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Oxidation/Reduction Reactions

Oxidation/Reduction Reactions. REDOX REACTONS!. All chemical reactions fall into two categories those that are redox and those that are not redox ! Redox reactions involve the transfer of electrons. What is oxidation and reduction?. Oxidation

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Oxidation/Reduction Reactions

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  1. Oxidation/Reduction Reactions • REDOX REACTONS! • All chemical reactions fall into two categories those that are redox and those that are not redox! • Redox reactions involve the transfer of electrons.

  2. What is oxidation and reduction? • Oxidation • Early chemists thought of oxidation as the combining of an element with oxygen, as in combustion. • Now we refer to oxidation as the loss of electrons. • Reduction • Early chemists thought of reduction as the loss of oxygen from a compound. • Now we refer to reduction as the gain of electrons.

  3. “LEO the lion says GER” • Oxidation • 4Fe(s) + 3O2(g)  2Fe2O3(s) - combustion or rusting • Fe  Fe3+ + 3e- - loss of electrons • Reduction • 2Fe2O3(s) + 3C(s)  4Fe(s) + 3CO2(g) -loss of oxygen • Fe3+ + 3e- Fe -gain of electrons

  4. Oxidation and Reduction must occur simultaneously: • Example: 2AgNO3(aq) + Cu(s)  2Ag(s) + Cu(NO3)2(aq) • Net Ionic: 2Ag+ + Cu  2Ag + Cu2+ • Nitrates ions are spectator ions! • Half Reactions: • 2Ag+ + 2e- 2Ag Reduction • Cu  Cu2+ + 2e- Oxidation • These reactions occur simultaneously and the electrons lost must equal the electrons gained.

  5. Reducing – Oxidizing Agents • Reducing Agent • The substance that loses electrons. • Cu is oxidized in the previous reaction but it is the REDUCING AGENT. • Oxidizing Agent • The substance that gains electrons. • Ag+ is reduced in the previous reaction but it is the OXIDIZING AGENT.

  6. Oxidation Numbers: • Elements always have an oxidation number of zero. • K -K is 0 • N2 -N is 0 • Oxygen is always -2 unless it is in a peroxide and then it is -1. • HNO3 - O is -2 • H2O2 - O is -1 • The charge on a monatomic ion is the oxidation number. • Fe3+ is +3 • Br1- is -1 • Hydrogen is always +1 unless it is in a metal hydride and then it is -1. • HNO3 - H is +1 • CaH2 - H is -1

  7. Determining Oxidation Numbers • K2CrO4 • O is -2 • K is +1 • Cr must be +6 • Cr2O3 • O is -2 • Cr must be +3 • Note that elements can have more than one oxidation number. • The sum of oxidation numbers in a neutral compound must be zero. • The sum of oxidation numbers in a polyatomic ions must equal the charge on the ion. • NO31- • O is -2 • N is +5

  8. Balancing Redox Reactions: • Example: • Sn + Ag+ Sn2+ + Ag • 1. Assign oxidation numbers. • 0 +1 +2 0 • Sn + Ag+ Sn2+ + Ag

  9. 2. Oxidation occurs when the oxidation number increases. Reduction occurs when the oxidation number decreases. Write the two half reactions. • Oxidation: Sn Sn2+ • Reduction: Ag+ Ag • 3. Use electrons to balance the charges in the half reactions. In oxidation the electrons appear on the right. In reduction the electrons appear on the left.

  10. Oxidation: SnSn2+ + 2e- • Reduction: Ag+ + 1e-  Ag • 4. If the number of electrons transferred is not equal multiple by a whole number so that the number of electrons lost equals the number gained. • Oxidation: Sn Sn2+ + 2e- • Reduction: (Ag+ + 1e- Ag) x2

  11. 5. Add the half reactions: • Oxidation: SnSn2+ + 2e- • Reduction: 2Ag+ + 2e-  2Ag • Net Balanced Redox Reaction: • Sn + 2 Ag+ Sn2+ + 2Ag

  12. Redox reactions are usually too complex to use trial and error method. • Balance the following example that is in an acidic solution (assume the presence of H2O and H+): • HNO3 + Fe2+ Fe3+ + NO2 • 1. Assign oxidation numbers. • +1+5-2 +2 +3 +4 -2 • HNO3 + Fe2+ Fe3+ + NO2

  13. 2. Write the half reactions: • Oxidation: Fe 2+ Fe3+ • Reduction: HNO3 NO2 • 3. Balance the half reactions. Use water to balance the oxygen (a), then hydrogen ions to balance the hydrogen(b), then electrons to balance the charges(c). • a) • Oxidation: Fe 2+ Fe3+ • Reduction: HNO3NO2 + H2O

  14. b) • Oxidation: Fe 2+ Fe3+ • Reduction: HNO3 + H+ NO2 + H2O • c) • Oxidation: Fe 2+ Fe3+ + 1e- • Reduction: HNO3 + H++ 1e- NO2 + H2O

  15. 4. Multiply the oxidation and reduction equations by whole numbers so that the number of electrons transferred is equal. • Oxidation: Fe 2+ Fe3+ + 1e- • Reduction: HNO3 + H+ + 1e- NO2 + H2O • 5. Add the reactions so that electrons cancel. • Net: Fe 2++ HNO3 + H+  Fe3+ + NO2 + H2O • 6. Check the equation is balance by charge and atom.

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