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Ch. 6 Oxidation-Reduction Reactions. Index. 5.1. Oxidation-reduction reactions involve electron transfer   5.2. The ion-electron method creates balanced net ionic equations for redox reactions   5.3. Metals are oxidized when they react with acids  

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index
Index

5.1. Oxidation-reduction reactions involve electron transfer  

5.2. The ion-electron method creates balanced net ionic equations for redox reactions  

5.3. Metals are oxidized when they react with acids  

5.4. A more active metal will displace a less active one from its compounds  

5.5. Molecular oxygen is a powerful oxidizing agent  

5.6. Redox reactions follow the same stoichiometric principles as other reactions

5.1. Oxidation-reduction reactions involve electron transfer

oxidation reduction reactions
Involve two processes:

Oxidation – the loss of electrons, and

Reduction – the gain of electrons

Ca(s) + 2 H2O(l)→ Ca(OH)2(s) + H2(g)

Oxidizer = Oxidizing agent = received the electrons and is reduced during the reaction

Reducer = Reducing agent = donated the electrons and is oxidized during the reaction

Oxidation –Reduction Reactions

5.1. Oxidation-reduction reactions involve electron transfer

guidelines for redox reactions
Guidelines For Redox Reactions
  • Oxidation and reduction always occur together
  • Total number of e- lost by one substance is the same as the total number of e- gained by the other
  • For a redox reaction to occur, something must accept the e- that are lost by another substance

5.1. Oxidation-reduction reactions involve electron transfer

oxidation numbers
Oxidation Numbers
  • Oxidation number:
    • of any free element is zero
    • of any simple, monoatomic ion is equal to the charge on the ion
    • Fluorine in compounds is –1
    • Group 1A in compounds is +1
    • Oxygen in compounds is -2
  • the sum of all oxidation numbers of the atoms in a molecule or polyatomic ion must equal the charge on the particle

5.1. Oxidation-reduction reactions involve electron transfer

special rules
If there is a conflict between two rules apply the rule listed first

Note that fractional values of oxidation numbers are allowed

Special Rules

5.1. Oxidation-reduction reactions involve electron transfer

assign oxidation states to all atoms
Assign Oxidation States To All Atoms:
  • Fe2O3
    • O: -2; Fe: +3
  • Na2CO3
    • Na: +1; O: -2; C: +4
  • V(OH)3
    • O: -2; H: +1; V: +3
  • K2Cr2O7
    • K: +1; O: -2; Cr: +6

5.1. Oxidation-reduction reactions involve electron transfer

your turn
Your Turn!

What is the oxidation number of F atoms in F2O?

  • -1
  • -2
  • -3
  • -4
  • none of these

5.1. Oxidation-reduction reactions involve electron transfer

your turn9
Your Turn!

What is the oxidation number of P in H3PO3?

  • +1
  • +2
  • +3
  • +4
  • none of these

5.1. Oxidation-reduction reactions involve electron transfer

your turn10
Your Turn!

What is the oxidation number of O atoms inH2O2?

  • -1
  • -2
  • -3
  • -4
  • none of these

5.1. Oxidation-reduction reactions involve electron transfer

determining if a redox reaction is occurring
Determining if a Redox reaction is occurring
  • If the oxidation numbers change, a redox reaction is occurring
    • Increasing oxidation # denotes oxidation
    • Decreasing oxidation # denotes reduction
  • If the oxidation numbers remain static, no redox is occurring

5.2 The ion–electron method creates balanced net ionic equations for redox reactions

balancing redox reactions ion electron method
Balancing Redox Reactions: Ion-Electron Method
  • Identify the half-reactions
  • Balance each atom in the half reaction, saving H and O for last
  • Balance O by adding 1 water molecule for each needed O
  • Balance H by adding 1 H+ ion for each needed H

5.2 The ion–electron method creates balanced net ionic equations for redox reactions

balancing redox reactions cont
Balancing Redox Reactions: (Cont.)
  • Balance charges by adding electrons to the more positive side
  • Find the least common multiple of electrons for the two half- reactions. Multiply each reaction by the factor needed to achieve the LCM of electrons
  • Add the half reactions, canceling like substances that appear on both sides

5.2 The ion–electron method creates balanced net ionic equations for redox reactions

balancing basic reactions
Balancing Basic Reactions
  • The simplest way to balance reactions in basic solution is to first balance them as if they were in acidic solution, then “convert” to basic solution
  • Additional Steps for Basic Solutions

8) To both sides of the equation, add the same number of OH- ions as there are H+.

