3+ + 8OH-  CrO H2O + 3e-

24. Group Work Balance the following equation in acidic solution:
MnO4- + Cl-  Mn2+ + ClO3-
Answer:
MnO4- + 8H+ + 5e-  Mn2+ + 4H2O
Cl- + 3H2O  ClO3- + 6H+ + 6e-
6MnO4- + 5Cl- + 18H+ 
6Mn2+ + 5ClO3- + 9H2O

25. Group Quiz 3V, 1Cr, 7O, 2H, 6 + charges on each side
Balance the following chemical equation in acidic solution:
VO2+ + CrO42-  VO2+ + Cr3+
3VO2+ + CrO H+  3VO2+ + Cr3+ + H2O
3V, 1Cr, 7O, 2H, 6 + charges on each side