, Solution manual for The Science and Engineering
of Materials, Enhanced Edition, 7th Edition Donald
R. Askeland
Notes
1- The file is chapter after chapter.
2- We have shown you few pages sample.
3- The file contains all Appendix and Excel
sheet if it exists.
4- We have all what you need, we make
update at every time. There are many new
editions waiting you.
5- If you think you purchased the wrong file
You can contact us at every time, we can
replace it with true one.
Our email:
, A SOLUTION AND ANSWER GUIDE TO ACCOMPANY
THE SCIENCE AND
ENGINEERING OF MATERIALS,
ENHANCED SEVENTH EDITION
DONALD R. ASKELAND AND WENDELIN J. WRIGHT
,© 2022, 2016, 2011 Cengage Learning® ISBN: 978-0-357-44789-5
Cengage
ALL RIGHTS RESERVED. No part of this work covered by the 200 Pier 4 Boulevard
copyright herein may be reproduced or distributed in any form or Boston, MA 02210
by any means, except as permitted by U.S. copyright law, without USA
the prior written permission of the copyright owner, except as may
be permitted by the license terms below. Cengage Learning is a leading provider of customized
learning solutions with employees residing in nearly 40
different countries and sales in more than 125 countries
For product information and technology assistance, contact us at
around the world. Find your local representative at
Cengage Learning Academic Resource Center, www.cengage.com.
1-800-423-0563.
For permission to use material from this text or product, submit
all requests online at www.cengage.com/permissions.
Further permissions questions can be emailed to
.
NOTE: UNDER NO CIRCUMSTANCES MAY THIS MATERIAL OR ANY PORTION THEREOF BE SOLD, LICENSED, AUCTIONED,
OR OTHERWISE REDISTRIBUTED EXCEPT AS MAY BE PERMITTED BY THE LICENSE TERMS HEREIN.
READ IMPORTANT LICENSE INFORMATION
Dear Professor or Other Supplement Recipient: unauthorized use, reproduction, or distribution. Your use of the
Supplement indicates your acceptance of the conditions set forth in
Cengage Learning has provided you with this product (the this Agreement. If you do not accept these conditions, you must
“Supplement”) for your review and, to the extent that you adopt return the Supplement unused within 30 days of receipt.
the associated textbook for use in connection with your course
(the “Course”), you and your students who purchase the All rights (including without limitation, copyrights, patents, and trade
textbook may use the Supplement as described below. secrets) in the Supplement are and will remain the sole and
Cengage Learning has established these use limitations in exclusive property of Cengage Learning and/or its licensors. The
response to concerns raised by authors, professors, and other Supplement is furnished by Cengage Learning on an “as is” basis
users regarding the pedagogical problems stemming from without any warranties, express or implied. This Agreement will be
unlimited distribution of Supplements. governed by and construed pursuant to the laws of the State of
New York, without regard to such State’s conflict of law rules.
Cengage Learning hereby grants you a nontransferable license
to use the Supplement in connection with the Course, subject to Thank you for your assistance in helping to safeguard the integrity
the following conditions. The Supplement is for your personal, of the content contained in this Supplement. We trust you find the
noncommercial use only and may not be reproduced, posted Supplement a useful teaching tool.
electronically or distributed, except that portions of the
Supplement may be provided to your students IN PRINT FORM
ONLY in connection with your instruction of the Course, so long
as such students are advised that they may not copy or
distribute any portion of the Supplement to any third party. Test
banks and other testing materials may be made available in the
classroom and collected at the end of each class session, or
posted electronically as described herein. Any material posted
electronically must be through a password-protected site, with
all copy and download functionality disabled, and accessible
solely by your students who have purchased the associated
textbook for the Course. You may not sell, license, auction, or
otherwise redistribute the Supplement in any form. We ask that
you take reasonable steps to protect the Supplement from
, SOLUTION AND ANSWER GUIDE
TO ACCOMPANY
The Science and
Engineering of Materials
Enhanced Seventh Edition
DONALD R. ASKELAND
University of Missouri—Rolla, Emeritus
WENDELIN J. WRIGHT
Bucknell University
, Contents
Chapter 1 .......................................................................................................................................... 1
Chapter 2 ........................................................................................................................................13
Chapter 3 ........................................................................................................................................29
Chapter 4 ........................................................................................................................................85
Chapter 5 ......................................................................................................................................109
Chapter 6 ......................................................................................................................................139
Chapter 7 ......................................................................................................................................181
Chapter 8 ......................................................................................................................................207
Chapter 9 ......................................................................................................................................243
Chapter 10 ....................................................................................................................................279
Chapter 11 ....................................................................................................................................309
Chapter 12 ....................................................................................................................................333
Chapter 13 ....................................................................................................................................371
Chapter 14 ....................................................................................................................................397
Chapter 15 ....................................................................................................................................411
Chapter 16 ....................................................................................................................................427
Chapter 17 ....................................................................................................................................443
Chapter 18 ....................................................................................................................................473
Chapter 19 ....................................................................................................................................483
Chapter 20 ....................................................................................................................................509
Chapter 21 ....................................................................................................................................529
Chapter 22 ....................................................................................................................................549
Chapter 23 ....................................................................................................................................563
, Chapter 23: Corrosion and Wear
23-1 Explain how it is possible to form a simple electrochemical cell using a lemon, a penny,
and a galvanized steel nail.
