Presentation is loading. Please wait.

Presentation is loading. Please wait.

Dry Cell Battery Anode (-) Zn ---> Zn 2+ + 2e- Cathode (+) 2 NH 4 + + 2e- ---> 2 NH 3 + H 2 Common dry cell Copyright © 1999 by Harcourt Brace & Company.

Published byGilbert Flynn Modified over 9 years ago

Similar presentations

Presentation on theme: "Dry Cell Battery Anode (-) Zn ---> Zn 2+ + 2e- Cathode (+) 2 NH 4 + + 2e- ---> 2 NH 3 + H 2 Common dry cell Copyright © 1999 by Harcourt Brace & Company."— Presentation transcript:

1 Dry Cell Battery Anode (-) Zn ---> Zn 2+ + 2e- Cathode (+) 2 NH 4 + + 2e- ---> 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 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 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 4 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Lead Storage Battery Anode (-) E o = +0.36 V Pb + HSO 4 - ---> PbSO 4 + H + + 2e- Cathode (+) E o = +1.68 V PbO 2 + HSO 4 - + 3 H + + 2e- ---> PbSO 4 + 2 H 2 O

5 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 -

6 Electrolysis of Aqueous NaOH Anode (+) E o = -0.40 V 4 OH - ---> O 2 (g) + 2 H 2 O + 2e- Cathode (-) E o = -0.83 V 4 H 2 O + 4e- ---> 2 H 2 + 4 OH - E o for cell = -1.23 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 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 8 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Electrolysis of Molten NaCl Anode (+) E o = -1.36 V 2 Cl - ---> Cl 2 (g) + 2e- Cathode (-) E o = -2.71 V Na + + e- ---> Na E o for cell = -4.07 V External energy needed because E o is (-). Note that signs of electrodes are reversed from batteries.

9 9 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Electrolysis of Aqueous NaCl Anode (+) E o = -1.36 V 2 Cl - ---> Cl 2 (g) + 2e- Cathode (-) E o = -0.83 V 2 H 2 O + 2e- ---> H 2 + 2 OH - E o for cell = -2.19 V Note that H 2 O is more easily reduced than Na +.

10 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 1995 25.1 x 10 9 lb Cl 2 25.1 x 10 9 lb Cl 2 26.1 x 10 9 lb NaOH 26.1 x 10 9 lb NaOH Also the source of NaOCl for use in bleach.

11 11 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Electrolysis of Aqueous CuCl 2 Anode (+) E o = -1.36 V 2 Cl - ---> Cl 2 (g) + 2e- Cathode (-) E o = +0.34 V Cu 2+ + 2e- ---> Cu E o for cell = -1.02 V Note that Cu is more easily reduced than either H 2 O or Na +.

12 12 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Producing Aluminum 2 Al 2 O 3 + 3 C ---> 4 Al + 3 CO 2 Charles Hall (1863-1914) developed electrolysis process. Founded Alcoa.

13 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 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 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 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 10 -19 C/e-)(6.02 x 10 23 e-/mol) = 96,500 C/mol e- = 1 Faraday Quantitative Aspects of Electrochemistry

17 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 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 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 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 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 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 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 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 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 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 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 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 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

Download ppt "Dry Cell Battery Anode (-) Zn ---> Zn 2+ + 2e- Cathode (+) 2 NH 4 + + 2e- ---> 2 NH 3 + H 2 Common dry cell Copyright © 1999 by Harcourt Brace & Company."

Similar presentations

1 © 2006 Brooks/Cole - Thomson Balancing Equations for Redox Reactions Some redox reactions have equations that must be balanced by special techniques.

1 © 2006 Brooks/Cole - Thomson Balancing Equations for Redox Reactions Some redox reactions have equations that must be balanced by special techniques.
1 Electrochemistry Chapter 18, 19-20. 2 Electrochemical processes are oxidation-reduction reactions in which: the energy released by a spontaneous reaction.

1 Electrochemistry Chapter 18, Electrochemical processes are oxidation-reduction reactions in which: the energy released by a spontaneous reaction.
Jeffrey Mack California State University, Sacramento Chapter 20 Principles of Chemical Reactivity: Electron Transfer Reactions.

