1. OXIDATION-REDUCTION
REACTIONS
•
•
•
•
21-1
Oxidation State/Numbers
Redox Reactions
Balancing Redox Reactions
Chapters 4.4; 21.1 Silberberg

2. Chapter 21
Electrochemistry:
Chemical Change and Electrical Work
21-2

3. Electrochemistry:
Chemical Change and Electrical Work
21.1 Redox Reactions and Electrochemical Cells
21-3

4. Goals & Objectives
• See these Learning Objectives on
page 170 & 894.
• Understand these Concepts:
• 4.8-10; 21.1-3.
• Master these Concepts:
• 4.7-10; 21.1
21-4

5. Oxidation-Reduction
Reactions
• Reactions in which substances
undergo changes in oxidation number
are called oxidation-reduction
reactions or simply redox reactions.
• Oxidation and reduction always occur
simultaneously in chemical reactions.
21-5

6. Basic Concepts
• Oxidation
– an algebraic increase in oxidation
number
– the process in which electrons are
– (or appear to be) lost.
• Reduction
21-6
– an algebraic decrease in oxidation
number
– the process in which electrons are
– (or appear to be) gained.

7. Pneumonic
• LEO the lion says GER
21-7

8. Basic Concepts
• Oxidizing agents
– substances that gain electrons and
oxidize other substances. Oxidizing
agents are always reduced.
• Reducing agents
– substances that lose electron and
reduce other substances. Reducing
agents are always oxidized.
21-8

9. Assigning Oxidation
Numbers
• Use a set of arbitrary rules to assign
the oxidation number of each element
in a compound or ion.
• Start with two simple rules:
21-9
– 1. The oxidation number of an element
uncombined with another element is
zero.
– 2. The sum of the oxidation numbers of
all the atoms in a species is equal to its
total charge.

10. Assigning Oxidation
Numbers
• Substance
•
H2
•
O2
•
P4
•
Ni
•
Fe3+
•
Br-1
21-10
Oxidation Number(ON)
0
0
0
0
+3
-1

11. Assigning Oxidation
Numbers
• Substance
•
Co2+
•
S2•
O22•
H-1
21-11
Oxidation Number(ON)
+2
-2
-1
-1

12. Assigning Oxidation
Numbers
• Additional Rules
– The oxidation number of H is +1 in
combination with nonmetals and -1 in
combination with metals.
• HCl ON of H = +1
• NaH ON of H = -1
21-12

13. Assigning Oxidation
Numbers
– The oxidation number of oxygen is -2 in
most of its compounds.
•
•
•
•
21-13
MgO
ON of O = -2
Na2SO4
ON of O = -2
H2O
ON of O = -2 ON of H = +1
H2O2 ON of H = +1
ON of O = -1

14. Assigning Oxidation
Numbers
– Group 1 metals in all their compounds
exhibit an oxidation number of +1
• NaCl
• LiOH
ON of Na = +1
ON of Li = +1
– Group 2 metals in all their compounds
exhibit an oxidation number of +2
• MgO
• BaSO4
21-14
ON of Mg = +2
ON of Ba = +2

15. Assigning Oxidation
Numbers
– Group 13 metals in most of their
compounds exhibit an oxidation number
of +3.
• Al2O3
• GaF3
ON of Al = +3
ON of Ga = +3
– Group16 elements in compounds with
metals and ammonium ion exhibit an
oxidation number of -2
• H2S
• (NH4)2Se
21-15
ON of S = -2
ON of Se = -2

16. Assigning Oxidation
Numbers
– Halogens (Group 17) in all their
compounds exhibit an oxidation number
of -1 except when in combination with
oxygen or another halogen higher in the
group.
21-16
•
•
•
•
•
•
NaCl
AlCl3
HClO
HClO2
HClO3
HClO4
ON of Cl = -1
ON of Cl = -1
ON of Cl = +1
ON of Cl = +3
ON of Cl = +5
ON of Cl = +7

17. Assigning Oxidation
Numbers
• Assign an oxidation number to the
underlined element in each of the
following:
• NaNO2
H3PO4
• K2Sn(OH)6
KO2
• SO3-2
HCO3-1
• Cr2O7-2
NH4+1
• HC2H3O2
21-17