Solution: The penny serves as the cathode (copper is the cathode). The steel nail with a zinc
coating serves as the anode (zinc is the anode), and the lemon serves as the
electrolyte. A voltmeter or wire connecting the penny and nail completes the circuit.
23-2 Silver can be polished by placing it in an aluminum pan containing a solution of baking
soda, salt, and water. Speculate as to how the polishing process works.
Solution:
The black tarnish on silver is silver sulfide. An electrochemical reaction occurs in which
the silver sulfide is reduced to silver plus aluminum sulfide according to
3 Ag2S + 2 Al → 6 Ag + Al2S3
The baking soda functions as an electrolyte. The silver is left behind on the part to be
cleaned.
23-3 A gray cast iron pipe is used in the natural gas distribution system for a city. The pipe
fails and leaks, even though no corrosion noticeable to the naked eye has occurred.
Offer an explanation for why the pipe failed.
Solution: Because no corrosion is noticeable, the corrosion byproduct apparently is still in
place on the pipe, hiding the corroded area. The circumstances suggest graphitic
corrosion, an example of a selective chemical attack. The graphite flakes in the gray
iron are not attacked by the corrosive soil, while the iron matrix is removed or
converted to an iron oxide or hydroxide trapped between the graphite flakes.
Although the pipe appears to be sound, the attacked area is weak, porous, and
spongy. The natural gas can leak through the area of graphitic corrosion and
accumulate, eventually leading to an explosion.
23-4 A brass plumbing fitting produced from a Cu-30% Zn alloy operates in the hot water
system of a large office building. After some period of use, cracking and leaking occur.
On visual examination, no metal appears to have been corroded. Offer an explanation
for why the fitting failed.
Solution: The high zinc brasses are susceptible to dezincification, particularly when the
temperature is increased, as in the hot water supply of the building. One of the
characteristics of dezincification is that copper is redeposited in the regions that are
attacked, obscuring the damage. The redeposited copper is spongy, brittle, and
weak, permitting the fitting to fail and leak. Therefore, dezincification appears to be
a logical explanation.
563
© 2022 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
,23-5 Plot the electronegativity of the elements listed in Table 23–1 as a function of electrode
potential. (Refer to Chapter 2 for electronegativity values.) What is the trend that you
observe in this data? Using an example, explain how the electronic structure of a metal
is consistent with its anodic or cathodic tendency.
Solution:
In general, as electronegativity increases, the tendency for the elements to corrode
decreases. Anodic elements are more electropositive (they have a low
electronegativity), meaning that they want to give up their electrons. Cathodic
elements are electronegative, i.e., they want to gain electrons. For example,
consider lithium. Li has the electronic structure [Li] = [He] 2s1. Lithium wants to give
up the 2s1 electron to achieve a stable configuration and has anodic tendencies.
23-6 Write the anodic or cathodic reaction (as appropriate) for aluminum, germanium and
chromium.
Solution:
𝐴𝐴𝐴𝐴 → 𝐴𝐴𝐴𝐴 3+ + 3𝑒𝑒 −
𝐺𝐺𝐺𝐺 → 𝐺𝐺𝐺𝐺 4+ + 4𝑒𝑒 −
𝐶𝐶𝐶𝐶 → 𝐶𝐶𝐶𝐶 3+ + 3𝑒𝑒 −
23-7 Suppose 10 g of Sn2+ are dissolved in 1000 mL of water to produce an electrolyte.
Calculate the electrode potential of the tin half-cell.
Solution: The concentration of the electrolyte is
C = 10 g/(118.69 g/mol)/(1 L) = 0.0843 M
The electrode potential from the Nernst equation is
E = –0.14 + (0.0592/2) log (0.0843) = –0.172 V
Note that the logarithm is to the base 10 in this equation.
