Transcript Chapter 6: Oxidation Reduction Reactions

Chapter 6:
Oxidation-Reduction
Reactions
Chemistry: The Molecular Nature
of Matter, 6E
Jespersen/Brady/Hyslop
Oxidation-Reduction Reactions
Electron transfer reactions
 Electrons transferred from one substance to
another
 Originally only combustion of fuels or reactions of
metal with oxygen
 Important class of chemical reactions that occur in
all areas of chemistry & biology
 Also called redox reactions
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Oxidation–Reduction Reactions
Involves 2 processes:
Oxidation = Loss of Electrons (LEO)
Na  Na+ + e
Oxidation Half-Reaction
Reduction = Gain of electrons (GER)
Cl2 + 2e  2Cl
Reduction Half-Reaction
Net reaction:
2Na + Cl2  2Na+ + 2Cl
 Oxidation & reduction always occur together
 Can't have one without the other
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Oxidation Reduction Reaction
Oxidizing Agent
 Substance that accepts e's
 Accepts e's from another substance
 Substance that is reduced
 Cl2 + 2e  2Cl–
Reducing Agent
 Substance that donates e's
 Releases e's to another substance
 Substance that is oxidized
 Na  Na+ + e–
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Redox Reactions
 Very common
 Batteries—car, flashlight,
cell phone, computer
 Metabolism of food
 Combustion
 Chlorine Bleach
 Dilute NaOCl solution
 Cleans through redox
reaction
 Oxidizing agent
 Destroys stains by oxidizing them
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Redox Reactions
Ex. Fireworks displays
Net: 2Mg + O2  2MgO
Oxidation:
Mg  Mg2+ + 2e
 Loses electrons = Oxidized
 Reducing agent
Reduction:
O2 + 4e  2O2
 Gains electrons = Reduced
 Oxidizing agent
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Your Turn!
Which species functions as the oxidizing agent in
the following oxidation-reduction reaction?
Zn(s) + Pt2+(aq)  Pt(s) + Zn2+(aq)
A. Pt(s)
B. Zn2+(aq)
C. Pt2+(aq)
D. Zn(s)
E. None of these, as this is not a redox reaction.
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Guidelines For Redox Reactions
 Oxidation & reduction always occur
simultaneously
 Total number of electrons lost by one
substance = total number of electrons gained
by second substance
 For a redox reaction to occur, something
must accept electrons that are lost by another
substance
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Oxidation Numbers
Bookkeeping Method
 Way to keep track of electrons
 Not all redox reactions contain O2 & give ions
 Covalent molecules & ions often involved
Ex. CH4, SO2, MnO4–, etc.
 Defined by set of rules
 How to divide up shared electrons in compounds
with covalent bonds
 Change in oxidation number of element during
reaction indicates redox reaction has occurred
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Hierarchy of Rules for Assigning
Oxidation Numbers
1. Oxidation numbers must add up to charge on
molecule, formula unit or ion.
2. Atoms of free elements have oxidation numbers
of zero.
3. Metals in Groups 1A, 2A, and Al have +1, +2,
and +3 oxidation numbers, respectively.
4. H & F in compounds have +1 & –1 oxidation
numbers, respectively.
5. Oxygen has –2 oxidation number.
6. Group 7A elements have –1 oxidation number.
