Chapter 4: OxidationReduction Reactions

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Chapter 4 Oxidation-Reduction Reactions

• Sections 9 10 Oxidation Numbers and Balancing
  of Redox Equations
• Oxidation Number hypothetical charge on an atom
  if shared (bonding) electrons are given to the
  more electronegative atom in a bond
• Ex H2O bonding es are assigned to O.
• Each H has no electrons, therefore 1 ox. state
• O atom has 8 electrons, therefore 2 ox. state
• CO2 Os have 8 es each, 2 oxidation state.
• C has no electrons, therefore 4 ox. state
  

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Rules for Assigning Oxidation Numbers

• 1) The sum of all oxidation numbers equals the
  net charge on the species.
• 2) The oxidation number of an atom in a pure
  element is zero.
• H 1 unless bonded to a less electronegative
  element (e.g., a metal), in which case it is 1.
• O 2 except in peroxides (1) or when bonded to
  F (2).
• The oxidation number equals the charge on a
  monatomic ion.
• Group 1 1 Group 2 2.

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Oxidation-Reduction (Redox) Reactions

• Reactions involving electron transfer.
• Ex Cu(s) 2 Ag(aq) ? Cu2(aq) 2
  Ag(s)
• Oxidation electron loss or increase in
  oxidation number
• (0)
  (2)
• Cu(s) ? Cu2(aq) 2 e
• Reduction electron gain or decrease in
  oxidation number
• (1)
  (0)
• Ag(aq) e ? Ag(s)

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Oxidizing and Reducing Agents

• Oxidizing Agent takes electrons away from
  another substance (becomes reduced).
• Reducing Agent gives electrons to another
  substance (becomes oxidized).
• Question In the previous reaction, which is the
  oxidizing agent and which is the reducing agent?
• Answer Ag(aq) is the oxidizing agent, and
  Cu(s) is the reducing agent.

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Balancing of Redox Equations

• This normally can not be done simply by
  inspec-tion, as in the case of non-redox
  reactions. Instead, a set procedure is used.
• Two methods may be used
• Half-Reaction (Ion-Electron) Method
• Oxidation-Number Method
• We will use the half-reaction method.

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A. For Reactions in Acidic Solution

• Abundant H and H2O are present.
• In general
• First split into half-reactions.
• Balance principal atoms (non-O, non-H)
• Balance O
• Balance H
• Balance charge with electrons.
• Make e gain equal to e loss.
• Recombine and cancel as necessary.

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• Ex. H2C2O4(aq) MnO4(aq) ? Mn2(aq)
  CO2(g)
• Split into half-reactions.
• H2C2O4 ? CO2
• MnO4 ? Mn2
• Balance principal atoms.
• H2C2O4 ? 2 CO2
• Balance oxygen using H2O.
• MnO4 ? Mn2 4 H2O

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• Balance hydrogen using H.
• H2C2O4 ? 2 CO2 2 H
• 8 H MnO4 ? Mn2 4 H2O
• Balance charge using electrons.
• H2C2O4 ? 2 CO2 2 H 2 e
• 5 e 8 H MnO4 ? Mn2 4 H2O
• Make electron gain electron loss
• H2C2O4 ? 2 CO2 2 H 2 e x 5
• 5 e 8 H MnO4 ? Mn2 4 H2O x 2

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• Recombine and cancel as necessary.
• 5 H2C2O4 2 MnO4 16 H 10 e ?
• 10 CO2 2 Mn2 10 H 8 H2O 10 e
• The result
• 5 H2C2O4 (aq) 2 MnO4 (aq) 6 H (aq) ?
• 10 CO2 (g) 2
  Mn2 (aq) 8 H2O (l)
• When finished
• Make sure that the electrons have cancelled.
• Make sure that all atoms charges balance.

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B. Reactions in Basic Solution

• An abundance of OH and H2O is present.
• Ex. SnO22(aq) Bi(OH)3(s) ? Bi(s) SnO32(aq)
• Split into half-reactions.
• SnO22 ? SnO32
• Bi(OH)3 ? Bi
• 2) Balance principal elements.
• (Already done).

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• 3) Balance oxygen with 2 OH/H2O
• 2 OH SnO22 ? SnO32 H2O
• 3 H2O Bi(OH)3 ? Bi 6 OH
• Balance hydrogen with H2O/OH (no H!)
• 3 OH 3 H2O Bi(OH)3 ? Bi 6 OH 3 H2O
• Balance charge with electrons.
• 2 OH SnO22 ? SnO32 H2O 2 e
• 3 e Bi(OH)3 ? Bi 3 OH

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• Make electron gain electron loss
• 2 OH SnO22 ? SnO32 H2O 2 e
  x 3
• 3 e Bi(OH)3 ? Bi 3 OH x 2
• Recombine, then cancel.
• 6 OH 3 SnO22 6 e 2 Bi(OH)3 ?
• 3 SnO32 3 H2O 6 e 2 Bi
  6 OH
• The result
• 3 SnO22 (aq) 2 Bi(OH)3 (s) ?
• 3 SnO32(aq) 2
  Bi(s) 3 H2O(l)