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Department of Chemistry CHEM1010 General Chemistry *********************************************** Instructor: Dr. Hong Zhang Foster Hall, Room 221 Tel: 931-6325 Email: hzhang@tntech.edu.
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Department of Chemistry CHEM1010 General Chemistry *********************************************** Instructor: Dr. Hong Zhang Foster Hall, Room 221 Tel: 931-6325 Email: hzhang@tntech.edu
CHEM1010/General Chemistry_________________________________________Chapter 8. (L32)-Oxidation and Reduction • Today’s Outline ..Electrochemistry and electrochemical cells ..Cells and Batteries Dry cells Lead storage batteries Other batteries Fuel cells ..Corrosion Rusting of iron Protection of aluminum Silver tarnish
Chapter 8. (L32)-Oxidation and Reduction • Electrochemistry and electrochemical cells Oxidation and reduction reactions can produce electricity because of transfer or flow of electrons between reductants and oxidants Example: Zn + Cu2+ + SO42- = Zn2+ + Cu + SO42- oxidation half reaction: Zn = Zn2+ + 2e- Zn is oxidized reduction half reaction: Cu2+ + 2e- = Cu Cu is reduced
Chapter 8. (L32)-Oxidation and Reduction • Electrochemistry and electrochemical cells If the two half reactions are occurring in two separate compartments and connected with a wire between Zn and Cu, then electron will pass from Zn to Cu until Zn is exhausted. This is an electrochemical cell: {Cu2+ + 2e- = Cu}-{Zn = Zn2+ + 2e-} electrodes: anode (Zn): where oxidation occurs, e- leaves cathode (Cu): where reduction occurs, e- enters
Chapter 8. (L32)-Oxidation and Reduction • Electrochemistry and electrochemical cells Represent a redox reaction in half reactions: Mg + Cl2 = Mg2+ + 2Cl- Oxidation half reaction: Mg = Mg2+ + 2e- Reduction half reaction: Cl2 + 2e- = 2Cl- 2Al + 3Br2 = 2Al3+ + 6Br- Oxidation half reaction: 2Al = 2Al3+ + 6e- Reduction half reaction: Br2 + 6e- = 6Br-
Chapter 8. (L32)-Oxidation and Reduction • Electrochemistry and electrochemical cells Balance redox reactions from half reactions: Sn2+ = Sn4+ Bi3+ = Bi Oxidation half reaction: Sn2+ = Sn4+ + 2e- (1) Reduction half reaction: Bi3+ + 3e- = Bi (2) Electron balance: e- loss should match e- gain Tip: find the common number, which is 6, so 3×(1) + 2×(2) give rise to: 3Sn2+ = 3Sn4+ + 6e- 2Bi3+ + 6e- = 2Bi 3Sn2+ + 2Bi3+ = 3Sn4+ + 2Bi
Chapter 8. (L32)-Oxidation and Reduction • Cells and batteries Dry cells: Zinc-carbon cell Zn + 2MnO2(s) + H2O = Zn2+ + Mn2O3 + 2OH- Oxidation half reaction: Zn = Zn2+ + 2e- Reduction half reaction: 2MnO2(s) + H2O + 2e- = Mn2O3 + 2OH- Zn: anode C: cathode Alkaline cells: KOH with MnO2
Chapter 8. (L32)-Oxidation and Reduction • Lead storage batteries A lead storage batter is a series of 2V cells each with its own electrode setup, connected together in series with total 12V of output Oxidation half reaction (anode): Pb(s) + SO42- = PbSO4(s) + 2e- Reduction half reaction (cathode): PbO2(s) + 4H+ + SO42- + 2e- = PbSO4(s) + H2O Overall discharge reaction: Pb + PbO2 + 2H2SO4 = 2PbSO4 + 2H2O Recharging reaction: Opposite 2PbSO4 + 2H2O = Pb + PbO2 + 2H2SO4
Chapter 8. (L32)-Oxidation and Reduction • Other batteries ..Lithium batteries Li-SO2: submarines, rockets Li-I2: pacemakers Li-FeS2 (Energizer): cameras, radios, compact disc players ..Rechargeable Ni batteries Ni-Cad cell (Cd anode, NiO cathode) Ni-hydride cell (ZrNi2 anode-NiO cathode, LaNi5 anode-NiO cathode) ..Button cells Zn-HgO or Zn-Air
Chapter 8. (L32)-Oxidation and Reduction • Corrosion: Oxidation of metals $100 billion cost each year in USA Rusting of iron in humid or moist air, rust reaction: 2Fe + O2 + 2H2O = 2Fe(OH)2 Oxidation half: Fe = Fe2+ + 2e- Reduction half: O2 + 2H2O + 4e- = 4OH- 4Fe(OH)2 + O2 + 2H2O = 4Fe(OH)3 or Fe2O3•3H2O, red rust
Chapter 8. (L32)-Oxidation and Reduction • Protection of aluminum Why Al seems “inert”, no corrosion, although it is more reactive than iron? Al + O2 = Al2O3 a film of Al oxide forms as a coating, hard, tough, protecting Al metal But, some salt solutions can enhance oxidation of Al by interfering with the coating.