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Hierarchy of Rules for Assigning
Oxidation Numbers
7. Group 6A elements have –2 oxidation number.
8. Group 5A elements have –3 oxidation number.
9. When there is a conflict between 2 of these
rules or ambiguity in assigning an oxidation
number, apply rule with lower oxidation
number & ignore conflicting rule.
Oxidation State
 Used interchangeably with oxidation number
 Indicates charge on monatomic ions
 Iron (III) means +3 oxidation state of Fe or Fe3+
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Ex. Assigning Oxidation Number
1. Li2O
Li (2 atoms) × (+1) = +2 (Rule 3)
O (1 atom) × (–2) = –2 (Rule 5)
sum = 0
(Rule 1)
+2 –2 = 0 so the charges are balanced to zero
2. CO2
C (1 atom) × (x) = x
O (2 atoms) × (–2) = –4
sum = 0
x  4 = 0 or x = +4
(Rule 5)
(Rule 1)
C is in +4 oxidation state
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Learning Check
Assign oxidation numbers to all atoms:
Ex. ClO4
O (4 atoms) × (–2) = –8
Cl (1 atom) × (–1) = –1
(molecular ion) sum ≠ –1
(violates Rule 1)
Rule 5 for O comes before Rule 6 for halogens
O (4 atoms) × (–2) = –8
Cl (1 atom) × (x) = x
sum = –1 (Rule 1)
–8 + x = –1 or x = 8 –1
So x = +7; Cl is oxidation state +7
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Learning Check
Assign Oxidation States To All Atoms:
 MgCr2O7
Mg =+2; O = –2; and Cr = x (unknown)
+2 + 2x + {7 × (–2)} = 0
2x – 12 = 0
x = +3
Cr is oxidation # of +3
 KMnO4
K =+1; O = – 2; so Mn = x
+1 + x + {4 × (–2)} = 0
x–7=0
x = +7
Mn is oxidation # of +7
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Your Turn!
What is the oxidation number of each atom in
H3PO4?
A. H = –1; P = +5; O = –2
B. H = 0; P = +3; O = –2
C. H = +1; P = +7; O = –2
D. H = +1; P = +1; O = –1
E. H = +1; P = +5; O = –2
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Redefine Oxidation-Reduction in
Terms of Oxidation Number
 A redox reaction occurs when there is a
change in oxidation number.
Oxidation
 Increase in oxidation number
 e loss
Reduction
 Decrease in oxidation number
 e gain
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Using Oxidation Numbers to
Recognize Redox Reactions
 Sometimes literal electron transfer:
decrease
reduction
increase oxidation
+2
Cu
2+
0
+ Zn
+2
Zn
2+
0
+ Cu
Cu: oxidation number decreases by 2
 reduction
Zn: oxidation number increases by 2
 oxidation
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Using Oxidation Numbers to
Recognize Redox Reactions
 Sometimes electron transferred in "formal"
O: decrease
reduction
sense.
C: increase
-4 +1
0
oxidation
CH4 + 2O2
+4 -2
+1 -2
CO2 + 2H2O
 O: oxidation number decreases by 2
 reduction
 C: oxidation number increases by 8
 oxidation
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Ion Electron Method
 Way to balance redox equations
 Must balance both mass & charge
 Write skeleton equation
 Only ions & molecules involved in reaction
 Break into 2 half-reactions
 Oxidation
 Reduction
 Balance each half-reaction separately
 Recombine to get balanced net ionic
equation
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Balancing Redox Reactions
Some Redox reactions are simple:
Ex. 1 Cu2+(aq) + Zn(s)  Cu(s) + Zn2+(aq)
Break into half-reactions
Zn(s)  Zn2+(aq) + 2e oxidation

