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Enduring Understanding 3.B.3: Redox Reactions

  • In oxidation-reduction (redox) reactions, there is a net transfer of electrons from one reactant to another.
  • The species that loses electrons is oxidized, and the species that gains electrons is reduced.
  • Since redox reactions always involve a species gaining or losing electrons, they can be written as half-reactions with the electrons explicitly written, as in these half-reactions for the oxidation of copper metal, and the reduction of silver (I) ions:
    • Cu(s) → Cu2+(aq) + 2e-
    Ag+(aq) + e- → Ag(s)
  • Multiplying the bottom reaction by 2, and adding the reactions, gives a balanced redox reaction:
    • Cu(s) + 2 Ag+(aq) → Cu2+(aq) + 2 Ag(s)

  • The oxidation number is the effective charge on an atom in a compound. When an atom is oxidized its oxidation number increases, and when it is reduced its oxidation number decreases. It is calculated from a set of rules, e.g. H is usually +1, O is usually -2, alkali metals are +1, etc... The sum of all the oxidation numbers of atoms in a species must equal the overall charge of the species.

  • Example. In which compound, MnO2 or KMnO4, does the manganese atom have the higher oxidation number?
  • MnO2 is uncharged overall. Each O has an oxidation number of -2, for a total of -4. The charge on the Mn must therefore be +4.
  • In KMnO4, the K has a charge of +1. Each O is -2, and there are 4, so charge from all the O's is -8. The charge of Mn must be -(-8+1) or +7.
  • KMnO4 has the higher oxidation number, of +7.
  • Redox reactions can be used in titrations, often in the determination of the concentration of metals and other species.
  • Example: Given the redox reaction 5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O, what was the original concentration of Fe2+ if 40 mL of the iron solution was titrated to endpoint with 20.0 mL of 0.20M KMnO4 solution?
  • Amount of permanganate (MnO4-) added = 0.020L x 0.20M = 0.0040 mol
  • 5 Fe2+ for every MnO4- added, so amount of Fe2+ = 0.0040 x 5 = 0.020 mol
  • 0.020 mol / 0.040 L = 0.50 mol/L

  • Redox equations can be balanced by a sequence of balancing oxygens, hydrogens, and electrons. The procedure can be illustrated by the following example:
  • Balance the following redox equation: Cr2O72- + HNO2 → Cr3+ + NO3-
  • First, consider the reduction half-reaction, balanced for chomium:
    • Cr2O72- → 2 Cr3+
  • Balance it for oxygens:
    • Cr272- → 2 Cr3+ + 7 H2O
  • Now balance for hydrogens:
    • 14H+ + Cr2O72- → 2 Cr3+ + 7 H2O
  • Add electrons to balance charges on both sides:
    • 6 e- + 14H+ + Cr2O72- → 2 Cr3+ + 7 H2O
  • Now consider the oxidation half reaction:
    • HNO2 → NO3-
  • Balance it for oxygens, then for hydrogens:
    • HNO2 + H2O → NO3- + 3H+
  • Add electrons to balance charges:
    • HNO2 + H2O → NO3- + 3H+ + 2e-
  • Reduction half-reaction has 6 e- on left, oxidation has 2 e- on right. Multiply oxidation by 3:
    • 6 e- + 14H+ + Cr2O72- → 2 Cr3+ + 7 H2O
    3 HNO2 + 3 H2O → 3 NO3- + 9H+ + 6e-
  • And add, and cancel electrons:
    • 6e- + 14H+ + Cr2O72- + 3 HNO2 + 3 H2O - → 2 Cr3+ + 3 NO3- + 7 H2O + 9 H++ 6e-
  • The balanced reaction is
    • Cr2O72- + 3 HNO2 + 5 H+ → 2 Cr3+ + 3 NO3- + 4 H2O



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