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Chemistry 12 Unit 5. Chapter 21:. Electrochemistry. Unit V: Electrochemistry. 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).
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Chemistry 12 Unit 5 Chapter 21: Electrochemistry
Unit V: Electrochemistry Oxidation – Reduction Reactions: Oxidation: A substance losing electrons. Reduction: A substance gaining electrons. eg: Al(s) in CuCl2(aq) Al(s) + CuCl2(aq) Cu(s) + AlCl3(aq) (not balanced) Net Ionic Equation: Al(s) + Cu2+(aq) Cu(s) + Al3+(aq) Reduction Half-Reaction: Cu2+(aq) + 2e- Cu(s) • The reaction shows that Cu2+ wants electrons more than Al3+ Oxidation Half-Reaction: Al(s) Al3+(aq) + 3e-
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: 3 ( ) Cu2+(aq) + 2e- Cu(s) 2 ( ) Al(s) Al3+(aq) + 3e- • This shows the same number of electrons gained as lost: 2Al(s) + 3Cu2+(aq)+ 6e- 2Al3+(aq) + 3Cu(s)+ 6e- Balanced Redox Equation: 2Al(s) + 3Cu2+(aq) 2Al3+(aq) + 3Cu(s)
Always check that the total charge on each side of the equation is equal. 2Al(s) + 3Cu2+(aq) 2Al3+(aq) + 3Cu(s) 3(2+) = 2(3+) Terminology: In this example: Cu2+ is the oxidizing agent. Al is the reducing agent. Cu2+ is reduced to Cu. Al is oxidized to Al3+.
Redox Practice I: For each of the following, which substance is oxidized and which substance is reduced? Substance Oxidized Substance Reduced • Zn + CuSO4 ZnSO4 + Cu • CrCl3 + Sn Cr + SnCl4 • Fe + CuSO4 FeSO4 + Cu • MgCl2 + Cr Mg + CrCl3 • Al + Pb(NO3)2 Al(NO3)3 + Pb
Redox Practice II: For each of the following, which substance is the oxidizing agent and which is the reducing agent? Oxidizing Agent Reducing Agent • CuSO4 + Zn ZnSO4 + Cu • Sn + CrCl3 Cr + SnCl4 • Fe + CuSO4 FeSO4 + Cu • Mg + CrCl3 Cr + MgCl2 • Al(NO3)3 + Pb Pb(NO3)2+ Al