Chapter 6: Oxidation-Reduction Reactions

1
Chapter 6 Oxidation-Reduction Reactions

• Chemistry The Molecular Nature of Matter, 6E
• Jespersen/Brady/Hyslop

2
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

3
OxidationReduction 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

4
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

5
Redox Reactions

• Very common
• Batteriescar, flashlight, cell phone, computer
• Metabolism of food
• Combustion
• Chlorine Bleach
• Dilute NaOCl solution
• Cleans through redox
• reaction
• Oxidizing agent
• Destroys stains by oxidizing them

6
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

7
Your Turn!

• Which species functions as the oxidizing agent in
  the following oxidation-reduction reaction?
• Zn(s) Pt2(aq) ?? Pt(s) Zn2(aq)
• Pt(s)
• Zn2(aq)
• Pt2(aq)
• Zn(s)
• None of these, as this is not a redox reaction.

8
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

9
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

• Group 6A elements have 2 oxidation number.
• Group 5A elements have 3 oxidation number.
• 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

• 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
• CO2
• C (1 atom) (x) x
• O (2 atoms) (2) 4 (Rule 5)
• sum 0 (Rule 1)
• x ? 4 0 or x 4
• 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
• 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" sense.
  
• 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? oxidation
• Cu2(aq) 2e? ?? Cu(s) 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 Fe 2 Cr 3 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
  half-reactions
• 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
• Right side has 7 O atoms
• Left side has none
• Add 7 H2O to left side
• Fe2 ?? Fe3
• No O to balance

7 H2O
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Ex. 2 Ion-Electron Method in Acid

• 4. Balance H by adding H to side that needs H
• 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

14H
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Ex. 2 Ion-Electron Method in Acid

• 5. Balance net charge by adding electrons.
• 14H Cr2O72? ?? 2Cr3 7H2O
• 6 electrons must be added to reactant side
• Fe2 ?? Fe3
• 1 electron must be added to product side
• Now both half-reactions balanced for mass charge

6e?
Net Charge 2(3)7(0) 6
Net Charge 14(1) (2) 12
e?
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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
• Fe2 ?? Fe3 e?
• 7. Cancel anything that's the same on both sides

6?

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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.
• 9. Combine H and OH to form H2O.
• 10. Cancel any H2O that you can

14 OH
14 OH
7
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Your Turn!

• Which of the following is a correctly balanced
  reduction half-reaction?
• Fe3 e ?? Fe
• 2Fe 6HNO3 ?? 2Fe(NO3)3 3H2
• Mn2 4H2O ?? MnO4 8H 5e
• 2O2 ?? O2 4e
• 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
• HSO3? ?? SO42?
• 4. Add H to balance H
• MnO4? ?? Mn2 4H2O
• H2O HSO3? ?? SO42?

4H2O
H2O
8H
3H
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Ex. 3 Ion-Electron Method

• 5. Balance net charge by adding e.
• 8H MnO4? ?? Mn2 4H2O
• 8?(1) (1) 7 2 0 2
• Add 5 e to reactant side
• H2O HSO3? ?? SO42? 3H
• 0 (1) 1 2 3(1) 1
• Add 2 e to product side

5e
2 e
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Ex. 3 Ion-Electron Method

• 6. Make e gain equal e loss
• 5e 8H MnO4? ?? Mn2 4H2O
• 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

2?

5?

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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
• 7. Cancel anything that is the same on both
  sides.
• Balanced in acid.

3
1
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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
• 9. Combine H and OH to form H2O
• 10. Cancel any H2O that you can
• 2MnO4? 5HSO3? ? 2Mn2 2H2O OH?
  5SO42?

OH
OH
2
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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)
• 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 dont 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
• 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, concd, with strong reducing agent
• Hot, concentrated

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Nitrate Ion as Oxidizing Agent

• A. Concentrated HNO3
• 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
• Ex.

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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, concd, 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?
• H2SO4 can behave as either an oxidizing or
  nonoxidizing acid, depending on the solution
  conditions.
• Oxidizing acids can oxidize metals that are less
  active than hydrogen.
• The anions of oxidizing acids are reduced in
  their reactions with metals.
• Most strong acids are oxidizing acids.
• 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
• Metal A can replace metal B
• If A is more active metal, or
• 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)

50
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!

51
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

53
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 gt Zn gt H gt 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)

57
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

58
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) ??
• Au(s) Ca2(aq) ??
• Sn(s) Na(aq) ??
• Mn(s) Co2(aq) ??
• Cu(s) H(aq) ??

2Au(s) 3Ca2(aq) rxn occurs
NO reaction
NO reaction
Co(s) Mn2(aq) rxn occurs
NO reaction
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Your Turn!

• The activity series of metals is
• Au lt Ag lt Cu lt Sn lt Cd lt Zn lt Al lt Mg lt Na lt Cs
• (least active) (most active)
• Based on this list, which element would undergo
  reduction most readily?
• Ag
• Al
• Cu
• Cd
• Zn

61
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

62
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
• Sootlamp black
• Component of air pollution

63
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

64
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)

65
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

66
Learning Check Complete Following Reactions

• Aluminum metal and oxygen gas forms aluminum
  oxide solid
• Solid sulfur (S8) burns in oxygen gas to make
  gaseous sulfur trioxide
• Copper metal is heated in oxygen to form black
  copper(II) oxide solid

4Al(s) 3O2(g) ? 2Al2O3(s)
S8(s) 12O2(g) ? 8SO3(g)
2Cu(s) O2(g) ? 2CuO(s)
67
Your Turn!

• Which of the following reactions is not a redox
  reaction?
• Na2S(aq) MnCl2(aq) ?? 2NaCl(aq) MnS(s)
• CH4(g) O2(g) ?? C(s) 2H2O
• 2Zn(s) O2(g) ?? 2ZnO(s)
• Cu(s) 4H(aq) 2NO3(aq) ? Cu2(aq) 2NO2(g)
  2H2O
• Sr(s) 2H(aq) ?? Sr2(aq) H2(g)

68
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

69
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

70
Stoichiometry Example (cont)

• grams K2Cr2O7 ? moles K2Cr2O7 ? moles Cr2O72?
  (aq)
• moles Cr2O72? (aq) ? moles 3SO32? (aq)
• moles SO32? (aq) ? moles Na2SO3 ? g Na2SO3

71
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

72
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

73
Redox Titration Example (cont)

• 2. Balance Equation
• Note we are in acidic solution
• 2I?(aq) ?? I2(s) 2e?
• Not done as not lowest whole number coefficients
• 5I?(aq) IO3?(aq) 6H(aq) ? 3I2(s) 3H2O

5 ?
2IO3?(aq) 12H(aq) 10e? ?? I2(s) 6H2O
10I?(aq) 2IO3?(aq) 12H(aq) ? 6I2(s) 6H2O
74
3. Now for the Calculations

• Calculate mmol of I titrated
• Convert to mmol of IO3 present
• Convert to M of IO3 solution

0.0561 M IO3
75
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?
• 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

3Sn2(aq) 2MnO4?(aq) 8H(aq) ??
3Sn4(aq) 2MnO2(s) 4H2O
76
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
• Mol Sn2 ? g/mol g Sn in original sample
• Sn g Sn/ g sample ? 100

0.606 mmol Sn2
0.07194 g Sn
23.97 Sn
77
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?
• 0.771 g
• 0.386 g
• 0.0771 g
• 386 g
• 154 g

0.386 g H2O2