COURSE GUIDE
Course: Chemistry for Engineers
Semester: First
S.Y.: 2020-2021
Class Schedule: 9:00-12:00 MTWTh
Instructor: Engr. Dan R Vinas
7:30 – 10:30 F
Course Description:
This course provides students with core concepts of chemistry that are
important in the practice of engineering profession.
Course Outline
SCHEDULE
AUGUST, SEPTEMBER,
OCTOBER and MIDTERM
EXAM
OCTOBER, NOVEMBER,
DECEMBER and FINAL
EXAM
TOPIC
Numbers and Measurement in Chemistry
Molecules, Moles, and Chemical Equation
Stoichiometry
Electrochemistry
Nuclear Chemistry
Atmospheric Chemistry and Air Pollution
Aquatic Chemistry and Water Pollution
Soil Chemistry and Pollution
Course Learning Outcomes
1. Discuss the application of chemistry in relation to the generation of energy
2. Discuss the chemical processes that takes place in the environment
3. Identify key chemistry concepts related to the specific field of engineering
Course Requirements:
Mid term:
1. Accomplish all quizzes covered from each chapter (chapters 1 – 4) in an
intermediate yellow pad.
2. Placed it inside a long brown envelope and submit that during the written
midterm exam.
Final term
3. Accomplish all quizzes covered from each chapter (chapters 5 – 8) in an
intermediate yellow pad.
4. Placed it inside a long brown envelope and submit that during the written
final exam.
Midterm and final exams will be a “face-to-face” mode of administering.
Coverage for the midterm exam will be chapters 1-4, while chapters 5-8 will be
covered on the final exam.
Grading System:
Grades will be computed as follows:
Midterm grade:
50% quizzes + 50% midterm exam
Final grade:
50% quizzes + 50% final exam
GRADE : (midterm grade + final grade) / 2
Consultation:
You should join an exclusive chemistry for engineers’ chat group for your
section. Consultation and announcements will be done on this chat group.
Chapter 1
Numbers and Measurements in Chemistry
Introduction
Calculations play a major role in the practice of chemistry and its application to real
world issues and problems. And engineering routinely relies on a tremendous number
of calculations. This chapter, though focused on chemistry, will provide practice with
techniques that can also be used in engineering applications.
Learning Outcomes
After mastering this chapter, you should be able to:
✓ Convert measurements from one unit to another using appropriate ratios.
✓ Solve contextual problems using the correct number of significant figures
LESSON 1
MEASUREMENT OF MATTER
Mass and Weight
Matter is defined as anything that has mass and
takes up space. Mass is the quantity of matter in a
particular sample of matter. The mass of a body is
constant and does not change, regardless of where it is
measured.
The weight of a body, however, is the
gravitational force of attraction between the body’s mass
Figure 1- Engineering students
and the mass of the planet on which it is weighed. Thus, measuring the mass of a sample
the weight of a body varies, depending on where it is using a triple beam balance
being weighed, whereas the mass does not. The mass of a body can be measured on
a balance.
The terms mass and weigh are unfortunately often used interchangeably;
however, you should recognize the difference in the terms.
Length and Volume
Length is simply measured with a ruler, divided into inches or centimeters, or a
meter stick.
Volume is measured in several ways. A liquid can be measured in a graduated
cylinder, in a volumetric flask, or with a buret or pipet. To find the volume of a regular
solid we can measure the dimensions of it and multiply them together. To find the
volume of an irregular solid, we can place it in water and measure the amount of water
that is displaced.