Jeffrey Mack California State University, Sacramento Chapter 20 Principles of Chemical Reactivity: Electron Transfer Reactions.
Electrolysis & Understanding Electrolytic Cells : When a non-spontaneous redox reaction is made to occur by putting electrical energy into the system.

Electrolysis & Understanding Electrolytic Cells : When a non-spontaneous redox reaction is made to occur by putting electrical energy into the system.
Chemistry 1011 Slot 51 Chemistry 1011 TOPIC Electrochemistry TEXT REFERENCE Masterton and Hurley Chapter 18.

Chemistry 1011 Slot 51 Chemistry 1011 TOPIC Electrochemistry TEXT REFERENCE Masterton and Hurley Chapter 18.
Chapter 21: Electrochemistry

Chapter 21: Electrochemistry
Electrochemistry Batteries. Batteries Lead-Acid Battery A 12 V car battery consists of 6 cathode/anode pairs each producing 2 V. Cathode: PbO 2 on a metal.

Electrochemistry Batteries. Batteries Lead-Acid Battery A 12 V car battery consists of 6 cathode/anode pairs each producing 2 V. Cathode: PbO 2 on a metal.
Chapter 20 Electrochemistry.

Chapter 20 Electrochemistry.
20-5 Batteries: Producing Electricity Through Chemical Reactions

20-5 Batteries: Producing Electricity Through Chemical Reactions
Lecture 284/3/06 Seminar today. Secondary Batteries (rechargeable) Lead Acid battery E° = 2.04 V Anode:Pb(s) + HSO 4 - (aq)  PbSO 4 (s) + H + + 2e -

Lecture 284/3/06 Seminar today. Secondary Batteries (rechargeable) Lead Acid battery E° = 2.04 V Anode:Pb(s) + HSO 4 - (aq)  PbSO 4 (s) + H + + 2e -
Lecture 263/30/07. E° F 2 (g) + 2e - ↔ 2F - +2.87 Ag + + e - ↔ Ag (s)+0.80 Cu 2+ + 2e - ↔ Cu (s)+0.34 Zn 2+ + 2e - ↔ Zn (s)-0.76 Quiz 1. Consider these.

Lecture 263/30/07. E° F 2 (g) + 2e - ↔ 2F Ag + + e - ↔ Ag (s)+0.80 Cu e - ↔ Cu (s)+0.34 Zn e - ↔ Zn (s)-0.76 Quiz 1. Consider these.
Lecture 294/13/05. Counting electrons 96,500 C/mol e-

Lecture 294/13/05. Counting electrons 96,500 C/mol e-
Lecture 25 3/28/07.

Lecture /28/07.
Copyright©2000 by Houghton Mifflin Company. All rights reserved. 1 Electrochemistry The study of the interchange of chemical and electrical energy.

Copyright©2000 by Houghton Mifflin Company. All rights reserved. 1 Electrochemistry The study of the interchange of chemical and electrical energy.
Representing electrochemical cells The electrochemical cell established by the following half cells: Zn(s) --> Zn 2+ (aq) + 2 e - Cu 2+ (aq) + 2 e - -->

Representing electrochemical cells The electrochemical cell established by the following half cells: Zn(s) --> Zn 2+ (aq) + 2 e - Cu 2+ (aq) + 2 e - -->
Predicting Spontaneous Reactions

Predicting Spontaneous Reactions
CHEM 160 General Chemistry II Lecture Presentation Electrochemistry December 1, 2004 Chapter 20.

CHEM 160 General Chemistry II Lecture Presentation Electrochemistry December 1, 2004 Chapter 20.
ELECTROCHEMISTRY REDOX REVISITED! 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)1.

ELECTROCHEMISTRY REDOX REVISITED! 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)1.
Section 18.1 Electron Transfer Reactions 1.To learn about metal-nonmetal oxidation–reduction reactions 2.To learn to assign oxidation states Objectives.

Section 18.1 Electron Transfer Reactions 1.To learn about metal-nonmetal oxidation–reduction reactions 2.To learn to assign oxidation states Objectives.
ELECTROCHEMISTRY CHARGE (Q) – A property of matter which causes it to experience the electromagnetic force COULOMB (C) – The quantity of charge equal to.

ELECTROCHEMISTRY CHARGE (Q) – A property of matter which causes it to experience the electromagnetic force COULOMB (C) – The quantity of charge equal to.

Similar presentations

Ads by Google