OxidationReduction Reactions

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OxidationReduction Reactions

• Redox reactions are those involving the oxidation
  and reduction of species.
• Oxidation and reduction must occur together. They
  cannot exist alone.

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OxidationReduction Reactions

• Oxidation
• Is
• Loss (of electrons)
• Anode Oxidation
• Reducing Agent

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OxidationReduction Reactions

• Reduction
• Is
• Gain (of electrons)
• Cathode Reduction
• Oxidizing Agent

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OxidationReduction Reactions

• Assigning Oxidation Numbers All atoms have an
  oxidation number regardless of whether it
  carries an ionic charge.
• 1. An atom in its elemental state has an
  oxidation number of zero.
• 2. An atom in a monatomic ion has an oxidation
  number identical to its charge.

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• 3. An atom in a polyatomic ion or in a molecular
  compound usually has the same oxidation number it
  would have if it were a monatomic ion.
• A. Hydrogen can be either 1 or 1.
• B. Oxygen usually has an oxidation number of 2.
• In peroxides, oxygen is 1.
• C. Halogens usually have an oxidation number of
  1.
• When bonded to oxygen, chlorine, bromine, and
  iodine have positive oxidation numbers.

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OxidationReduction Reactions

• 4. The sum of the oxidation numbers must be zero
  for a neutral compound and must be equal to the
  net charge for a polyatomic ion.
• A. H2SO4 2(1) (?) 4(2) 0 net charge
• ? 0 2(1) 4(2) 6
• B. ClO4 (?) 4(2) 1 net charge
• ? 1 4(2) 7

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OxidationReduction Reactions

• 5. Whenever one atom loses electrons (is
  oxidized), another atom must gain those electrons
  (be reduced).
• A substance which loses electrons (oxidized) is
  called a reducing agent. Its oxidation number
  increases.
• A substance which gains electrons (reduced) is
  called the oxidizing agent. Its oxidation number
  decreases.

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OxidationReduction Reactions

• Assign oxidation numbers to each atom in the
  following substances
• A. CdS B. AlH3 C. Na2Cr2O7 D. SnCl4
• E. CrO3 F. VOCl3 G. V2O3 H. HNO3
• I. FeSO4 J. Fe2O3 K. H2PO4 L. MnO4
• M. Cr2O72

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OxidationReduction Reactions

• For each of the following, identify which species
  is the reducing agent and which is the oxidizing
  agent.
• Ca(s) 2 H(aq) ? Ca2(aq) H2(g)
• 2 Fe2(aq) Cl2(aq) ? 2 Fe3(aq) 2 Cl(aq)
• SnO2(s) 2 C(s) ? Sn(s) 2 CO(g)
• Sn2(aq) 2 Fe3(aq) ? Sn4(aq) 2 Fe2(aq)

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Activity Series of Elements
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Activity series looks at the relative reactivity
of a free metal with an aqueous cation.

• Fe(s) Cu2(aq) ? Fe2(aq) Cu(s)
• Zn(s) Cu2(aq) ? Zn2(aq) Cu(s)
• Cu(s) 2 Ag(aq) ? 2 Ag(s) Cu2(aq)
• Mg(s) 2 H(aq) ? Mg2(aq) H2(g)

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Given the following three reactions, determine
the activity series for Cu, Zn, Fe.

• Fe(s) Cu2(aq) ? Fe2(aq) Cu(s)
• Zn(s) Cu2(aq) ? Zn2(aq) Cu(s)
• Fe(s) Zn2(aq) ? NR

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Balancing Redox Reactions

• Half-Reaction Method Allows you to focus on the
  transfer of electrons. This is important when
  considering batteries and other aspects of
  electrochemistry.
• The key to this method is to realize that the
  overall reaction can be broken into two parts, or
  half-reactions.

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• Balancing for an acidic solution
• MnO4(aq) Br(aq) ? Mn2(aq) Br2(aq)
• The steps involved follow the same basic
  procedure as described for the oxidation-number
  method.
• 1. Determine oxidation and reduction
  half-reactions
• Oxidation half-reaction Br(aq) ? Br2(aq)
• Reduction half-reaction MnO4(aq) ? Mn2(aq)

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• 2. Balance for atoms other than H and O
• Oxidation 2 Br(aq) ? Br2(aq)
• Reduction MnO4(aq) ? Mn2(aq)
• 3. Balance for oxygen by adding H2O
• Oxidation 2 Br(aq) ? Br2(aq)
• Reduction MnO4(aq) ? Mn2(aq) 4 H2O(l)

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• 4. Balance for hydrogen by adding H
• Oxidation 2 Br(aq) ? Br2(aq)
• Reduction MnO4(aq) 8 H(aq) ? Mn2(aq)
  4H2O(l)
• 5. Balance for charge by adding electrons (e)
• Oxidation 2 Br(aq) ? Br2(aq) 2 e
• Reduction MnO4(aq) 8 H(aq) 5 e ?
  Mn2(aq) 4 H2O(l)

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• 6. Balance for numbers of electrons by
  multiplying
• Oxidation 52 Br(aq) ? Br2(aq) 2 e
• Reduction 2MnO4(aq) 8 H(aq) 5 e ?
  Mn2(aq) 4 H2O(l)
• 7. Combine and cancel to form one equation
• Oxidation 10 Br(aq) ? 5 Br2(aq) 10 e
• Reduction 2 MnO4(aq) 16 H(aq) 10 e ? 2
  Mn2(aq) 8 H2O(l)
• 2 MnO4(aq) 10 Br(aq) 16 H(aq) ? 2 Mn2(aq)
  5 Br2(aq) 8 H2O(l)

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• Balancing for Basic Solution
• MnO4(aq) SO32(aq) ? MnO42(aq) SO42(aq)
• 7. Balance for acidic solution.
• 2 MnO4(aq) SO32(aq) H2O(l) ? 2
  MnO42(aq) SO42(aq) 2 H(aq)

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• 8. Add OH to neutralize the H ions from the
  acidic balancing.
• 2 MnO4(aq) SO32(aq) H2O(l) 2 OH(aq) ?
• 2 MnO42(aq) SO42(aq) 2 H(aq) 2 OH(aq)
• 9. Reduce to simplest form.
• 2 MnO4(aq) SO32(aq) 2 OH(aq) ?
• 2 MnO42(aq) SO42(aq) H2O(l)

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• Balance the following for acidic and basic
  solution
• ClO(aq) Cr(OH)4(aq) ? CrO42(aq) Cl(aq)
• NO3(aq) Cu(s) ? NO(g) Cu2(aq)
• Fe(OH)2(s) O2(g) ? Fe(OH)3(s)
• MnO4(aq) IO3(aq) ? MnO2(s) IO4(aq)
• Cr2O72(aq) Fe2(aq) ? CrO42(aq) Cl(aq)

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