LESSON 2
THE METRIC SYSTEM AND THE INTERNATIONAL SYSTEM OF UNITS
The metric system has as its basic units the gram (g) for mass (weight), the
meter (m) for length, the liter (l) for volume, and the second (s) for time. The units are
related in multiples of 10, 100, 1000, and so on, like our monetary system. Table 1
lists the prefixes used in the metric system and gives an example of each. It is
important that you become familiar with these prefixes since you will use this system
of measurement for all calculations
Table Error! No text of specified style in document.-1- Prefixes used in the metric system
Prefix (Abbreviation)
Value
mega (m)
kilo (k)
hecto (h)
deka (da)
deci (d)
1,000,000 (106)
1,000 (103)
100 (102)
10 (101)
1/10 (10-1)
centi (c)
1/100 (10-2)
milli (m)
1/1,000 (10-3)
micro (µ)
1/1,000,00 (10-6)
nano (n)
1/109 (10-9)
pico (p)
1/1012 (10-12)
Example
1 megagram (Mg) = 106 g
1 kilogram = 103 g
1 hectoliter = 100 l
1 decaliter (dal) = 10 l
1 decimeter (dm) = 10-1 m
or 1 m = 10 dm
1 centimeter (cm) = 10-2 m
or 1 m = 102 cm
1 millimeter (mm) = 10-3 m
or 1 m = 103 mm
1 micrometer (µm) = 10-6
m or 1 m = 106 µ
1 nanometer (nm) = 10-9 m
or 1 m = 109 nm
1 picometer (pm) = 10-12 m
or 1 m = 1012 pm
The unit angstrom (Å) is another unit of length that is often used to discuss the size of
atoms; [1 Å = 10-8 cm = 10-10 m], or 1 m = 1010 Å. The unit nanometer (nm), however,
which is a similar size unit (10 Å = 1 nm) is preferred and is often used in place of Å.
A cube 1 m on a side will hold a volume of exactly 1 ml. Therefore, [1 ml = 1 cm3 (cc)].
These units, since they are exactly the same, are often used interchangeably.
SI Units
The International System of Units (SI units) are now being adopted throughout
the world. This system is basically the same as the metric system, except that the
standard unit of mass is the kilogram (kg) and the unit of volume is the cubic meter
(m3). This system also includes units for energy, force, pressure, and so forth. All of
these are derived from the six basic units listed in Table 2. Some of the more common
derived units, are listed in Table 3. Since the unit of volume is the m3, the liter (l) is
defined as 1 dm3, and therefore 1 ml = 1 cm3 (cc). That is, a cube 1 cm on a side
would hold a volume of exactly 10-3 l or 1 ml.
Table Error! No text of specified style in document.-2- Basic SI units
MEASUREMENT
mass
length
time
temperature
amount
electric current
UNIT (ABBREVIATION)
kilogram (kg
meter (m)
second (s)
kelvin (K)
mole (mol)
amperes (A)
Metric-English Equivalents
Table 4 lists a metric-English equivalent for each of the three units of
measurement. You must learn these three equivalents to convert from one system to
the other. There are also other equivalents that can be used but if you know these
three and the prefixes used in the metric system (as listed in Table 1) you can do any
conversions between the two systems.
Table Error! No text of specified style in document.-3- Some derived units used in the SI system
MEASUREMENT
volume
density
velocity
force
energy
UNIT
m3
kg/m3 or kg m-3
m/s or m s-1
newton (N) ; N = kg m/s2
Joule (J) ; J = kg m2/s2
Table Error! No text of specified style in document.-4 - Metric-English unit equivalents
DIMENSION
Mass
Volume
Length
Time
ENGLISH UNIT
1 pound (lb)
1.06 quarts (qt)
1 inch (in.)
1 mile (mi)
1 second (s)
METRIC EQUIVALENT
454 grams (g)
1 liter (l)
2.54 centimeters (cm)
1.6 kilometers (km)
1 second (s)
LESSON 3
CONVERSION FACTORS
In making conversions within a given system or from one system to another, a
very simple and logical method is to use what is generally called conversion factors.
These are factors that change the units in a number without changing the value;
therefore, it is essential that the conversion factor always be equal to one.
For example, since 1 m = 100 cm, dividing both sides of this equation by 100
cm gives,
1𝑚
100 𝑐𝑚
=1
or dividing both sides by 1 m gives
100 𝑐𝑚
1𝑚
=1
These are referred to as conversion factors. The first thing to remember about
a conversion factor is that it must be equal to one. The next thing to keep in mid when
using these to solve problems is that they must be used properly so that the units come
out as desired.