18. 21-18

19. Overview of Redox Reactions
Oxidation is the loss of electrons and reduction is the
gain of electrons. These processes occur simultaneously.
Oxidation results in an increase in O.N. while reduction
results in a decrease in O.N.
The oxidizing agent takes electrons from the substance
being oxidized. The oxidizing agent is therefore reduced.
The reducing agent takes electrons from the substance
being oxidized. The reducing agent is therefore oxidized.
21-19

20. Figure 21.1 A summary of redox terminology, as applied to the
reaction of zinc with hydrogen ion.
0
Zn(s) +
21-20
+1
2H+(aq)
→
+2
Zn2+(aq)
0
+ H2(g)

21. Redox Reactions
• Redox reactions must involve a
change in oxidation number.
– H2 + Cl2
=
2HCl
– Zn + HNO3 = Zn(NO3)2 + NO2 + H2O
– Ag+1 + Cl-1 = AgCl
21-21

22. 21-22

23. Half-Reaction Method for
Balancing Redox Reactions
The half-reaction method divides a redox reaction into
its oxidation and reduction half-reactions.
- This reflects their physical separation in electrochemical cells.
This method does not require assigning O.N.s.
The half-reaction method is easier to apply to reactions in
acidic or basic solutions.
21-23

24. Steps in the Half-Reaction Method
• Divide the skeleton reaction into two half-reactions, each
of which contains the oxidized and reduced forms of one
of the species.
• Balance the atoms and charges in each half-reaction.
– First balance atoms other than O and H, then O, then H.
– Charge is balanced by adding electrons (e-) to the left side in
the reduction half-reaction and to the right side in the
oxidation half-reaction.
• If necessary, multiply one or both half-reactions by an
integer so that
– number of e- gained in reduction = number of e- lost in oxidation
• Add the balanced half-reactions, and include states of
matter.
21-24

25. Balancing Redox Reactions in Acidic Solution
Cr2O72-(aq) + I-(aq) → Cr3+(aq) + I2(s)
Step 1: Divide the reaction into half-reactions.
Cr2O72- → Cr3+
I→ I2
Step 2: Balance the atoms and charges in each half-reaction.
For the Cr2O72-/Cr3+ half-reaction:
Balance atoms other than O and H:
Cr2O72- → 2Cr3+
Balance O atoms by adding H2O molecules:
Cr2O72- → 2Cr3+ + 7H2O
21-25

26. Balance H atoms by adding H+ ions:
14H+ + Cr2O72- → 2Cr3+ + 7H2O
Balance charges by adding electrons:
6e- + 14H+ + Cr2O72- → 2Cr3+ + 7H2O
This is the reduction half-reaction. Cr2O72- is reduced, and is the
oxidizing agent. The O.N. of Cr decreases from +6 to +3.
For the I-/I2 half-reaction:
Balance atoms other than O and H:
2I- → I2
There are no O or H atoms, so we balance charges by adding electrons:
2I- → I2 + 2eThis is the oxidation half-reaction. I- is oxidized, and is the reducing
agent. The O.N. of I increases from -1 to 0.
21-26

27. Step 3: Multiply each half-reaction, if necessary, by an integer so that
the number of e- lost in the oxidation equals the number of e- gained
in the reduction.
The reduction half-reaction shows that 6e- are gained; the oxidation
half-reaction shows only 2e- being lost and must be multiplied by 3:
3(2I- → I2 + 2e-)
6I- → 3I2 + 6eStep 4: Add the half-reactions, canceling substances that appear on
both sides, and include states of matter. Electrons must always cancel.
6e- + 14H+ + Cr2O72- → 2Cr3+ + 7H2O
6I- → 3I2 + 6e6I-(aq) + 14H+(aq) + Cr2O72-(aq) → 3I2(s) + 7H2O(l) + 2Cr3+(aq)
21-27

28. Ion-Electron Method
• Use the ion electron method to write
and balance the net ionic equation for
the following:
• Sn2+ + Br2
= Sn4+ + Br-1
21-28