23-8 Use the Nernst equation to determine the concentration of chromium in water if the
difference from standard potential is +0.016 volts.
564
© 2022 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
,Solution:
The Nernst equation:
0.0592
𝐸𝐸 = 𝐸𝐸0 + log 𝐶𝐶ion
𝑛𝑛
With data from the problem, and for chromium:
0.0592
+0.016 𝑉𝑉 = log 𝐶𝐶ion
3
Rearranging and solving:
0.811 = log 𝐶𝐶ion
𝐶𝐶ion = 6.5 M
23-9 A half-cell produced by dissolving copper in water produces an electrode potential of
+0.32 V. Calculate the amount of copper that must have been added to 1000 mL of
water to produce this potential.
Solution: From the Nernst equation, with E0 = 0.34 V and the molecular weight of copper of
63.54 g/mol, we can find the number of grams “x” added to 1000 ml of the solution.
For Cu2+, n = 2.
0.32 = 0.34 + (0.0592/2) log (x/63.54)
log (x/63.54) = (–0.02)(2)/0.0592 = –0.67568
x /63.54 0.211
= = or x 13.4 g Cu per 1000 mL H 2 O
23-10 An electrode potential in a platinum half-cell is 1.10 V. Determine the concentration of
Pt4+ ions in the electrolyte.
Solution: From the Nernst equation, with E0 = 1.20 V and the molecular weight of platinum of
195.09 g/mol, we can find the amount “x” of Pt added per 1000 mL of solution. For
Pt, n = 4.
1.10 = 1.20 + (0.0592/4) log (x/195.09)
log (x/195.09) = –6.7568
x/195.09 = 0.000000175
x = 0.000034 g Pt per 1000 mL H2O
23-11 Use Faraday’s equation to determine the time for a 5 A current to completely corrode
away a block of silver weighing 50 grams.
Solution:
Using Faraday’s equation:
𝐼𝐼𝐼𝐼𝐼𝐼
𝑤𝑤 =
𝑛𝑛F
Rearranged:
565
© 2022 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 𝑤𝑤𝑤𝑤F
= 𝑡𝑡
𝐼𝐼𝐼𝐼
Solving:
50 g 96,500 C 1 1 mol A ∙ s
𝑡𝑡 = � � � �
1 1 mol 5 A 107.868 g C
𝑡𝑡 = 8946 s = 2.5 h
23-12 A current density of 0.05 A/cm2 is applied to a 150 cm2 cathode. What period of time is
required to plate out a 1-mm-thick coating of silver onto the cathode?
Solution: The current in the cell is I = iA = (0.05 A/cm2)(150 cm2) = 7.5 A
The weight of silver that must be deposited to produce a 1 mm = 0.1 cm thick layer is
w = (150 cm2)(0.1 cm)(10.49 g/cm3) = 157.35 g
From the Faraday equation, with n = 1 for silver:
157.35 g = (7.5 A)(t )(107.868 g/mol)/[(1)(96,500 C)]
=t 18,
= 769 s 5.21 h
23-13 We wish to produce 100 g of platinum per hour on a 1000 cm2 cathode by electro
plating. What plating current density is required? Determine the current required.
Solution: In the Faraday equation, n = 4 for platinum, which has an atomic weight of 195.09
g/mol. The weight of platinum that must be deposited per second is 100 g/(3600
s/h) = 0.02778 g/s.
0.02778 g/s = (i )(1000 cm 2 )(195.09 g/mol)/[(4)(96,500 C)]
i = 0.055 A/cm 2
The current must then be
I = iA = (0.055 A/cm2)(1000 cm2) = 55 A
23-14 A 1-m-square steel plate is coated on both sides with a 0.005-cm-thick layer of zinc. A
current density of 0.02 A/cm2 is applied to the plate in an aqueous solution. Assuming
that the zinc corrodes uniformly, determine the length of time required before the steel
is exposed.
Solution: The surface area includes both sides of the plate:
A = (2 sides) (100 cm) (100 cm) = 20,000 cm2
The weight of zinc that must be removed by corrosion is
w = (20,000 cm2)(0.005 cm)(7.133 g/cm3) = 713.3 g
From Faraday’s equation, where n = 2 for zinc:
(0.02 A/cm 2 )(20, 000 cm 2 )(t )(65.38 g/cm3 )
713.3 g =
(2)(96,500 C)
= t 5264
= s 1.462 h
566
© 2022 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
of Materials, Enhanced Edition, 7th Edition Donald
R. Askeland
Notes
1- The file is chapter after chapter.