For example, if you want to convert 2.5 centimeters to meters, multiply the
original number (2.5 cm) by the proper conversion factor to give the unit meter
2.5 cm x
1𝑚
100 𝑐𝑚
= 0.025 m
PROBLEM EXAMPLE 1-1
Convert 55 mg to kilograms
SOLUTION
Recall that:
1 kg
= 1000 g
1g
= 1000 mg
Using these relationships and setting up the proper conversion factors gives
55 mg x
=
1𝑔
1000 𝑚𝑔
55
1 𝑥 106
x
1 𝑘𝑔
1000 𝑔
=
55
kg
(1000)(1000)
kg = 55 x 10-6 kg = 5.5 x 10-5 kg
PROBLEM EXAMPLE 1-2
Convert 15.0 in2 to m2
SOLUTION
Here we must convert square inches to square centimeters, and square centimeters
to square meters. To do these conversions we use the relationships
(1 in)2 = (2.54 cm)2 or 12 in2 = (2.54)2 cm2
And
(1 m)2 = (100 cm)2 or 12 m2 = (100)2 cm2
Keep in mind with conversions of this type that both the number and the unit must be
squared. Therefore, the conversion is
15.0 in2 x
(2.54)2 𝑐𝑚2
12 𝑖𝑛2
x
12 𝑚2
2
(100)2 𝑐𝑚2
=
(15)(2.54) 𝑚2
(100)
2
= 9.68 x 10-3 m2
QUIZ #1-1
Convert the following
•
•
•
•
8.5 kg to g
0.15 mm to km
24.8 lb to mg
8.5 qt to ml
•
•
•
•
1.85 m to µm
3.20 Å to nm
4.5 ft to m
80.0 lb/ft3 to g/ml
•
•
•
•
0.25 dg to g
0.022 cm3 to l
52.0 cm/s to ft/min
0.020 nm2 to ft2
LESSON 4
TEMPERATURE
There are three commonly used temperature scales. These are the Fahrenheit
scale (oF), the Celsius scale (oC) (which is also sometimes referred to as the centigrade
scale), and the kelvin scale (oK) (which is sometimes referred to as the absolute scale.
In the SI system, the degree sign is not used, and thus the unit of temperature is simply
K, as 273 K.
To convert a given temperature from degrees Celsius to degrees Fahrenheit or
degrees Fahrenheit to degrees Celsius, we use these formulas:
To convert degrees Celsius to degrees Fahrenheit
o
F=
o
9
5
o
C + 32
F = 1.8 oC + 32
To convert degrees Fahrenheit to degrees Celcius
5
o
C=
9
(oF – 32)
𝑜𝐹−32
o
C=
1.8
To convert from oC to oK we need only to add 273
To convert oC to oK
o
K = oC + 273
PROBLEM EXAMPLE 1-3
Convert 68oF (room temperature) to oC
SOLUTION
Using the formula
o
C =
5
9
(oF – 32)
Substitute the given
o
C =
=
5
9
5
9
(68 – 32)
(36)
= 20oC
PROBLEM EXAMPLE 1-4
Convert -25oC to oF
SOLUTION
Using the formula
o
F = 1.8 oC + 32
Substitute the given
o
F = 1.8 (-25) + 32
= -45 + 32
= -13oF
QUIZ #1-2
A. Convert each of the following temperatures to oF and oK
1) 40.0 oC
2) -80 oC
3) -120 oC
4) 500 oC
B. Convert each of the following temperatures to oC and oK
1) 86.0 oF
2) -20 oF
3) 450 oF
4) -120 oF
LESSON 5
DENSITY
The density of a substance is defined as the mass of a substance occupying
a unit volume.
Density =
𝑚𝑎𝑠𝑠
𝑣𝑜𝑙𝑢𝑚𝑒
We know that if we compare the same volume of various substances, some
will be heavier than others. For example, a lead brick is much heavier than a piece of
wood of the same size. Hence, we say that the lead is denser, or has a greater density
than the wood.
In the metric system the density of liquids and solids is generally measured in
grams per milliliter or grams per cubic centimeter. In SI units, density is expressed as
kilogram per cubic meter. It is important that the units are included when expressing
densities. For example, the density of water is 1.00 g/ml. It can also be expressed,
however, as 62.4 lb/ft3 or 8.35 lb/gal.