29. 21-29

30. Ion-Electron Method
• In acid solutions use H2O to balance
O and H+ to balance H.
• Use the ion-electron method to write
and balance the net ionic equation for
the following:
– Fe2+ + Cr2O72- = Fe3+ + Cr3+ (acidic)
– Zn + NO3-1 = Zn+2 + N2 (acidic solution)
21-30

31. 21-31

32. 21-32

33. Balancing Redox Reactions in Basic Solution
An acidic solution contains H+ ions and H2O. We use H+
ions to balance H atoms.
A basic solution contains OH- ions and H2O. To balance
H atoms, we proceed as if in acidic solution, and then
add one OH- ion to both sides of the equation.
For every OH- ion and H+ ion that appear on the same
side of the equation we form an H2O molecule.
Excess H2O molecules are canceled in the final step,
when we cancel electrons and other common species.
21-33

34. Sample Problem 21.1
Balancing a Redox Reaction in Basic
Solution
PROBLEM: Permanganate ion reacts in basic solution with oxalate
ion to form carbonate ion and solid manganese dioxide.
Balance the skeleton ionic equation for the reaction
between NaMnO4 and Na2C2O4 in basic solution:
MnO4-(aq) + C2O42-(aq) → MnO2(s) + CO32-(aq) [basic solution]
PLAN: We follow the numbered steps as described in the text, and
proceed through step 4 as if this reaction occurs in acidic
solution. Then we add the appropriate number of OH- ions
and cancel excess H2O molecules.
SOLUTION:
Step 1: Divide the reaction into half-reactions.
MnO4- → MnO2
C2O42- → CO32-
21-34

35. Sample Problem 21.1
Step 2: Balance the atoms and charges in each half-reaction.
Balance atoms other than O and H:
MnO4- → MnO2
C2O42- → 2CO32-
Balance O atoms by adding H2O molecules:
MnO4- → MnO2 + 2H2O
2H2O + C2O42- → 2CO32Balance H atoms by adding H+ ions:
4H+ + MnO4- → MnO2 + 2H2O
2H2O + C2O42- → 2CO32- + 4H+
Balance charges by adding electrons:
3e- + 4H+ + MnO4- → MnO2 + 2H2O 2H2O + C2O42- → 2CO32- + 4H+ + 2e[reduction]
[oxidation]
21-35

36. Sample Problem 21.1
Step 3: Multiply each half-reaction, if necessary, by an integer so that
the number of e- lost in the oxidation equals the number of e- gained
in the reduction.
x2
6e- + 8H+ + 2MnO4- → 2MnO2 + 4H2O
x3
6H2O + 3C2O42- → 6CO32- + 12H+ + 6eStep 4: Add the half-reactions, canceling substances that appear on
both sides.
6e- + 8H+ + 2MnO4- → 2MnO2 + 4H2O
2 6H2O + 3C2O42- → 6CO32- +412H+ + 6e2MnO4- + 2H2O + 3C2O42- → 2MnO2 + 6CO32- + 4H+
21-36

37. Sample Problem 21.1
Basic. Add OH- to both sides of the equation to neutralize H+, and
cancel H2O.
2MnO4- + 2H2O + 3C2O42- + 4OH- → 2MnO2 + 6CO32- + [4H+ + 4OH-]
2MnO4- + 2H2O + 3C2O42- + 4OH- → 2MnO2 + 6CO32- + 2 4H2O
Including states of matter gives the final balanced equation:
2MnO4-(aq) + 3C2O42-(aq) + 4OH-(aq) → 2MnO2(s) + 6CO32-(aq) + 2H2O(l)
21-37

38. Ion-Electron Method
21-38
• In basic solution, balance the
equation as if it were acidic and then
neutralize the H+ with OH-1 to
produce water.
• Use the ion-electron method to write
and balance the net ionic equation for
the following:
• Br2 = Br-1 + BrO3-1 (basic solution)
• S-2 + Cl2 = SO4 -2 + Cl-1 (basic

39. 21-39

40. 21-40