Dry Cell Battery Anode (-) Zn ---> Zn e- Cathode (+) 2 NH e- ---> 2 NH 3 + H 2 Common dry cell Copyright © 1999 by Harcourt Brace & Company All rights reserved. Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida

2 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Nearly same reactions as in common dry cell, but under basic conditions. Alkaline Battery

3 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Anode: Zn is reducing agent under basic conditions Cathode: HgO + H 2 O + 2e- ---> Hg + 2 OH - Mercury Battery

4 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Lead Storage Battery Anode (-) E o = V Pb + HSO > PbSO 4 + H + + 2e- Cathode (+) E o = V PbO 2 + HSO H + + 2e- ---> PbSO H 2 O

5 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Ni-Cad Battery Anode (-) Cd + 2 OH - ---> Cd(OH) 2 + 2e- Cathode (+) NiO(OH) + H 2 O + e- ---> Ni(OH) 2 + OH -

Electrolysis of Aqueous NaOH Anode (+) E o = V 4 OH - ---> O 2 (g) + 2 H 2 O + 2e- Cathode (-) E o = V 4 H 2 O + 4e- ---> 2 H OH - E o for cell = V Electric Energy ----> Chemical Change Copyright © 1999 by Harcourt Brace & Company All rights reserved. Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida

7 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Electrolysis Electric Energy ---> Chemical Change Electrolysis of molten NaCl. Here a battery “pumps” electrons from Cl - to Na +.

8 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Electrolysis of Molten NaCl Anode (+) E o = V 2 Cl - ---> Cl 2 (g) + 2e- Cathode (-) E o = V Na + + e- ---> Na E o for cell = V External energy needed because E o is (-). Note that signs of electrodes are reversed from batteries.

9 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Electrolysis of Aqueous NaCl Anode (+) E o = V 2 Cl - ---> Cl 2 (g) + 2e- Cathode (-) E o = V 2 H 2 O + 2e- ---> H OH - E o for cell = V Note that H 2 O is more easily reduced than Na +.

10 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Electrolysis of Aqueous NaCl Cells like these are the source of NaOH and Cl 2. In x 10 9 lb Cl x 10 9 lb Cl x 10 9 lb NaOH 26.1 x 10 9 lb NaOH Also the source of NaOCl for use in bleach.

11 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Electrolysis of Aqueous CuCl 2 Anode (+) E o = V 2 Cl - ---> Cl 2 (g) + 2e- Cathode (-) E o = V Cu e- ---> Cu E o for cell = V Note that Cu is more easily reduced than either H 2 O or Na +.

12 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Producing Aluminum 2 Al 2 O C ---> 4 Al + 3 CO 2 Charles Hall ( ) developed electrolysis process. Founded Alcoa.

Quantitative Aspects of Electrochemistry Consider electrolysis of aqueous silver ion. Ag + (aq) + e- ---> Ag(s) 1 mol e----> 1 mol Ag If we could measure the moles of e-, we could know the quantity of Ag formed. But how to measure moles of e-? Copyright © 1999 by Harcourt Brace & Company All rights reserved. Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida

14 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Quantitative Aspects of Electrochemistry Consider electrolysis of aqueous silver ion. Ag + (aq) + e- ---> Ag(s) 1 mol e----> 1 mol Ag If we could measure the moles of e-, we could know the quantity of Ag formed. But how to measure moles of e-?

15 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Quantitative Aspects of Electrochemistry Consider electrolysis of aqueous silver ion. Ag + (aq) + e- ---> Ag(s) 1 mol e----> 1 mol Ag If we could measure the moles of e-, we could know the quantity of Ag formed. But how to measure moles of e-?