PROBLEM EXAMPLE 1-5
Calculate the density in grams per milliliter of a piece of metal that has a mass of 12 g
and occupies a volume of 1.6 ml.
SOLUTION
12 𝑔
1.6 𝑚𝑙
= 7.5 g/ml
PROBLEM EXAMPLE 1-6
A cube of lead measures 3.00 cm on each edge and has a mass of 308 g. Calculate
the density in g/cm3.
SOLUTION
The volume of the cube of lead is
3.00 cm x 3.00 cm x 3.00 cm = 27.0 cm3
And the density is
308 𝑔
27.0 𝑐𝑚3
= 11.4 g/cm3
PROBLEM EXAMPLE 1-7
The density of a liquid is 1.20 g/ml. Calculate
(a) the mass in kilograms of 10.0 ml and
(b) the volume in liters occupied by 10.0 g
SOLUTION
a) 10.0 ml x
b) 10.0 g x
1.20 𝑔
1 𝑚𝑙
1 𝑚𝑙
1.20 𝑔
x
x
1 𝑘𝑔
1000 𝑔
1𝑙
1000 𝑚𝑙
= 0.0120 kg
= 8.33 x 10-3 l
QUIZ #1-3
Calculate the density of each of the following in grams per cubic centimeter:
1. A piece of metal measuring 1.0 cm by 0.10 dm by 25 mm having a mass of
5.0 g.
2. A substance with a mass of 425 kg occupying a volume of 0.23 m3
3. A substance with a mass of 83.5 kg occupying a volume of 0.015 m3
LESSON 6
SIGNIFICANT FIGURES
We often encounter very small and very large numbers in chemistry problems.
For example, pesticide production in the world exceeds million of tons, whereas
pesticide residues that may harm animals or humans can have masses as small as
nanograms. For either type of number, scientific notation is useful. Numbers written
using scientific notation factor out all powers of ten and write them separately. Thus,
the number 54,000 is written as 5.4 x 104. This notation is equivalent to 5.4 x 10,000,
which is clearly 54,000. Small numbers can also be written in scientific notation using
negative powers of ten because 10-2 is identical to 1/10x. The number 0.000042 is 4.2
x 10-5 in scientific notation.
When numbers are derived from observations of nature, we need to report
them with the correct number of significant figures. Significant figures are used to
indicate the amount of information that is reliable when discussing a measurement.
“Pure” numbers can be manipulated in a mathematical sense without accounting for
how much information is reliable.
All measurement are approximations – no measuring device can give perfect
measurements without experimental uncertainty. The purpose of significant figures is
to indicate the approximate uncertainty in a measurement. The number of significant
figures in a measurement is simply the number of figures that are known with some
degree of reliability.
Generally, the last significant digit is considered to be uncertain. For example,
a mass measured to 13.2 g is said to have an absolute uncertainty of 0.1 g and is said
to have been measured to the nearest 0.1 g. In other words, we are somewhat
uncertain about that last digit – it could be a “2”; then again, it could be a “1” or a “3”.
The mass could be 13.1 g, 13.2 g, or 13.3 g. A mass of 13.25 g indicates an absolute
uncertainty of 0.01 g. The mass could be 13.24 g, 13.25 g, or 13.26 g.
Rules for deciding the number of significant figures in a measured quantity.
1. All non-zero digits are significant.
12.34 g has 4 significant figures
5.6 has 2 significant figures
2. Zeroes between non-zero digits are significant
8901 kg has 4 significant figures
2.02 ml has 3 significant figures
3. Zeroes to the left of the first non-zero digits are not significant; such zeroes
merely indicate the position of the decimal point.
0.01
o
C has only 1 significant figure
0.024 g has 2 significant figures
4. Zeroes to the right of a decimal point (after the first non-zero digit) are
significant.
0.02030 ml has 4 significant figures
0.200 g has 3 significant figures
5. When a number ends in zeroes that are not to the right of a decimal point, the
zeroes are not necessarily significant.