2- We have shown you few pages sample.
3- The file contains all Appendix and Excel
sheet if it exists.
4- We have all what you need, we make
update at every time. There are many new
editions waiting you.
5- If you think you purchased the wrong file
You can contact us at every time, we can
replace it with true one.
Our email:
, A SOLUTION AND ANSWER GUIDE TO ACCOMPANY
THE SCIENCE AND
ENGINEERING OF MATERIALS,
ENHANCED SEVENTH EDITION
DONALD R. ASKELAND AND WENDELIN J. WRIGHT
,© 2022, 2016, 2011 Cengage Learning® ISBN: 978-0-357-44789-5
Cengage
ALL RIGHTS RESERVED. No part of this work covered by the 200 Pier 4 Boulevard
copyright herein may be reproduced or distributed in any form or Boston, MA 02210
by any means, except as permitted by U.S. copyright law, without USA
the prior written permission of the copyright owner, except as may
be permitted by the license terms below. Cengage Learning is a leading provider of customized
learning solutions with employees residing in nearly 40
different countries and sales in more than 125 countries
For product information and technology assistance, contact us at
around the world. Find your local representative at
Cengage Learning Academic Resource Center, www.cengage.com.
1-800-423-0563.
For permission to use material from this text or product, submit
all requests online at www.cengage.com/permissions.
Further permissions questions can be emailed to
.
NOTE: UNDER NO CIRCUMSTANCES MAY THIS MATERIAL OR ANY PORTION THEREOF BE SOLD, LICENSED, AUCTIONED,
OR OTHERWISE REDISTRIBUTED EXCEPT AS MAY BE PERMITTED BY THE LICENSE TERMS HEREIN.
READ IMPORTANT LICENSE INFORMATION
Dear Professor or Other Supplement Recipient: unauthorized use, reproduction, or distribution. Your use of the
Supplement indicates your acceptance of the conditions set forth in
Cengage Learning has provided you with this product (the this Agreement. If you do not accept these conditions, you must
“Supplement”) for your review and, to the extent that you adopt return the Supplement unused within 30 days of receipt.
the associated textbook for use in connection with your course
(the “Course”), you and your students who purchase the All rights (including without limitation, copyrights, patents, and trade
textbook may use the Supplement as described below. secrets) in the Supplement are and will remain the sole and
Cengage Learning has established these use limitations in exclusive property of Cengage Learning and/or its licensors. The
response to concerns raised by authors, professors, and other Supplement is furnished by Cengage Learning on an “as is” basis
users regarding the pedagogical problems stemming from without any warranties, express or implied. This Agreement will be
unlimited distribution of Supplements. governed by and construed pursuant to the laws of the State of
New York, without regard to such State’s conflict of law rules.
Cengage Learning hereby grants you a nontransferable license
to use the Supplement in connection with the Course, subject to Thank you for your assistance in helping to safeguard the integrity
the following conditions. The Supplement is for your personal, of the content contained in this Supplement. We trust you find the
noncommercial use only and may not be reproduced, posted Supplement a useful teaching tool.
electronically or distributed, except that portions of the
Supplement may be provided to your students IN PRINT FORM
ONLY in connection with your instruction of the Course, so long
as such students are advised that they may not copy or
distribute any portion of the Supplement to any third party. Test
banks and other testing materials may be made available in the
classroom and collected at the end of each class session, or
posted electronically as described herein. Any material posted
electronically must be through a password-protected site, with
all copy and download functionality disabled, and accessible
solely by your students who have purchased the associated
textbook for the Course. You may not sell, license, auction, or
otherwise redistribute the Supplement in any form. We ask that
you take reasonable steps to protect the Supplement from
, SOLUTION AND ANSWER GUIDE
TO ACCOMPANY
The Science and
Engineering of Materials
Enhanced Seventh Edition
DONALD R. ASKELAND
University of Missouri—Rolla, Emeritus
WENDELIN J. WRIGHT
Bucknell University
, Contents
Chapter 1 .......................................................................................................................................... 1
Chapter 2 ........................................................................................................................................13
Chapter 3 ........................................................................................................................................29
Chapter 4 ........................................................................................................................................85
Chapter 5 ......................................................................................................................................109
Chapter 6 ......................................................................................................................................139
Chapter 7 ......................................................................................................................................181
Chapter 8 ......................................................................................................................................207
Chapter 9 ......................................................................................................................................243
Chapter 10 ....................................................................................................................................279
Chapter 11 ....................................................................................................................................309
Chapter 12 ....................................................................................................................................333
Chapter 13 ....................................................................................................................................371
Chapter 14 ....................................................................................................................................397
Chapter 15 ....................................................................................................................................411
Chapter 16 ....................................................................................................................................427
Chapter 17 ....................................................................................................................................443
Chapter 18 ....................................................................................................................................473
Chapter 19 ....................................................................................................................................483
Chapter 20 ....................................................................................................................................509
Chapter 21 ....................................................................................................................................529
Chapter 22 ....................................................................................................................................549
Chapter 23 ....................................................................................................................................563
, Chapter 23: Corrosion and Wear
23-1 Explain how it is possible to form a simple electrochemical cell using a lemon, a penny,
and a galvanized steel nail.