LEO
Reducing agent
Cu2+(aq) + 2e  Cu(s) reduction

GER
Oxidizing agent
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Example 1
Zn(s)  Zn2+(aq) + 2e
Cu2+(aq) + 2e  Cu(s)
oxidation
reduction
 Each half-reaction is balanced for atoms
 Same # atoms of each type on each side
 Each half-reaction is balanced for charge
 Same sum of charges on each side
 Add both equations algebraically, canceling e’s
 NEVER have e's in net ionic equation
Cu2+(aq) + Zn(s)  Cu(s) + Zn2+(aq)
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Balancing Redox Equations in
Aqueous Solutions
 Many redox reactions in aqueous solution
involve H2O and H+ or OH
 Balancing the equation cannot be done by
inspection.
 Need method to balance equation correctly
 Start with acidic solution then work to basic
conditions
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Redox in Aqueous Solution
Ex. 2 Mix solutions of K2Cr2O7 & FeSO4
 Dichromate ion, Cr2O72–, oxidizes Fe2+ to Fe3+
 Cr2O72– is reduced to form Cr3+
 Acidity of mixture decreases as H+ reacts with
oxygen to form water
Skeletal Eqn. Cr2O72– + Fe2+  Cr3+ + Fe3+
Ox. # Cr = +6
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Cr = +3
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Fe = +3
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Ion-Electron Method – Acidic
Solution
1. Divide equation into 2 half-reactions
2. Balance atoms other than H & O
3. Balance O by adding H2O to side that needs O
4. Balance H by adding H+ to side that needs H
5. Balance net charge by adding e–
6. Make e– gain equal e– loss; then add halfreactions
7. Cancel anything that is the same on both
sides
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Ion Electron Method
Ex. 2 Balance in Acidic Solution
Cr2O72– + Fe2+  Cr3+ + Fe3+
1. Break into half-reactions
Cr2O72  Cr3+
Fe2+  Fe3+
2. Balance atoms other than H & O
Cr2O72  2Cr3+
 Put in 2 coefficient to balance Cr
Fe2+  Fe3+
 Fe already balanced
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Ex. 2 Ion-Electron Method in Acid
3. Balance O by adding H2O to the side that
needs O.
Cr2O72  2Cr3+ + 7 H2O
 Right side has 7 O atoms
 Left side has none
 Add 7 H2O to left side
Fe2+  Fe3+
 No O to balance
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Ex. 2 Ion-Electron Method in Acid
4. Balance H by adding H+ to side that needs H
14H+ + Cr2O72  2Cr3+ + 7H2O
 Left side has 14 H atoms
 Right side has none
 Add 14 H+ to right side
Fe2+  Fe3+
 No H to balance
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Ex. 2 Ion-Electron Method in Acid
5. Balance net charge by adding electrons.
6e + 14H+ + Cr2O72  2Cr3+ + 7H2O
Net Charge =
14(+1) (–2) = 12
Net Charge =
2(+3)+7(0) = 6
 6 electrons must be added to reactant side
Fe2+

Fe3+ +
e
 1 electron must be added to product side
 Now both half-reactions balanced for mass &
charge
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Ex. 2 Ion-Electron Method in Acid
6. Make e– gain equal e– loss; then add half-
reactions
6e + 14H+ + Cr2O72–  2Cr3+ + 7H2O
6[ Fe2+  Fe3+ + e
]
3+ + 2Cr3+
6e + 6Fe2+
6Fe

+
2
+ 14H + Cr2O7
+ 7H2O + 6e
7. Cancel anything that's the same on both
sides
6Fe2+ + 14H+ 
+ Cr2O72
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6Fe3+ + 2Cr3+
+ 7H2O
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Ion-Electron in Basic Solution
 The simplest way to balance an equation in
basic solution
Use steps 1-7 above, then
8. Add the same number of OH– to both sides
of the equation as there are H+.
9. Combine H+ & OH– to form H2O
10. Cancel any H2O that you can from both
sides
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Ex.2 Ion-Electron Method in Base
Returning to our example of Cr2O72 & Fe2+
8. Add to both sides of equation the same number
of OH– as there are H+.
6Fe2++ 14H+  6Fe3+ + 2Cr3+
+ 7H2O + 14 OH–
+ Cr2O72 + 14 OH–
9. Combine H+ and OH– to form H2O.
7
2+
6Fe + 14H2O  6Fe3+ + 2Cr3+
+ Cr2O72
+ 7H2O + 14OH 
10. Cancel any H2O that you can
6Fe2+ + 7H2O  6Fe3+ + 2Cr3+
+ Cr2O72
+ 14OH 
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Your Turn!
Which of the following is a correctly balanced
reduction half-reaction?
A. Fe3+ + e–  Fe°
B. 2Fe + 6HNO3  2Fe(NO3)3 + 3H2
C. Mn2+ + 4H2O  MnO4– + 8H+ + 5e–
D. 2O2–  O2 + 4e–
E. Mg2+ + 2e–  Mg°
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Ex. 3 Ion-Electron Method
Balance the following equation in basic solution:
MnO4– + HSO3–  Mn2+ + SO42
1. Break it into half-reactions
MnO4–  Mn2+
HSO3–  SO42–
2. Balance atoms other than H & O
MnO4  Mn2+

