0
OXIDATION-REDUCTION
REACTIONS
•
•
•
•

21-1

Oxidation State/Numbers
Redox Reactions
Balancing Redox Reactions
Chapters 4....
Chapter 21
Electrochemistry:
Chemical Change and Electrical Work

21-2
Electrochemistry:
Chemical Change and Electrical Work
21.1 Redox Reactions and Electrochemical Cells

21-3
Goals & Objectives
• See these Learning Objectives on
page 170 & 894.
• Understand these Concepts:
• 4.8-10; 21.1-3.
• Mas...
Oxidation-Reduction
Reactions
• Reactions in which substances
undergo changes in oxidation number
are called oxidation-red...
Basic Concepts
• Oxidation
– an algebraic increase in oxidation
number
– the process in which electrons are
– (or appear t...
Pneumonic
• LEO the lion says GER

21-7
Basic Concepts
• Oxidizing agents
– substances that gain electrons and
oxidize other substances. Oxidizing
agents are alwa...
Assigning Oxidation
Numbers
• Use a set of arbitrary rules to assign
the oxidation number of each element
in a compound or...
Assigning Oxidation
Numbers
• Substance
•
H2
•
O2
•
P4
•
Ni
•
Fe3+
•
Br-1
21-10

Oxidation Number(ON)
0
0
0
0
+3
-1
Assigning Oxidation
Numbers
• Substance
•
Co2+
•
S2•
O22•
H-1

21-11

Oxidation Number(ON)
+2
-2
-1
-1
Assigning Oxidation
Numbers
• Additional Rules
– The oxidation number of H is +1 in
combination with nonmetals and -1 in
c...
Assigning Oxidation
Numbers
– The oxidation number of oxygen is -2 in
most of its compounds.
•
•
•
•

21-13

MgO
ON of O =...
Assigning Oxidation
Numbers
– Group 1 metals in all their compounds
exhibit an oxidation number of +1
• NaCl
• LiOH

ON of...
Assigning Oxidation
Numbers
– Group 13 metals in most of their
compounds exhibit an oxidation number
of +3.
• Al2O3
• GaF3...
Assigning Oxidation
Numbers
– Halogens (Group 17) in all their
compounds exhibit an oxidation number
of -1 except when in ...
Assigning Oxidation
Numbers
• Assign an oxidation number to the
underlined element in each of the
following:
• NaNO2
H3PO4...
21-18
Overview of Redox Reactions
Oxidation is the loss of electrons and reduction is the
gain of electrons. These processes occ...
Figure 21.1 A summary of redox terminology, as applied to the
reaction of zinc with hydrogen ion.
0

Zn(s) +

21-20

+1

2...
Redox Reactions
• Redox reactions must involve a
change in oxidation number.
– H2 + Cl2
=
2HCl
– Zn + HNO3 = Zn(NO3)2 + NO...
21-22
Half-Reaction Method for
Balancing Redox Reactions
The half-reaction method divides a redox reaction into
its oxidation an...
Steps in the Half-Reaction Method
• Divide the skeleton reaction into two half-reactions, each
of which contains the oxidi...
Balancing Redox Reactions in Acidic Solution
Cr2O72-(aq) + I-(aq) → Cr3+(aq) + I2(s)

Step 1: Divide the reaction into hal...
Balance H atoms by adding H+ ions:
14H+ + Cr2O72- → 2Cr3+ + 7H2O
Balance charges by adding electrons:
6e- + 14H+ + Cr2O72-...
Step 3: Multiply each half-reaction, if necessary, by an integer so that
the number of e- lost in the oxidation equals the...
Ion-Electron Method
• Use the ion electron method to write
and balance the net ionic equation for
the following:
• Sn2+ + ...
21-29
Ion-Electron Method
• In acid solutions use H2O to balance
O and H+ to balance H.
• Use the ion-electron method to write
a...
21-31
21-32
Balancing Redox Reactions in Basic Solution
An acidic solution contains H+ ions and H2O. We use H+
ions to balance H atoms...
Sample Problem 21.1

Balancing a Redox Reaction in Basic
Solution

PROBLEM: Permanganate ion reacts in basic solution with...
Sample Problem 21.1
Step 2: Balance the atoms and charges in each half-reaction.
Balance atoms other than O and H:
MnO4- →...
Sample Problem 21.1
Step 3: Multiply each half-reaction, if necessary, by an integer so that
the number of e- lost in the ...
Sample Problem 21.1
Basic. Add OH- to both sides of the equation to neutralize H+, and
cancel H2O.
2MnO4- + 2H2O + 3C2O42-...
Ion-Electron Method

21-38

• In basic solution, balance the
equation as if it were acidic and then
neutralize the H+ with...
21-39
21-40
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  1. 1. OXIDATION-REDUCTION REACTIONS • • • • 21-1 Oxidation State/Numbers Redox Reactions Balancing Redox Reactions Chapters 4.4; 21.1 Silberberg
  2. 2. Chapter 21 Electrochemistry: Chemical Change and Electrical Work 21-2
  3. 3. Electrochemistry: Chemical Change and Electrical Work 21.1 Redox Reactions and Electrochemical Cells 21-3
  4. 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. 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. 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. 7. Pneumonic • LEO the lion says GER 21-7
  8. 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. 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. 10. Assigning Oxidation Numbers • Substance • H2 • O2 • P4 • Ni • Fe3+ • Br-1 21-10 Oxidation Number(ON) 0 0 0 0 +3 -1
  11. 11. Assigning Oxidation Numbers • Substance • Co2+ • S2• O22• H-1 21-11 Oxidation Number(ON) +2 -2 -1 -1
  12. 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. 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. 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. 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. 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. 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. 18. 21-18
  19. 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. 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. 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. 22. 21-22
  23. 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. 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. 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. 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. 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. 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. 29. 21-29
  30. 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. 31. 21-31
  32. 32. 21-32
  33. 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. 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. 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. 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. 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. 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. 39. 21-39
  40. 40. 21-40
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