Chapter 8. (L32)-Oxidation and Reduction • Silver tarnish ..Tarnish on silver results from oxidation of silver surface by H2S to form black Ag2S Ag = Ag+ ..Chemical recovery of tarnished silverware 3Ag+ + Al = 3Ag + Al3+ Method: Al is placed in contact with Al foil and covered with a solution of baking soda (NaHCO3). Sliver is thus recovered by sacrificing Al.
Chapter 8. (L32)-Oxidation and Reduction Quiz Time Which are the right half reactions for the following reaction: 2H2 + O2 = 2H2O? (a) oxdiation half: O2 + 4H+ + 4e- = 2H2O reduction half: 2H2 = 4H+ + 4e-; (b) oxdiation half: H2 + 4e- = 4H+ reduction half: 2O2 = 4O2- + 4e-; (c) oxdiation half: 2H2 = 4H+ + 4e- reduction half: O2 + 4H+ + 4e- = 2H2O; (d) none no above is right.
Chapter 8. (L32)-Oxidation and Reduction Quiz Time Which is the right balanced reaction for the following redox half reactions: Zn = Zn2+ H2SO4 = H2 + SO42-? (a) Zn + H2O = Zn2+; (b) Zn + SO42- = ZnSO4; (c) Zn + H2SO4 = Zn2+ + H2 + SO42-; (d) H2SO4 = 2H+ + SO42-.
Chapter 8. (L32)-Oxidation and Reduction Quiz Time Which is the right balanced reaction for the following redox half reactions: Oxidation half reaction: Al = Al3+ Reduction half reaction: Br2 = Br- (a) 2Al3+ + 6Br- = 2Al + 3Br2; (b) Al + Br- = Al3+ + Br2; (c) Al + Br2 = Al3+ + Br-; (d) 2Al + 3Br2 = 2Al3+ + 6Br-.
Chapter 8. (L32)-Oxidation and Reduction Quiz Time In an electrochemical cell, the anode is where: (a) reduction occurs and e- enters; (b) neutralization occurs; (c) neither reduction nor oxidation occurs; (d) oxidation occurs and e- leaves.
Chapter 8. (L32)-Oxidation and Reduction Quiz Time In an electrochemical cell, the cathode is where: (a) reduction occurs and e- enters; (b) neutralization occurs; (c) neither reduction nor oxidation occurs; (d) oxidation occurs and e- leaves.
Chapter 8. (L32)-Oxidation and Reduction Quiz Time Which is the reaction occurring in a car battery? (a) 3Ag+ + Al = 3Ag + Al3+; (b) 2H2 + O2 = 2H2O; (c) Zn + H2SO4 = Zn2+ + H2 + SO42-; (d) Pb + PbO2 + 2H2SO4 = 2PbSO4 + 2H2O
Chapter 8. (L32)-Oxidation and Reduction Quiz Time Which is the the reaction(s) responsible for formation of iron rust? (a) 3Ag+ + Al = 3Ag + Al3+; (b) 2H2 + O2 = 2H2O; (c) Al + O2 = Al2O3; (d) 2Fe + O2 + 2H2O = 2Fe(OH)2 and 4Fe(OH)2 + O2 + 2H2O = 4Fe(OH)3
Chapter 8. (L32)-Oxidation and Reduction Quiz Time Which is the reaction responsible for protection of aluminum foil or pot? (a) 3Ag+ + Al = 3Ag + Al3+; (b) 2H2 + O2 = 2H2O; (c) Al + O2 = Al2O3; (d) 2Fe + O2 + 2H2O = 2Fe(OH)2 and 4Fe(OH)2 + O2 + 2H2O = 4Fe(OH)3
Chapter 8. (L32)-Oxidation and Reduction Quiz Time Which is the reaction responsible for removal of silver tarnish? (a) 3Ag+ + Al = 3Ag + Al3+; (b) 2H2 + O2 = 2H2O; (c) Al + O2 = Al2O3; (d) 2Fe + O2 + 2H2O = 2Fe(OH)2 and 4Fe(OH)2 + O2 + 2H2O = 4Fe(OH)3