Balanced for Mn
HSO3  SO42

Balanced for S
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Ex. 3 Ion-Electron Method
3. Add H2O to balance O
MnO4  Mn2+ + 4H2O
H2O + HSO3  SO42
4. Add H+ to balance H

+
8H + MnO4  Mn2+ + 4H2O

H2O + HSO3  SO42+ 3H+
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Ex. 3 Ion-Electron Method
5. Balance net charge by adding e–.
5e– + 8H+ + MnO4  Mn2+ + 4H2O
8(+1) + (–1) = +7
+2 + 0 = +2
Add 5 e– to reactant side
H2O + HSO3  SO42 + 3H+ + 2 e–
0 + (–1) = –1
–2 + 3(+1) = +1
Add 2 e– to product side
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Ex. 3 Ion-Electron Method
6. Make e– gain equal e– loss
2[5e– + 8H+ + MnO4  Mn2+ + 4H2O ]
5[H2O + HSO3  SO42 + 3H+ + 2e– ]
 Must multiply Mn half-reaction by 2
 Must multiply S half-reaction by 5
 Now have 10 e– on each side
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Ex. 3 Ion-Electron Method
6. Then add the two half-reactions
10e– + 16H+ + 2MnO4  2Mn2+ + 8H2O
5H2O + 5HSO3  5SO42 + 15H+ + 10e–
1
3

+
2+
+ 16H + 2MnO4
2Mn + 8H2O +


+ 5H2O + 5HSO3
5SO42 + 15H+ + 10e 
10e–
7. Cancel anything that is the same on both
sides.
H+ + 2MnO4
+ 5HSO3
2Mn2+ + 3H2O

+ 5SO42
Balanced in acid.
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Ex.3 Ion-Electron Method in Base
8. Add same number of OH– to both sides of
equation as there are H+
H+ + 2MnO4
2Mn2+ + 3H2O
 
–
+ 5SO42 + OH–
+ OH + 5HSO3
9. Combine H+ and OH– to form H2O
H2O + 2MnO4
+ 5HSO3
2
+ 3H2O

+ 5SO42 + OH
10. Cancel any H2O that you can
2MnO4 + 5HSO3  2Mn2+ + 2H2O + OH
+ 5SO42
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2Mn2+
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Your Turn!
Balance each equation in Acid & Base using the
Ion Electron Method.
MnO4– + C2O42–  MnO2 + CO32–
Acid: 2MnO4– + 3C2O42– + 2H2O  2MnO2 + 4H+ + 6CO32–
Base: 2MnO4– + 3C2O42– + 4OH–  2MnO2 + 2H2O + 6CO32–
ClO– + VO3–  ClO3– + V(OH)3
Acid: ClO– + 2H2O + 2VO3– + 2H+  ClO3–+ 2V(OH)3
Base: ClO– + 4H2O + 2VO3–  ClO3–+ 2V(OH)3 + 2OH–
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Acids as Oxidizing Agents
 Metals often react with acid
 Form metal ions &
 Molecular hydrogen gas
Molecular Equation
Zn(s) + 2HCl(aq) H2(g) + ZnCl2(aq)
Net Ionic Equation
Zn(s) + 2H+(aq) H2(g) + Zn2+(aq)




