Redox Reactions

1
Redox Reactions

• AP Chemistry Unit 3

2
Titration volumetric analysis to determine
concentration lab work!

• Stoichiometric point equivalence point
• Moles of acid (H) moles of base (OH-) in
  acid-base titrations
• How many mL of 0.610 M sodium hydroxide solution
  are needed to neutralize 20.0 mL of 0.245 M
  sulfuric acid solution?
• Write a balanced equation
• 2. Identify moles of one reactant.
• 3. Use mole ratios to predict moles of other
  reactant.
• 4. Solve for Molarity or mass.

3
2NaOH H2SO4 ? 2 H2O Na2SO4

• 0.245 M ___x____
• 0.0200 L
• 0.00490 moles

• 0.00980 moles
• 0.00980 moles 0.610 M
• x
• 0.0161 L 16.1 mL
• since we are finding molarity of entire
  reactant (not net ionic) use the molecular
  equationusually common in titration problems

4
Redox reaction transfer of electrons

• Reduction process of gaining electrons becoming
  more negative
• LEO the lion

• Oxidation process of losing electrons becoming
  more positive
• goes GER

5
Oxidation and Reduction

• Reactant reduced is the oxidizing agent.
• H oxidizes Zn by taking electrons from it. (e-
  acceptor)
• Reactant oxidized is the reducing agent.
• Zn reduces H by giving it electrons. (e- donor)

6
Oxidation Numbers

• In order to keep track of what loses electrons
  and what gains them, we assign oxidation numbers.

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Oxidation numbers charge of an atomCa2 F-1
Mn7
-2
-2 x 4 -8

• K2SO4
• CO3-2
• ___ 3(-2) -2
• Fe3O4
• 3(__) 4(-2) 0

• Rules
• Elements 0
• Hydrogen 1 except in metal hydrides (NaH), then
  it is -1. Placement is the key!
• Oxygen -2 except in peroxides (H2O2 O -1)
• In a compound, oxidation numbers must balance out
  to net charge of zero.

2(1) ___ (-8) 0
8
Balancing simple redox reactions

• Cu2 Al ? Cu Al3
• 1. Cu2 ? Cu Al ? Al3
• 2. no changes
• 3. 2e- Cu2 ? Cu
• Al ? Al3 3e-
• 4. (2e- Cu2 ? Cu)3
• (Al ? Al3 3e-)2
• 5. 6e- 3Cu2 ? 3Cu
• 2Al ? 2Al3 6e-
• 2Al 3Cu2 ? 3Cu 2Al3
• 2(0) 3(2) 3(0) 2(3)

• Half reaction method
• Write ½ reactions.
• Balance atoms, if necessary (except H O)
• Balance charges by adding electrons.
• Multiply (if needed) to cancel electrons.
• Check work charges atoms

9
Half-Reaction Method

• First, we assign oxidation numbers.

Since the manganese goes from 7 to 2, it is
reduced and changes color.
Since the carbon goes from 3 to 4, it is
oxidized.
10
Balancing redox reactions in acidic solutions
MnO4-(aq) C2O42-(aq) ??? Mn2(aq) CO2(aq)

• 1. C2O42- ??? CO2
• MnO4- ??? Mn2
• 2. C2O42- ??? 2 CO2
• 3. MnO4- ??? Mn2 4 H2O
• 4. 8 H MnO4- ??? Mn2 4 H2O
• 5. C2O42- ??? 2 CO2 2 e-
• 5 e- 8 H MnO4- ?? Mn2 4 H2O
• 6. 5 C2O42- ??? 10 CO2 10 e-
• 10 e- 16 H 2 MnO4- ? 2 Mn2 8 H2O
• 16 H 2 MnO4- 5 C2O42- ???
• 2 Mn2 8 H2O 10 CO2

• Write ½ reactions
• Balance all atoms except H O, if needed.
• Balance oxygen atoms by adding H2O to one side
  since aqueous.
• Balance hydrogen atoms by adding H
• Balance electric charge by adding electrons
• Make electrons equal, to cancel
• Check atoms charge.

11
Balancing redox reactions in basic solutions

• Zn VO3-1 ? VO2 Zn2
• Zn ? Zn2
• VO3-1 ? VO2
• 3. VO3-1 ? VO2 2H2O
• 4 H VO3-1 ? VO2 2H2O
• 2 (e- 4 H VO3-1 ? VO2 2H2O)
• Zn ? Zn2 2 e-
• 6. 2e- 8 H 2VO3-1 ?2VO2 4H2O
• Zn ? Zn2 2 e-
• Zn 2 VO3-1 8H ? 2VO2 4H2O Zn2
• 8. 8 OH- 8 OH-
• 8 H2O
• Zn 2 VO3-1 8H2O ? 2VO2 4H2O Zn2 8
  OH-
• Zn 2 VO3-1 4H2O ? 2VO2 Zn2 8 OH-

• Write ½ reactions
• Balance all atoms except H O, if needed.
• Balance oxygen atoms by adding H2O to one side
  since aqueous.
• Balance hydrogen atoms by adding H
• Balance electric charge by adding electrons
• Make electrons equal, to cancel
• Check atoms charge.
• Now add OH- to both sides to eliminate H

12
Half-Reaction Method

• Consider the reaction between MnO4- and C2O42-
• MnO4-(aq) C2O42-(aq) ??? Mn2(aq) CO2(aq)
• This is an example of a redox titrationnext
  lab!