160 miles may be 2 or 3 significant figures
30,200 calories may be 3, 4, or 5 significant figures
The potential ambiguity in the last rule can be avoided by the use of scientific
notation. For example, depending on whether 3, 4, or 5 significant figures is correct,
we could write 30,200 calories as
3.02 x 104 calories (3 significant figures)
3.020 x 104 calories (4 significant figures)
3.0200 x 104 calories (5 significant figures)
Some numbers are exact because they are known with complete certainty.
Most exact numbers are integers: exactly 12 inches are in a foot, there might be exactly
32 students in a class. Exact numbers are often found as conversion factors or as
counts of objects.
Exact numbers can be considered to have an infinite number of significant
figures. Thus, number of apparent significant figures in any exact number can be
ignored as a limiting factor in determining the number of significant figures in the result
of a calculation.
Rules for mathematical operations.
Multiplication and Division
To determine the number of significant digits in your final answer when multiplying or
dividing, first do the calculation. Then round the answer to the same number of
significant digits that is in the number with the least number of significant digits in your
calculation
PROBLEM EXAMPLE 1-8
Convert 116 pounds to kilograms. Use the approximation 1 kg = 2.2 pounds.
SOLUTION
116 ÷ 2.2 = 52.727272….
116 has 3 significant digits
2.2 has 2 significant digits
So the final answer will have 2 significant digits, making it no more precise than
the least precise.
Final answer: 53 kg
PROBLEM EXAMPLE 1-9
A child measures 25.75 inches tall. Convert this to centimeters.
SOLUTION
25.75 x 2.54 = 65.405
25.75 has 4 significant digits
2.54 has 3 significant digits.
So the final answer will have 3 significant digits
Final answer: 65.4 cm
Adding and Subtracting
When adding of subtracting, first do the calculation. Then round the answer to
the same number of decimal places (places to the right of the decimal point) as the
number in the calculation with the fewest decimal places.
PROBLEM EXAMPLE 1-10
Find the sum of 4.89 ft + 1.9 ft + 3.506 ft
SOLUTION
4.89 ft + 1.9 ft + 3.506 ft = 10.296 ft
4.89 has 2 decimal places
1.9 has 1 decimal place
3.506 has 3 decimal places
The fewest decimal places is 1
So the final answer is 10.3 ft
PROBLEM EXAMPLE 1-10
Subtract: 1.268 liters from 2.5 liters
SOLUTION
2.5 liters – 1.268 liters = 1.232 liters
2.5 has 1 decimal place
1.268 has 3 decimal places
The fewest decimal places is 1
So the final answer is 1.2 liters
QUIZ #1-4
A. Determine the number of significant digits
1. 0.530
2. 410.0
3. 1.00
4. 43.00240
5. 3,000
B. Round to the number of significant digits indicated
1. 5.67498 to 1 significant digit
2. 0.04102 to 3 significant digits
3. 2.998 to 2 significant digits
4. 26,384 to 2 significant digits
5. 37.446 to 3 significant digits
C. Multiply or divide as indicated. Round the final answer to the appropriate
number of significant digits.
1. 27.3 x 4.5
2. 4.68 x 400
3. 323 x 0.0002
4. 4008 ÷ 2.763
5. 69 ÷ 7.0
D. Add or subtract as indicated. Round to the appropriate number of decimal
places.
1. 5.72 + 2
2. 500 – 79.4
3. 0.006 + 0.04
4. 84.3 – 0.009
5. 66.3 + 27.008
Chapter 2
Molecules, Moles, and Chemical Equation
Introduction
Atoms and molecules are the building blocks of chemistry. You’ve probably
been hearing this since junior high school, so the existence of atoms is not something
that you are likely to question or challenge. Chances are that you rarely think about
atoms or molecules when you come across items in your day-to-day life. When
chemists want to understand some aspect of the world around them, they focus their
attention at the level of atoms and molecules. So an important part of studying
chemistry is learning how to interpret nature by thinking about what atoms and
molecules are doing.
Learning Outcomes
At the end of this chapter you should be able to:
✓ Determine the molecular weight of a chemical formula.