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M  oxidized
H+  reduced
H+  oxidizing reagent
Zn  reducing reagent
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Oxidation of Metals by Acids
 Ease of oxidation process depends on metal
 Metals that react with HCl or H2SO4
 Easily oxidized by H+
 More active than hydrogen (H2)
Ex. Mg, Zn, alkali metals
Mg(s) + 2H+(aq)  Mg2+(aq) + H2(g)
2Na(s) + 2H+(aq)  2Na+(aq) + H2(g)
 Metals that don’t react with HCl or H2SO4
 Not oxidized by H+
 Less active than H2
Ex. Cu, Pt
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Anion Determines Oxidizing Power
 Acids are divided into 2 classes:
1.Nonoxidizing Acids
 Anion is weaker oxidizing agent than H3O+
 Only redox reaction is
 2H+ + 2 e–  H2 or
 2H3O+ + 2 e–  H2 + 2H2O
 HCl(aq), HBr(aq), HI(aq)
 H3PO4(aq)
 Cold, dilute H2SO4(aq)
 Most organic acids (e.g., HC2H3O2)
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2. Oxidizing Acids
 Anion is stronger oxidizing agent than H3O+
 Used to react metals that are less active than H2
 No H2 gas formed
 HNO3(aq)
 Concentrated
 Dilute
 Very dilute, with strong reducing agent
 H2SO4(aq)
 Hot, conc’d, with strong reducing agent
 Hot, concentrated
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Nitrate Ion as Oxidizing Agent
A. Concentrated HNO3