9) Combine H+ and OH- to form H2O

10) Cancel H2O molecules that are on both sides of the reaction.

5.2 The ion–electron method creates balanced net ionic equations for redox reactions

balance using the ion electron method
Balance, Using the Ion-Electron Method
  • MnO4- + C2O42-→ MnO2 + CO32- (acidic)
    • MnO4- + 4H+ +3e-→ MnO2 + 2H2O
    • C2O42- + 2H2O → 2CO32- + 4H+ +2e-
    • 2MnO4- + 8H+ +6e-+ 3C2O42- + 6H2O → 2MnO2 + 4H2O + 6CO32- + 12H+ +6e-
    • 2MnO4- + 3C2O42- + 2H2O → 2MnO2 + 6CO32- + 4H+
  • ( )×2
  • ( )×3

5.2 The ion–electron method creates balanced net ionic equations for redox reactions

balance using the ion electron method16
Balance, Using the Ion-Electron Method
  • Mn(s) + Cu2+(aq)→ Cu(s) + MnO2(s) (basic)
    • Mn + 2H2O → MnO2 + 4H+ + 4e-
    • Cu2+(aq) + 2e-→ Cu(s)
    • Mn + 2H2O + 2Cu2+(aq) + 4e-→ MnO2 + 4H+ + 4e-+ 2Cu(s)
    • Mn + 2H2O + 2Cu2+(aq)→ MnO2 + 4H+ + 2Cu(s)
    • Mn + 2H2O + 2Cu2+(aq) +4OH-(aq)→ MnO2 + 4H+ + 4OH-(aq)+ 2Cu(s)
    • Mn + 2Cu2+(aq) +4OH-(aq)→ MnO2 + 2H2O (aq)+ 2Cu(s)
  • ( )×1
  • ( )×2

5.2 The ion–electron method creates balanced net ionic equations for redox reactions

balance using the ion electron method17
Balance, Using the Ion-Electron Method
  • H2SO3(aq) + H2CO3(aq) →H2C2O4(aq) + SO42- (aq)
    • H2SO3(aq) + H2O → SO42- (aq)+ 4H+(aq) + 2e-
    • 2 H2CO3(aq) + 2 H+ + 2e- →H2C2O4(aq) + 2H2O
    • H2SO3(aq) + 2 H2CO3(aq) →H2C2O4(aq) + SO42- (aq) + H2O + 2 H+

5.2 The ion–electron method creates balanced net ionic equations for redox reactions

balance using the ion electron method18
Balance, Using the Ion-Electron Method
  • ClO- + VO3-(aq) →ClO3-(aq)+ V(OH)3(s) (basic)
    • ClO- + 2H2O →ClO3-(aq) + 4H+ + 4e-
    • 3H+ + 2e- + VO3- (aq) → V(OH)3(s)
    • ClO- + 2H2O + 2H+ + 2VO3- (aq) → ClO3-(aq) + 2V(OH)3(s)
    • ClO- + 4H2O + 2VO3- (aq) → ClO3-(aq) + 2V(OH)3(s) + 2OH-

5.2 The ion–electron method creates balanced net ionic equations for redox reactions

single displacement replacement reaction
A + BC → AC + B

Metal A can replace metal B if it is a more active metal, or

Nonmetal A can replace nonmetal C if it is more active than C.

An activity series arranges metals according to their ease of oxidation (Table 5.2)

Single Displacement (Replacement) Reaction

5.4 A more active metal will displace a less active one from its compounds

activity series
Activity Series

5.4 A more active metal will displace a less active one from its compounds

interactions between different metals galvanic cells
Interactions between different metals: Galvanic Cells

5.1. Oxidation-reduction reactions involve electron transfer

interactions between different metals galvanic corrosion
Interactions between different metals: Galvanic Corrosion

5.1. Oxidation-reduction reactions involve electron transfer

interactions between different metals sacrificial anodes
Interactions between different metals: Sacrificial Anodes

5.1. Oxidation-reduction reactions involve electron transfer

your turn24
Your Turn!

