# 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

 Related Links: Chemistry Chemistry Quizzes AP Chemistry Notes Electrochemistry

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