Ex.
NO3– more powerful oxidizing agent than H+
NO2 is product
Partial reduction of N (+5 to +4)
NO3–(aq) + 2H+(aq) + e–  NO2(g) + H2O
oxidation
reduction
0
+5
+2
+4
Cu(s) + 2NO3–(aq) + 4H+(aq)  Cu2+(aq) + 2NO2(g) + 2H2O
Reducing Oxidizing
agent
agent
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Nitrate Ion as Oxidizing Agent
B. Dilute HNO3
 NO3– is more powerful oxidizing agent than H+
 NO is product
 Partial reduction of N (+5 to +2)
 NO3–(aq) + 4H+(aq) + 3e–  NO(g) + 2H2O
 Used to react metals that are less active than H2
Ex. Reaction of copper with dilute nitric acid
3Cu(s) + 8HNO3(dil, aq)  3Cu(NO3)2(aq) + 2NO(g) + 4H2O
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Reactions of Sulfuric Acid
A. Hot, Concentrated H2SO4
 Becomes potent oxidizer
 SO2 is product
 Partial reduction of S (+6 to +4)
 SO42– + 4H+ + 2e–  SO2(g) + 2H2O
Ex. Cu + 2H2SO4(hot, conc.)  CuSO4 + SO2 + 2H2O
B. Hot, conc’d, with strong reducing agent
 H2S is product
 Complete reduction of S (+6 to –2)
 SO42– + 10H+ + 8e–  H2S(g) + 4H2O
Ex. 4Zn + 5H2SO4(hot, conc.)  4ZnSO4 + H2S + 4H2O
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Your Turn!
Which of the following statements about oxidizing
acids is false?
A. H2SO4 can behave as either an oxidizing or
nonoxidizing acid, depending on the solution
conditions.
B. Oxidizing acids can oxidize metals that are less
active than hydrogen.
C. The anions of oxidizing acids are reduced in their
reactions with metals.
D. Most strong acids are oxidizing acids.
E. Oxidizing acids are acids whose anions are stronger
oxidizing agents than H+.
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Redox Reactions of Metals
 Acids reacting with metal
 Special case of more general phenomena
Single Replacement Reaction
 Reaction where one element replaces another
 A + BC → AC + B
1.Metal A can replace metal B
 If A is more active metal, or
2.Nonmetal A can replace nonmetal C
 If A is more active than C
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Single Replacement Reaction
 Left = Zn(s) + CuSO4(aq)
 Center = Cu2+(aq) reduced to Cu(s);
Zn(s) oxidized to Zn2+(aq)
 Right = Cu(s) plated out on Zn bar
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
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Single Replacement Reaction
 Zn2+ ions take place of
Cu2+ ions in solution
 Cu atoms take place of
Zn atoms in solid
 Cu2+ oxidizes Zn° to Zn2+
 Zn° reduces Cu2+ to Cu°
 More active Zn° replaces
less active Cu2+
 Zn° is easier to oxidize!
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Activity Series of Metals
 Cu less active, can't replace Zn2+
 Can't reduce Zn2+
 Cu(s) + Zn2+(aq)  No reaction
 General phenomenon
 Element that is more easily oxidized will displace
one that is less easily oxidized from its compounds
Activity Series (Table 6.3)
 Metals at bottom more easily oxidized (more active)
than those at top
 This means that given element will be displaced
from its compounds by any metal below it in table
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How Activity Series Generated
2H+(aq) + Sr(s)  Sr2+(aq) + H2(g)
 H+ oxidizes Sro to Sr2+
 Sro reduces H+ to H2
 More active Sro replaces less active H+
 Sro is easier to oxidize!
H2 (g) + Sr2+(aq)  NO REACTION!
 Why?
 H2 less active, can't replace Sr2+
 Can't reduce Sr2+
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Learning Check: Metal Activity
 Using the following observations, rank these
metals from most reactive to least reactive:
 Cu(s) + HCl(aq) → no reaction
 Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
 Mg(s) + ZnCl2(aq) → MgCl2(aq) + Zn(s)
Mg > Zn > H > Cu
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Table 6.3 Activity Series of Some Metals
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Reactivity Varies by Metal
 M at very bottom of Table
 Very strong reducing agents
 Very easily oxidized
 Na down to Cs
 Alkali & alkaline earth metals
 React with H2O as well as H+
2Na(s) + 2H2O  H2(g) + 2NaOH(aq)
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Reactivity Varies by Metal
 Ag = no reaction (top of activity series)
 2HCl(aq) + Ag(s)  2AgCl(aq) + H2(g)
 Zn= somewhat reactive (middle of activity series)
 2HCl(aq) + Zn(s)  ZnCl2(aq) + H2(g)
 Mg = very reactive (bottom of activity series)
 2HCl(aq) + Mg(s)  MgCl2(aq) + H2(g)
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Using Activity Series to Predict
Reactions
If M is below H
 Can displace H from solutions containing H+
2H+  H2(g)
If M is above H