Solution: The penny serves as the cathode (copper is the cathode). The steel nail with a zinc
coating serves as the anode (zinc is the anode), and the lemon serves as the
electrolyte. A voltmeter or wire connecting the penny and nail completes the circuit.
23-2 Silver can be polished by placing it in an aluminum pan containing a solution of baking
soda, salt, and water. Speculate as to how the polishing process works.
Solution:
The black tarnish on silver is silver sulfide. An electrochemical reaction occurs in which
the silver sulfide is reduced to silver plus aluminum sulfide according to
3 Ag2S + 2 Al → 6 Ag + Al2S3
The baking soda functions as an electrolyte. The silver is left behind on the part to be
cleaned.
23-3 A gray cast iron pipe is used in the natural gas distribution system for a city. The pipe
fails and leaks, even though no corrosion noticeable to the naked eye has occurred.
Offer an explanation for why the pipe failed.
Solution: Because no corrosion is noticeable, the corrosion byproduct apparently is still in
place on the pipe, hiding the corroded area. The circumstances suggest graphitic
corrosion, an example of a selective chemical attack. The graphite flakes in the gray
iron are not attacked by the corrosive soil, while the iron matrix is removed or
converted to an iron oxide or hydroxide trapped between the graphite flakes.
Although the pipe appears to be sound, the attacked area is weak, porous, and
spongy. The natural gas can leak through the area of graphitic corrosion and
accumulate, eventually leading to an explosion.
23-4 A brass plumbing fitting produced from a Cu-30% Zn alloy operates in the hot water
system of a large office building. After some period of use, cracking and leaking occur.
On visual examination, no metal appears to have been corroded. Offer an explanation
for why the fitting failed.
Solution: The high zinc brasses are susceptible to dezincification, particularly when the
temperature is increased, as in the hot water supply of the building. One of the
characteristics of dezincification is that copper is redeposited in the regions that are
attacked, obscuring the damage. The redeposited copper is spongy, brittle, and
weak, permitting the fitting to fail and leak. Therefore, dezincification appears to be
a logical explanation.
563
© 2022 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
,23-5 Plot the electronegativity of the elements listed in Table 23–1 as a function of electrode
potential. (Refer to Chapter 2 for electronegativity values.) What is the trend that you
observe in this data? Using an example, explain how the electronic structure of a metal
is consistent with its anodic or cathodic tendency.
Solution:
In general, as electronegativity increases, the tendency for the elements to corrode
decreases. Anodic elements are more electropositive (they have a low
electronegativity), meaning that they want to give up their electrons. Cathodic
elements are electronegative, i.e., they want to gain electrons. For example,
consider lithium. Li has the electronic structure [Li] = [He] 2s1. Lithium wants to give
up the 2s1 electron to achieve a stable configuration and has anodic tendencies.
23-6 Write the anodic or cathodic reaction (as appropriate) for aluminum, germanium and
chromium.
Solution:
𝐴𝐴𝐴𝐴 → 𝐴𝐴𝐴𝐴 3+ + 3𝑒𝑒 −
𝐺𝐺𝐺𝐺 → 𝐺𝐺𝐺𝐺 4+ + 4𝑒𝑒 −
𝐶𝐶𝐶𝐶 → 𝐶𝐶𝐶𝐶 3+ + 3𝑒𝑒 −
23-7 Suppose 10 g of Sn2+ are dissolved in 1000 mL of water to produce an electrolyte.
Calculate the electrode potential of the tin half-cell.
Solution: The concentration of the electrolyte is
C = 10 g/(118.69 g/mol)/(1 L) = 0.0843 M
The electrode potential from the Nernst equation is
E = –0.14 + (0.0592/2) log (0.0843) = –0.172 V
Note that the logarithm is to the base 10 in this equation.
