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Reduction / Oxidation Reactions

Reduction / Oxidation Reactions. History. Oxidation reactions involved the union of oxygen with another substance Eg . 2Mg(s) + O 2 (g)  2MgO Reduction reactions involved the breaking down (reduction) of an ore to the pure element Eg . CuO (s) + H 2 (g)  Cu(s) + H 2 O(g).

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Reduction / Oxidation Reactions

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  1. Reduction / Oxidation Reactions

  2. History • Oxidation reactions involved the union of oxygen with another substance Eg. 2Mg(s) + O2(g)  2MgO • Reduction reactions involved the breaking down (reduction) of an ore to the pure element Eg. CuO(s) + H2(g)  Cu(s) + H2O(g)

  3. Oxidation Numbers • Oxidation numbers are hypothetical charges assigned to atoms to describe redox reactions in terms of electrons lost or gained

  4. Oxidation Number Rules • A pure element has an oxidation number of 0. - Na(s) = ox. Number is 0 - Br in Br2 = ox number is 0 - P in P4 = 0 Notation will be a superscript number. (Na0)

  5. Oxidation Number Rules (cont.) 2. The oxidation number of an element in a monatomic ion equals the charge of the ion. Al3+ - ox. Number is 3+ Se2- - ox. Number is 2- 3. The oxidation number of hydrogen in its compounds is 1+ (except in metal hydrides, where it is 1-) CH4 = H1+ H2S = H1+ CaH2 = H1-

  6. Oxidation Number Rules (cont.) 4. The oxidation number of oxygen in its compounds is 2- (except in peroxides where it is -1, and the compound OF2 where it is 2+) Li2O - O2- H2O2 - O1- (exception to the rule) 5. In covalent compounds that do not contain hydrogen or oxygen, the more electronegative element is assigned an oxidation number that equals the negative charge it normally has in its ionic compounds PCl3 - Cl-1

  7. Oxidation Number Rules (cont.) 6. The sum of all the oxidation numbers in a compound is 0. CF4 - F1- - C4+ 7. The sum of all the oxidation numbers in a polyatomic ion equals the charge of the ion. NO2- - O2- - N3+

  8. Oxidation Number Rules (cont.) Oxidation numbers can be used to determine whether electrons have been lost or gained in a chemical reaction.

  9. Oxidation • Chemists recognized that non-metals, for example chlorine, behave in a similar way to oxygen. H2(g) + Cl2 (g)  2HCl(g) Mg(s) + Cl2 (g)  MgCl2 (s) Consider: Mg + O2  MgO Mg + Cl2  MgCl2 Mg  Mg2+ +2e- Mg  Mg2+ + 2e-

  10. Oxidation (cont.) • An underlying pattern is evident. The element being oxidized is losing one or more electrons during the reactions. OXIDATION IS THE LOSS OF ELECTRONS The molecule responsible for the removal of the electrons is called the oxidizing agent.

  11. Reduction • Chemists found that free elements can be formed in other ways than were originally considered. 2Fe + 3CuSO4 Fe2 (SO4) 3 + 3Cu 2NaCl  2Na + Cl2 Consider: Cu2+ + 2e-  Cu Na+ + e-  Na

  12. Reduction (cont.) • In each case, the element being reduced is gaining one or more electrons. REDUCTION IS THE GAIN OF ELECTRONS A reducing agent donates electrons to the element being reduced.

  13. Mnemonic devices: • LEO the lion says GER • Loss of electrons is oxidation • Gain of electrons is reduction • OIL RIG • Oxidation is loss, reduction is gain

  14. Note: • Electrons lost = electrons gained • Oxidation and reduction are taking place simultaneously • The substance oxidized is the reducing agent (ie. It donates its electrons to the substance reduced) • The substance reduced is the oxidizing agent (ie. It steals the electrons from the substance oxidized) • Not all reactions are redox reactions – only ones involving atoms gaining or losing electrons.

  15. Examples... • Begin with assigning oxidation numbers. Then identify what is oxidized, reduced, oxidizing agent and reducing agent. 1. 2Sb- + 3Cl2 2SbCl3 2. FeCl2 + Na2S  FeS + 2NaCl 3. 4Fe + 3Se2  2Fe2Se3

  16. Balancing Redox Equations • Summary: We will separate the given equation into two half-reactions (one oxidation and one reduction), balance each for atoms and electrons, then recombine the half reactions. Example: Balance the following – I2 + HClO + H2O  HIO3 + HCl • Step 1: • Write the half-reactions, one oxidation and one reduction, then balance each for elements other than oxygen and hydrogen

  17. Step 2: • Add H2O to balance the oxygens and add H+ to balance the hydrogens • Step 3: • Multiply each equation by a number (to make the lowest common multiple) so the electrons lost equals the electrons gained. • Step 4: • The two half reactions are recombined. Like terms on the same side of the half reactions are added and like terms on opposite sides are cancelled.

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