Which of the following will react with Cu(s)?

  • HNO3(conc)
  • ZnCl2(aq)
  • AgNO3(aq)
  • None of these

5.4 A more active metal will displace a less active one from its compounds

oxygen reacts with many substances
Oxygen Reacts With Many Substances

The products depend, in part, on how much oxygen is available

  • Combustion of hydrocarbons
    • O2 plentiful; CH4 + 2O2→ CO2 + 2H2O
    • O2 limited: 2CH4 + 3O2 → 2 CO + 4H2O
    • O2 scant: CH4 + O2 → C + 2H2O
  • Organic compounds containing O also produce carbon dioxide and water
    • C2H5OH+ 3O2→ 2CO2 + 3H2O
  • Organic compounds containing S produce sulfur dioxide
    • 2C2H5SH + 9O2→4CO2 + 6H2O + 2 SO2

5.5 Molecular oxygen is a powerful oxidizing agent

other oxygen reactions
Other Oxygen Reactions
  • Organic compounds containing S produce sulfur dioxide
    • 2C2H5SH + 9O2→4CO2 + 6H2O + 2 SO2
  • Many metals corrodeor tarnish when exposed to oxygen
    • 4Fe + 3O2→2Fe2O3
    • 4Ag + O2 →2Ag2O
  • Most nonmetals react with oxygen directly
    • Plentiful: C + O2→CO2
    • Limited: 2C + O2 →CO

5.5 Molecular oxygen is a powerful oxidizing agent

your turn27
Your Turn!

What is the coefficient on O2 when octane, C8H18 is combusted with scant oxygen?

  • 1
  • 2
  • 3
  • 4
  • none of these

2 C8H18 + 9 O2→16 C + 18 H2O

5.5 Molecular oxygen is a powerful oxidizing agent

ore analysis
Ore Analysis

A 0.3000 g sample of tin ore (containing tin (II)) is titrated with 8.08 mL of 0.0500 M KMnO4 to oxidize the all the tin(II) to tin(IV) and reducing the MnO4- to MnO2 under acid conditions. What was the percentage tin in the original sample?

5.6 Redox reactions follow the same stoichiometric principles as other reactions

your turn29
Your Turn!

A 25.0 g sample of granite contains a vein of copper. What is the % of Cu present if 25.00 mL of concentrated 15 M HNO3 are reacted? The Cu is converted to Cu2+ and the NO3- ions are converted to NO2.

a) 12

b) 38

c) 48

d) 95

e) none of these

47.7 %

5.6 Redox reactions follow the same stoichiometric principles as other reactions

slide30

Key Points About Redox Reactions

  • Oxidation (electron loss) always accompanies reduction (electron gain).
  • The oxidizing agent is reduced, and the reducing agent is oxidized.
  • The number of electrons gained by the oxidizing agent always equals the number lost by the reducing agent in the balanced equation.
slide31

Zn(s) + 2H+(aq) Zn2+(aq) + H2(g)

Figure 21.1

A summary of redox terminology.

OXIDATION

One reactant loses electrons.

Zn loses electrons.

Reducing agent is oxidized.

Zn is the reducing agent and becomes oxidized.

Oxidation number increases.

The oxidation number of Zn increases from 0 to +2.

REDUCTION

Other reactant gains electrons.

Hydrogen ion gains electrons.

Oxidizing agent is reduced.

Hydrogen ion is the oxidizing agent and becomes reduced.

Oxidation number decreases.

The oxidation number of H decreases from +1 to 0.

slide32

Half-Reaction Method for Balancing Redox Reactions

  • Summary: This method divides the overall redox reaction into oxidation and reduction half-reactions.
  • Each reaction is balanced for mass (atoms) and charge.
  • One or both are multiplied by some integer to make the number of electrons gained and lost equal.
  • The half-reactions are then recombined to give the balanced redox equation.
  • Advantages:
  • The separation of half-reactions reflects actual physical separations in electrochemical cells.
  • The half-reactions are easier to balance especially if they involve acid or base.
  • It is usually not necessary to assign oxidation numbers to those species not undergoing change.
slide33

Cr2O72-(aq) + I-(aq) Cr3+(aq) + I2(aq)

Cr2O72- Cr3+

I-I2

Cr2O72- Cr3+

6e- +

14H+(aq) +

Cr2O72- Cr3+

2

+ 7H2O(l)

Balancing Redox Reactions in Acidic Solution

Cr2O72-(aq) + I-(aq) Cr3+(aq) + I2(aq)

1. Divide the reaction into half-reactions -

Determine the O.N.s for the species undergoing redox.

