Chapter 18, lecture 3

Chapter 18 lecture - MOLTEN sodium chloride plus the large amount of electricity needed ALUMINUM CHAPTER 18 LECTURE 3 Page 3 Example 18.13 Use of

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CHAPTER 18 LECTURE 3 Page 1 LEAD-ACID STORAGE BATTERY CATHODE: PbO 2 + 3 H + + HSO 4 - + 2 e - PbSO 4(s) + 2 H 2 O ANODE: Pb (s) + HSO 4(s) - PbSO 4(s) + H + + 2 e - CELL REACTION: Pb (s) + PbO 2(s) + 2 H + + 2 HSO 4 - 2 PbSO 4(s) + 2 H 2 O cell = (cathode) - Eº (anode) = 1.690 V - (-0.356 V) = 2.046 V Six cells connected in series produces a total of approximately 12 volts. Recharging Reaction: 2 PbSO 4(s) + 2 H 2 O Pb (s) + PbO 2 + 2 H + + 2 HSO 4 - ALKALINE BATTERIES (1.5 V) CATHODE: 2 MnO 2(s) + 2 H 2 O + 2 e - 2 MnO(OH) (s) + 2 OH - ANODE: Zn (s) + 2 OH - Zn(OH) 2(s) + 2 e -
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CHAPTER 18 LECTURE 3 Page 2 ELECTROLYTIC CELLS We have been studying electrochemical cells, cells that produce a flow of electrons from a spontaneous chemical process. The opposite process, use of electrical energy to force a non-spontaneous process to take place, is an ELECTROLYTIC CELL. An example: Oxidation: 2 Cl - Cl 2 + 2 e - Reduction: 2 Na + + 2 e - 2 Na Overall: 2 Cl - + 2 Na + 2 Na (l) + 2 Cl 2 Since this is a non-spontaneous process, electrical current must be supplied in order to make this a feasible reaction. One of the major problems with this process is the requirement that the reaction be carried out in
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Unformatted text preview: MOLTEN sodium chloride plus the large amount of electricity needed. ALUMINUM CHAPTER 18 LECTURE 3 Page 3 Example 18.13 Use of electrolysis to measure the amount of gold in a sample. Sample Prepn Acid Sample Au +3 The Au +3 is reduced by an electrical current of 1.50 amps for 1.00 hour.: Au +3 + 3 e-Au How much Au will be deposited over this time period ? 1. am’t of charge = 1.50 C/sec x 3600 C/hr = 5.40 x 10 3 C 2. Convert charge to moles of e-Moles of e-= 5.40 x 10 3 C x 1 mol e-/ 96,500 C = 0.0560 moles e-3. Convert moles e-to moles Au Moles of Au = 0.0560 moles e-x 1 mole Au / 3 mole e-= 0.0187 mole Au 4. Convert moles Au to g Au = 0.0187 Mole Au x 197.0 g Au / mole Au = 3.68 g Au CHAPTER 18 LECTURE 3 Page 4 CHAPTER 18 LECTURE 3 Page 5 CHAPTER 18 LECTURE 3 Page 6 CHAPTER 18 LECTURE 3 Page 7 CHAPTER 18 LECTURE 3 Page 8...
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This note was uploaded on 04/07/2008 for the course CHEM 1124 taught by Professor Mohan during the Fall '07 term at Seton Hall.

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Chapter 18 lecture - MOLTEN sodium chloride plus the large amount of electricity needed ALUMINUM CHAPTER 18 LECTURE 3 Page 3 Example 18.13 Use of

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