 Doesn't react with Nonoxidizing acids
 HCl, H3PO4, etc.
In general
 Metal below replaces ion above
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Uses of Activity Series
 Predictive tool for determining outcome of
single replacement reactions
 Given M & M'n+
 Look at chart & draw arrow from M to M'n+
 Arrow that points up from bottom left to
top right will occur
 Arrow that points down from top left to
bottom right will NOT occur
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Learning Check
2Au3+(aq) + 3Ca(s)  2Au(s) + 3Ca2+(aq)
rxn occurs
Au(s) + Ca2+(aq) 
NO reaction
Sn(s) + Na+(aq) 
NO reaction
Mn(s) + Co2+(aq) 
Co(s) + Mn2+(aq)
rxn occurs
Cu(s) + H+(aq) 
NO reaction
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Your Turn!
The activity series of metals is
Au < Ag < Cu < Sn < Cd < Zn < Al < Mg < Na < Cs
(least active)
(most active)
Based on this list, which element would undergo
reduction most readily?
A. Ag
B. Al
C. Cu
D. Cd
E. Zn
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Oxygen as an Oxidizing Agent
 Oxygen Reacts With Many Substances
Combustion
 Rapid reaction of substance with oxygen that gives
off both heat and light
 Hydrocarbons are important fuels
 Products depend on how much O2 is available
1. Complete Combustion
 O2 plentiful
 CO2 & H2O products
Ex. CH4(g) + 2 O2(g)  CO2(g) + 2 H2O
2 C8H18(g) + 25 O2(g)  16 CO2(g) + 18 H2O
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Oxidation of Organic Compounds
2. Incomplete Combustion
 Not enough O2
a. Limited O2 supply
 CO is carbon product
2CH4(g) + 3O2(g)  2CO(g) + 4H2O
b. Very limited O2
 C(s) is carbon product
CH4(g) + O2(g)  C(s) + 2H2O
 Gives tiny black particles
 Soot—lamp black
 Component of air pollution
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Oxidation of Organic Compounds
3. Combustion of Organics containing O
 Still produce CO2 & H2O
 Need less added O2
C12H22O11(s) + 12 O2(g)  12 CO2(g) + 11 H2O
4. Combustion of Organics containing S
 Produce SO2 as product
2C4H9SH + 15O2(g)  8CO2(g) + 10H2O + 2SO2(g)
 SO2 turns into acid rain when mixed with water
 SO2 oxidized to SO3
 SO3 reacts with H2O to form H2SO4
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B. Reaction of Metals with O2
 Corrosion
 Direct reaction of metals with O2
 Many metals corrode or tarnish when
exposed to O2
Ex.
2Mg(s) + O2(g)  2MgO(s)
4Al(s) + 3O2(g)  2Al2O3(s)
4Fe(s) + 3O2(g)  2Fe2O3(s)
4Ag(s) + O2(g)  2Ag2O(s)
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C. Reaction of Nonmetals with O2
 Many nonmetals react directly with O2 to form
nonmetal oxides
 Sulfur reacts with O2
 Forms SO2
S(s) + O2(g)  2SO2(g)
 Nitrogen reacts with O2
 Forms various oxides
 NO, NO2, N2O, N2O3, N2O4, and N2O5
 Dinitrogen oxide, N2O
 Laughing gas used by dentists
 Propellant in canned whipped cream
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Learning Check: Complete
Following Reactions
 Aluminum metal and oxygen gas forms
aluminum oxide solid
4Al(s) + 3O2(g) → 2Al2O3(s)
 Solid sulfur (S8) burns in oxygen gas to make
gaseous sulfur trioxide
S8(s) + 12O2(g) → 8SO3(g)
 Copper metal is heated in oxygen to form black
copper(II) oxide solid
2Cu(s) + O2(g) → 2CuO(s)
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Your Turn!
Which of the following reactions is not a redox
reaction?
A. Na2S(aq) + MnCl2(aq)  2NaCl(aq) + MnS(s)
B. CH4(g) + O2(g)  C(s) + 2H2O
C. 2Zn(s) + O2(g)  2ZnO(s)
D. Cu(s) + 4H+(aq) + 2NO3–(aq)  Cu2+(aq) + 2NO2(g)
+ 2H2O
E. Sr(s) + 2H+(aq)  Sr2+(aq) + H2(g)
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Stoichiometry in Redox Reactions
 Like any other stoichiometry problem
 Balance redox reaction
 Use stoichiometric coefficients to relate mole of 1
substance to moles of another
 Types of problems
 Start with mass or volume of one reactant & find
mass or volume of product
 Perform titrations
 Have limiting reactant calculations
 Calculate % yields
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Stoichiometry in Redox Reactions
Ex. How many grams of Na2SO3 (126.1 g/mol)
are needed to completely react with 12.4 g of
K2Cr2O7 (294.2 g/mol)?
 1st need balanced redox equation
8H+(aq) + Cr2O72(aq) + 3SO32(aq)  3SO42(aq)
+ 2Cr3+(aq) + 4H2O
 Then do calculations
1. g K2Cr2O7  moles K2Cr2O7  moles Cr2O72(aq)
2. moles Cr2O72(aq)  moles 3SO42(aq)
3. moles SO32(aq)  moles Na2SO3  g Na2SO3
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Stoichiometry Example (cont)
grams K2Cr2O7  moles K2Cr2O7  moles Cr2O72 (aq)
1 mol K 2 Cr2 O 7
1mol Cr2 O 27 
12.4 g K 2 Cr2 O 7 