16 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved But how is charge related to moles of electrons? Charge on 1 mol of e- = (1.60 x C/e-)(6.02 x e-/mol) = 96,500 C/mol e- = 1 Faraday Quantitative Aspects of Electrochemistry

17 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Quantitative Aspects of Electrochemistry 1.50 amps flow thru a Ag + (aq) solution for 15.0 min. What mass of Ag metal is deposited? Solution (a)Calc. charge Coulombs = amps x time = (1.5 amps)(15.0 min)(60 s/min) = 1350 C

18 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Quantitative Aspects of Electrochemistry 1.50 amps flow thru a Ag + (aq) solution for 15.0 min. What mass of Ag metal is deposited? Solution (a)Charge = 1350 C (b)Calculate moles of e- used

19 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Quantitative Aspects of Electrochemistry 1.50 amps flow thru a Ag + (aq) solution for 15.0 min. What mass of Ag metal is deposited? Solution (a)Charge = 1350 C (b)Calculate moles of e- used

20 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Quantitative Aspects of Electrochemistry 1.50 amps flow thru a Ag + (aq) solution for 15.0 min. What mass of Ag metal is deposited? Solution (a)Charge = 1350 C (b)Calculate moles of e- used (c)Calc. quantity of Ag

21 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Quantitative Aspects of Electrochemistry 1.50 amps flow thru a Ag + (aq) solution for 15.0 min. What mass of Ag metal is deposited? Solution (a)Charge = 1350 C (b)Calculate moles of e- used (c)Calc. quantity of Ag

22 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Quantitative Aspects of Electrochemistry The anode reaction in a lead storage battery is Pb(s) + HSO 4 - (aq) ---> PbSO 4 (s) + H + (aq) + 2e- If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last? Solution a)454 g Pb = 2.19 mol Pb b)Calculate moles of e-

23 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Quantitative Aspects of Electrochemistry The anode reaction in a lead storage battery is Pb(s) + HSO 4 - (aq) ---> PbSO 4 (s) + H + (aq) + 2e- If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last? Solution a)454 g Pb = 2.19 mol Pb b)Calculate moles of e-

24 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Quantitative Aspects of Electrochemistry The anode reaction in a lead storage battery is Pb(s) + HSO 4 - (aq) ---> PbSO 4 (s) + H + (aq) + 2e- If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last? Solution a)454 g Pb = 2.19 mol Pb b)Calculate moles of e- c)Calculate charge

25 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Quantitative Aspects of Electrochemistry The anode reaction in a lead storage battery is Pb(s) + HSO 4 - (aq) ---> PbSO 4 (s) + H + (aq) + 2e- If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last? Solution a)454 g Pb = 2.19 mol Pb b)Calculate moles of e- c)Calculate charge 4.38 mol e- 96,500 C/mol e- = 423,000 C 4.38 mol e- 96,500 C/mol e- = 423,000 C

26 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Quantitative Aspects of Electrochemistry The anode reaction in a lead storage battery is Pb(s) + HSO 4 - (aq) ---> PbSO 4 (s) + H + (aq) + 2e- If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last? Solution a)454 g Pb = 2.19 mol Pb b)Mol of e- = 4.38 mol c)Charge = 423,000 C

27 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Quantitative Aspects of Electrochemistry The anode reaction in a lead storage battery is Pb(s) + HSO 4 - (aq) ---> PbSO 4 (s) + H + (aq) + 2e- If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last? Solution a)454 g Pb = 2.19 mol Pb b)Mol of e- = 4.38 mol c)Charge = 423,000 C d)Calculate time

28 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Quantitative Aspects of Electrochemistry The anode reaction in a lead storage battery is Pb(s) + HSO 4 - (aq) ---> PbSO 4 (s) + H + (aq) + 2e- If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last? Solution a)454 g Pb = 2.19 mol Pb b)Mol of e- = 4.38 mol c)Charge = 423,000 C d)Calculate time

29 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Quantitative Aspects of Electrochemistry The anode reaction in a lead storage battery is Pb(s) + HSO 4 - (aq) ---> PbSO 4 (s) + H + (aq) + 2e- If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last? Solution a)454 g Pb = 2.19 mol Pb b)Mol of e- = 4.38 mol c)Charge = 423,000 C d)Calculate time About 78 hours