23-8 Use the Nernst equation to determine the concentration of chromium in water if the
difference from standard potential is +0.016 volts.
564
© 2022 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
,Solution:
The Nernst equation:
0.0592
𝐸𝐸 = 𝐸𝐸0 + log 𝐶𝐶ion
𝑛𝑛
With data from the problem, and for chromium:
0.0592
+0.016 𝑉𝑉 = log 𝐶𝐶ion
3
Rearranging and solving:
0.811 = log 𝐶𝐶ion
𝐶𝐶ion = 6.5 M
23-9 A half-cell produced by dissolving copper in water produces an electrode potential of
+0.32 V. Calculate the amount of copper that must have been added to 1000 mL of
water to produce this potential.
Solution: From the Nernst equation, with E0 = 0.34 V and the molecular weight of copper of
63.54 g/mol, we can find the number of grams “x” added to 1000 ml of the solution.
For Cu2+, n = 2.
0.32 = 0.34 + (0.0592/2) log (x/63.54)
log (x/63.54) = (–0.02)(2)/0.0592 = –0.67568
x /63.54 0.211
= = or x 13.4 g Cu per 1000 mL H 2 O
23-10 An electrode potential in a platinum half-cell is 1.10 V. Determine the concentration of
Pt4+ ions in the electrolyte.
Solution: From the Nernst equation, with E0 = 1.20 V and the molecular weight of platinum of
195.09 g/mol, we can find the amount “x” of Pt added per 1000 mL of solution. For
Pt, n = 4.
1.10 = 1.20 + (0.0592/4) log (x/195.09)
log (x/195.09) = –6.7568
x/195.09 = 0.000000175
x = 0.000034 g Pt per 1000 mL H2O
23-11 Use Faraday’s equation to determine the time for a 5 A current to completely corrode
away a block of silver weighing 50 grams.
Solution:
Using Faraday’s equation:
𝐼𝐼𝐼𝐼𝐼𝐼
𝑤𝑤 =
𝑛𝑛F
Rearranged:
565
© 2022 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 𝑤𝑤𝑤𝑤F
= 𝑡𝑡
𝐼𝐼𝐼𝐼
Solving:
50 g 96,500 C 1 1 mol A ∙ s
𝑡𝑡 = � � � �
1 1 mol 5 A 107.868 g C
𝑡𝑡 = 8946 s = 2.5 h
23-12 A current density of 0.05 A/cm2 is applied to a 150 cm2 cathode. What period of time is
required to plate out a 1-mm-thick coating of silver onto the cathode?
Solution: The current in the cell is I = iA = (0.05 A/cm2)(150 cm2) = 7.5 A
The weight of silver that must be deposited to produce a 1 mm = 0.1 cm thick layer is
w = (150 cm2)(0.1 cm)(10.49 g/cm3) = 157.35 g
From the Faraday equation, with n = 1 for silver:
157.35 g = (7.5 A)(t )(107.868 g/mol)/[(1)(96,500 C)]
=t 18,
= 769 s 5.21 h
23-13 We wish to produce 100 g of platinum per hour on a 1000 cm2 cathode by electro
plating. What plating current density is required? Determine the current required.
Solution: In the Faraday equation, n = 4 for platinum, which has an atomic weight of 195.09
g/mol. The weight of platinum that must be deposited per second is 100 g/(3600
s/h) = 0.02778 g/s.
0.02778 g/s = (i )(1000 cm 2 )(195.09 g/mol)/[(4)(96,500 C)]
i = 0.055 A/cm 2
The current must then be
I = iA = (0.055 A/cm2)(1000 cm2) = 55 A
23-14 A 1-m-square steel plate is coated on both sides with a 0.005-cm-thick layer of zinc. A
current density of 0.02 A/cm2 is applied to the plate in an aqueous solution. Assuming
that the zinc corrodes uniformly, determine the length of time required before the steel
is exposed.
Solution: The surface area includes both sides of the plate:
A = (2 sides) (100 cm) (100 cm) = 20,000 cm2
The weight of zinc that must be removed by corrosion is
w = (20,000 cm2)(0.005 cm)(7.133 g/cm3) = 713.3 g
From Faraday’s equation, where n = 2 for zinc:
(0.02 A/cm 2 )(20, 000 cm 2 )(t )(65.38 g/cm3 )
713.3 g =
(2)(96,500 C)
= t 5264
= s 1.462 h
566
© 2022 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
