department of chemistry chem1010 general chemistry ***********************************************
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Department of Chemistry CHEM1010 General Chemistry *********************************************** Instructor: Dr. Hong Zhang Foster Hall, Room 221 Tel: 931-6325 Email: [email protected]. - PowerPoint PPT PresentationTRANSCRIPT
Department of Chemistry
CHEM1010 General Chemistry***********************************************
Instructor: Dr. Hong ZhangFoster Hall, Room 221
Tel: 931-6325
Email: [email protected]
CHEM1010/General Chemistry_________________________________________
Chapter 8. (L32)-Oxidation and Reduction
• Today’s Outline..Electrochemistry and electrochemical cells..Cells and Batteries
Dry cellsLead storage batteriesOther batteriesFuel cells
..CorrosionRusting of ironProtection of aluminumSilver tarnish
Chapter 8. (L32)-Oxidation and Reduction
• Electrochemistry and electrochemical cellsOxidation and reduction reactions can produce electricity because of transfer or flow of electrons between reductants and oxidants
Example:
Zn + Cu2+ + SO42- = Zn2+ + Cu + SO4
2- oxidation half reaction: Zn = Zn2+ + 2e-
Zn is oxidizedreduction 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- gainTip: find the common number, which is 6, so3×(1) + 2×(2) give rise to:
3Sn2+ = 3Sn4+ + 6e- 2Bi3+ + 6e- = 2Bi 3Sn2+ + 2Bi3+ = 3Sn4+ + 2Bi
Chapter 8. (L32)-Oxidation and Reduction
• Cells and batteriesDry 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 batteriesNi-Cad cell (Cd anode, NiO cathode)
Ni-hydride cell (ZrNi2 anode-NiO cathode, LaNi5 anode-NiO cathode)
..Button cellsZn-HgO or Zn-Air
Chapter 8. (L32)-Oxidation and Reduction
• Corrosion: Oxidation of metals
$100 billion cost each year in USARusting of ironin 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 metalBut, 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