294.2 g K 2 Cr2 O 7 1 mol K 2 Cr2 O 7
 0.0421mol Cr2O27 
moles Cr2O72 (aq)  moles 3SO32 (aq)
0.0421mol Cr2 O 27  
3 mol SO 23 
1 mol Cr2 O 27 
 0.126 mol SO23 
moles SO32 (aq)  moles Na2SO3  g Na2SO3
0.126 mol
SO23 

1 mol Na2SO3
1 mol SO23 
126.1 g Na2SO3

1 mol Na2SO3
 15.9 g Na2SO3
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Redox Titrations
 Equivalence point reached when # of moles of
oxidizing & reducing agents have been mixed in
the correct stoichiometric ratio
 No simple indicators to detect endpoints
 3 very useful oxidizing agents that change color
1. KMnO4: Deep purple of MnO4 fades to almost
colorless Mn2+ (very pale pink)
2. K2Cr2O7: Bright yellow orange of Cr2O72 changes
to pale blue green of Cr3+
3. IO3 : When reduced to I2(s) in presence of I,
forms I3 which forms dark blue complex with
starch
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Redox Titration Example
I reacts with IO3 in acidic solution to form
I2(s). If 12.34 mL of 0.5678M I is needed to
titrate 25.00 mL of a solution containing IO3,
what is the M of the solution?
1. Write Unbalanced Equation
1
+5
0


I (aq) + IO3 (aq)  I2(s)

 I (aq) is oxidized to I2
 IO3(aq) is reduced to I2
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Redox Titration Example (cont)
2. Balance Equation
 Note: we are in acidic solution
5  [2I(aq) 
2IO3(aq) + 12H+(aq) + 10e 
10I(aq) + 2IO3(aq) + 12H+(aq) 
2
2
2
I2(s) + 2e ]
I2(s) + 6H2O
6I2(s) + 6H2O
2
2
 Not done as not lowest whole number coefficients
5I(aq) + IO3(aq) + 6H+(aq)  3I2(s) + 3H2O
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3. Now for the Calculations
 Calculate mmol of I– titrated

12.34 mL I 
0.5678mmol I

 7.007mmol I

1 mL I 
 Convert to mmol of IO3– present

1
mmol
IO
3
 1.401mmol IO3
7.007 mmol I  

5 mmol I
 Convert to M of IO3– solution

1.401mmol IO 3
25.00 mL IO 3
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= 0.0561 M IO3–
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Ore Analysis
A 0.3000 g sample of tin ore was dissolved in acid
solution converting all the tin to tin(II). In a titration,
8.08 mL of 0.0500 M KMnO4 was required to oxidize
the tin(II) to tin(IV). What was the percentage tin in
the original sample?
3Sn2+(aq) + 2MnO4(aq) + 8H+(aq) 
3Sn4+(aq) + 2MnO2(s) + 4H2O
M of KMnO4  V = mol KMnO4
mol KMnO4  mol Sn/mol KMnO4 = mol Sn2+
mol Sn2+  MM = g Sn2+ in sample
%Sn = g Sn/g sample  100 %
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Tin Ore Analysis Continued
M of KMnO4  V = mmol KMnO4
0.0500 M KMnO4  8.08 mL = 0.404 mmol KMnO4
mmol KMnO4  mmol MnO4  mmol Sn2+
1 mmol MnO 4
3 mmol Sn2 
0.404 mmol KMnO 4 ×
×
1 mmol KMnO 4
2 mmol MnO 4
= 0.606 mmol Sn2+
Mol Sn2+  g/mol = g Sn in original sample
0.606 mmol Sn2  
1 mmol Sn
1 mmol Sn2 
= 0.07194 g Sn

118.7 mg Sn
1g

1 mmol Sn 1000 mg
%Sn = g Sn/ g sample  100 %
0.07194 g Sn
 100 = 23.97% Sn
0.3000 g ore
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Your Turn!
The amount of hydrogen peroxide (H2O2, MM = 34.01
g/mol) in hair bleach was determined by titration with a
standard KMnO4 (MM = 158.0 g/mol) solution:
2MnO4–(aq) + 5H2O2(aq) + 6H+(aq)  5O2(g) + 2Mn2+(aq) + 8H2O
If 43.2 mL of 0.105 M MnO4– was needed to reach the
endpoint, how many grams of H2O2 are in the sample
of hair bleach?
A. 0.771 g
B. 0.386 g
C. 0.0771 g
D. 386 g
E. 154 g
0.105M
MnO 4
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5 mol H2 O2 34.01 g H2 O2
43.2 mL




1000 mL/1 L 2 mol MnO 4 1 mol H2 O2
= 0.386 g H2O2
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