Chemistry 12 Unit 5

I.Oxidation – Reduction Reactions: Oxidation:A substance losing electrons. Reduction:A substance gaining electrons. eg:Al (s) in CuCl 2(aq) Al (s) + CuCl 2(aq)  Cu (s) + AlCl 3(aq) (not balanced) Net Ionic Equation: Al (s) + Cu 2+ (aq)  Cu (s) + Al 3+ (aq) Reduction Half-Reaction: Cu 2+ (aq) + 2e -  Cu (s) Oxidation Half-Reaction: Al (s)  Al 3+ (aq) + 3e - The reaction shows that Cu 2+ wants electrons more than Al 3+

This type of reaction is called an “oxidation – reduction” reaction or a “redox” reaction. Aluminum metal is oxidized and copper (II) ions are reduced. The complete redox equation consists of some multiple of the two half-reactions: Cu 2+ (aq) + 2e -  Cu (s) Al (s)  Al 3+ (aq) + 3e - 3 ( ) 2 ( ) This shows the same number of electrons gained as lost: 2Al (s) + 3Cu 2+ (aq) + 6e -  2Al 3+ (aq) + 3Cu (s) + 6e - Balanced Redox Equation: 2Al (s) + 3Cu 2+ (aq)  2Al 3+ (aq) + 3Cu (s)

Always check that the total charge on each side of the equation is equal. 2Al (s) + 3Cu 2+ (aq)  2Al 3+ (aq) + 3Cu (s) 3(2+) = 2(3+) Terminology: In this example: Cu 2+ is the oxidizing agent. Al is the reducing agent. Cu 2+ is reduced to Cu. Al is oxidized to Al 3+.

For each of the following, which substance is oxidized and which substance is reduced? 1.Zn + CuSO 4  ZnSO 4 + Cu Redox Practice I: 2.CrCl 3 + Sn  Cr + SnCl 4 3.Fe + CuSO 4  FeSO 4 + Cu 4.MgCl 2 + Cr  Mg + CrCl 3 5.Al + Pb(NO 3 ) 2  Al(NO 3 ) 3 + Pb Substance Oxidized Substance Reduced

For each of the following, which substance is the oxidizing agent and which is the reducing agent? 1.CuSO 4 + Zn  ZnSO 4 + Cu Redox Practice II: 2.Sn + CrCl 3  Cr + SnCl 4 3.Fe + CuSO 4  FeSO 4 + Cu 4.Mg + CrCl 3  Cr + MgCl 2 5.Al(NO 3 ) 3 + Pb  Pb(NO 3 ) 2 + Al Oxidizing Agent Reducing Agent