✓ Calculate the number of moles in a compound.
✓ Explain the relationship between chemical equations and chemical reactions.
LESSON 1
MOLES AND MOLECULES
Chemical Formulas
Chemical formulas are the chemist’s “shorthand.” They are written
representations of a compound’s components but they do not depict the structure of
the molecules
.
Examples of chemical formulas include:
H2O (water)
NaOH (sodium hydroxide or caustic soda)
and CO2 (carbon dioxide).
In order to determine how much of a chemical need to be added to a solution
or how much will be required to create another compound, it is necessary to calculate
the molecular weight of compounds.
Molecular Weight
The molecular weight (or formula weight) of a substance is the sum of the
atomic weights of all the atoms in the molecule.
Some compounds contain more than one atom of an element. H2O is an
example. When a compound contains more than one atom of an element, all of the
atoms must be included in order to calculate the formula weight. The Periodic Table
lists the atomic weights of the elements.
PROBLEM EXAMPLE 2-1
What is the molecular weight of NaCl (sodium chloride or table salt) given that
the atomic weight of Na is 22.989 and the atomic weight of Cl is 35.453?
SOLUTION
The molecular weight of NaCl is:
1 Na atom = 22.2989
1 Cl atom = + 35.453
NaCl = 58.442 g/mole
PROBLEM EXAMPLE 2-2
What is the molecular weight of H2O (water), given that hydrogen has an atomic
weight of 1.0080 and oxygen has an atomic weight of 15.9994?
SOLUTION
The molecular weight of H2O is:
2 H atoms = 2.016
1 O atom = + 15.9994
H2O = 18.01
Mole
A mole is a method to discuss how many “molecular weights” of a compound
are present for a particular mass of the compound. A quantity of compound equal in
weight to its molecular weight is a mole. The concept of moles is an important
component necessary for the calculation and determination of concentrations of
chemical solutions. Given the extremely tiny size of molecules and the incredible
number of molecules which are present, the mole is the basic unit used to conveniently
describe how much of a chemical compound is present.
Moles work for any system of weights. For example, the molecular weight of
water (H2O) is 18.0099. To calculate the number of moles, divide the molecular
weight by the total number of grams. This means that 18.0099 grams of water is
equal to one mole of water, or, in other words:
moles =
moles =
𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 (𝑖𝑛 𝑔𝑟𝑎𝑚𝑠)
18.0099
18.0099
moles = 1 mol H2O
As another example, sodium chloride (NaCl), also known as table salt, is made
using one mole each of Na (sodium) and Cl (chlorine). By adding 22.9898 grams of
Na to 35.453 grams of Cl, NaCl is formed. Another option is to add 22.9898 pounds
of Na to 35.453 pounds of Cl. Again, NaCl is formed. In either scenario, the proportions
of the elements are the same, regardless of whether they are measured in grams or in
pounds.
The mass of 6.02 x 1023 (Avogadro’s number) formula units (atoms, molecules,
ions) is the formula weight expressed in grams. The mass of 1 mole of a substance is
sometimes referred to as the molar mass. That is, the molar mass is the mass of any
substance that contains Avogadro’s number of units, where the units can be atoms,
molecules, formula units, individual ions, and so on.
PROBLEM EXAMPLE 2-3
Calculate the number of moles of NaOH in 85.0 g of NaOH
SOLUTION
The formula weight (molecular weight) of NaOH is 40.0; therefore, there are
40.0 g of NaOH in 1 mole. Thus, the number of moles in 85.0 g of NaOH is calculated
as:
85.0 g of NaOH x
1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
40.0 𝑔 𝑁𝑎𝑂𝐻
= 2.12 mol NaOH
PROBLEM EXAMPLE 2-4
Calculate the mass in grams of 0.720 mol of Ca3(PO4)2
SOLUTION
The formula weight of Ca3(PO4)2 is (3 x 40.1) + (2 x 31.0) + (8 x 16.0) = 310.