+6

-1

+3

0

Cr is going from +6 to +3

I is going from -1 to 0

2. Balance atoms and charges in each half-reaction -

14H+(aq) +

2

+ 7H2O(l)

net: +6

Add 6e- to left.

net: +12

slide34

2

+ 2e-

X 3

I-I2

I-I2

I-I2

6e- +

6e- +

14H+ +

Cr2O72- Cr3+

2

+ 7H2O(l)

6

3

+ 6e-

14H+(aq) +

Cr2O72-(aq) + 6 I-(aq) 2Cr3+(aq) + 3I2(s)

+ 7H2O(l)

14H+(aq) +

Cr2O72- Cr3+

+ 7H2O(l)

Balancing Redox Reactions in Acidic Solution

continued

2

2

+ 2e-

Cr(+6) is the oxidizing agent and I(-1)is the reducing agent.

3. Multiply each half-reaction by an integer, if necessary -

4. Add the half-reactions together -

Do a final check on atoms and charges.

slide35

14H2O +

Cr2O72- + 6 I- 2Cr3+ + 3I2

+ 7H2O + 14OH-

7H2O +

Cr2O72- + 6 I- 2Cr3+ + 3I2

+ 14OH-

14H+(aq) +

Cr2O72-(aq) + 6 I-(aq) 2Cr3+(aq) + 3I2(s)

+ 7H2O(l)

Balancing Redox Reactions in Basic Solution

Balance the reaction in acid and then add OH- so as to neutralize the H+ ions.

+ 14OH-(aq)

+ 14OH-(aq)

Reconcile the number of water molecules.

Do a final check on atoms and charges.

slide36

Cr2O72-

I-

Figure 21.2

The redox reaction between dichromate ion and iodide ion.

Cr3+ + I2

slide37

PROBLEM:

Permanganate ion is a strong oxidizing agent, and its deep purple color makes it useful as an indicator in redox titrations. It reacts in basic solution with the oxalate ion to form carbonate ion and solid mangaese dioxide. Balance the skeleton ionic reaction that occurs between NaMnO4 and Na2C2O4 in basic solution:

MnO4-(aq) + C2O42-(aq) MnO2(s) + CO32-(aq)

C2O42- CO32-

MnO4- MnO2

MnO4- MnO2

C2O42- CO32-

2

MnO4- MnO2

C2O42-2CO32-

Sample Problem 21.1:

Balancing Redox Reactions by the Half-Reaction

Method

PLAN:

Proceed in acidic solution and then neutralize with base.

SOLUTION:

+7

+4

+3

+4

4H+ +

+ 2H2O

+ 2H2O

+ 4H+

+3e-

+2e-

slide38

3C2O42- + 6H2O 6CO32- + 12H+ + 6e-

C2O42- + 2H2O 2CO32- + 4H+ + 2e-

C2O42- + 2H2O 2CO32- + 4H+ + 2e-

4H+ + MnO4- +3e- MnO2+ 2H2O

4H+ + MnO4- +3e- MnO2+ 2H2O

8H+ + 2MnO4- +6e- 2MnO2+ 4H2O

8H+ + 2MnO4- +6e- 2MnO2+ 4H2O

X 3

X 2

3C2O42- + 6H2O 6CO32- + 12H+ + 6e-

2MnO2-(aq) + 3C2O42-(aq) + 2H2O(l) 2MnO2(s) + 6CO32-(aq) + 4H+(aq)

2MnO2-(aq) + 3C2O42-(aq) + 4OH-(aq) 2MnO2(s) + 6CO32-(aq) + 2H2O(l)

Sample Problem 21.1:

Balancing Redox Reactions by the Half-Reaction

Method

continued:

+ 4OH-

+ 4OH-