Therefore, 1 mol Ca3(PO4)2 = 310 g and the mass of 0.720 mol is calculated as
0.720 mol Ca3(PO4)2 x
310 𝑔 Ca3(PO4)2
1 𝑚𝑜𝑙 Ca3(PO4)2
= 223 g Ca3(PO4)2
PROBLEM EXAMPLE 2-5
Calculate the number of molecules of 24.5 g of CO2
SOLUTION
In the calculation, we must first calculate the number of moles, using the
formula weight (44.0 g CO2/mol) and then use the relationship between molecules and
moles to find the number of molecules.
24.5 g CO2 x
1 𝑚𝑜𝑙 CO2
44.0 𝑔CO2
x
6.02 𝑥 1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠
1 𝑚𝑜𝑙CO2
= 3.35 x 1023 molecules CO2
QUIZ #2-1
Calculate:
1.
2.
3.
4.
5.
6.
The mass in grams of 1 mole of CaSO4
The number of moles of 0.020 kg H2O
The number of moles of 8.30 x 1020 molecules H2O
The number of molecules I 2.5 liters of water
The mass in grams of 12.5 mmol NaCl
Calculate the volume, in cubic centimeters, occupied by 8.5 x 1024 molecule
CCl4 (density = 1.6 g/ml)
7. Calculate the molecular (formula) weight of 2.8 kg that contains 1.2 x 1022
molecules
8. Calculate the number of moles of oxygen atoms in 2.52 g Ca(NO3)2
LESSON 2
CHEMICAL EQUATIONS
Writing Chemical Equations
A chemical equation is a chemist’s shorthand expression for describing a
chemical change. As an example, consider what takes place when iron rusts. The
equation for this change is:
In this expression, the symbols and formulas of the reacting substances, called
the reactants, are written on the left side of the and the products of the reaction are
written on the right side. The arrow is read as “gives”, “yields”, or “forms” and the plus
(+) sign is read as “and”. When the plus (+) sign appears between the formulas for
two reactants, it can be read as “reacts with”. (The + sign does not imply mathematical
addition.)
The equation, above, can be read as iron reacts with oxygen to yield (or form)
iron (III) oxide.
Balancing A Chemical Equation
The equation indicates in a qualitative way what substances are consumed in
the reaction and what new substances are formed. In order to have quantitative
information about the reaction, the equation must be balanced so that it conforms to
the Law of Conservation of Matter. That is, there must be the same number of atoms
of each element on the right-hand side of the equations as there are on the left side.
If the number of atoms of each element in the equation above are counted, it
is observed that there are 1 atom of Fe and 2 atoms of O on the left side and 2 atoms
Fe and 3 atoms of O on the right.
The balancing of the equation is accomplished by introducing the proper
number or coefficient before each formula. To balance the number of O atoms, write
a 3 in front of the O2 and a 2 in front of the Fe2O3.
The equation above, now has 6 atoms of O on each side, but the Fe atoms are
not balanced. Since there is 1 atom of Fe on the left and 4 atoms of Fe on the right,
the Fe atoms can be balanced by writing a 4 in front of the Fe.
This equation is now balanced. It contains 4 atoms of Fe and 6 atoms of O on
each side of the equation. The equation is interpreted to mean that 4 atoms of Fe will
react with 3 molecules of O2 to form 2 molecules of Fe2O3.
It is important to note that the balancing of an equation is accomplished by
placing numbers in front of the proper atoms or molecules and not as subscripts. In
an equation, all chemical species appear as correct formula units. The addition (or
change) of a subscript change the meaning of the formula unit and of the equation.
Coefficients in front of a formula unit multiply that entire formula unit.
Another example of balancing an equation is:
Counting the atoms of each element in the equation it is found that there are
1 atom Al, 7 atoms O, 5 atoms H, and 1 atom S on the left side and 2 atoms Al, 13
atoms O, 2 atoms H, and 3 atoms S on the right side.
The counting can be simplified by observing that the S and O in the SO4
polyatomic ion acts as a single unbreakable unit in this equation. Recounting, using
the SO4 as a single unit, it is found that there are 1 atom Al, 3 atoms O, 5 atoms H,
and 1 SO4 polyatomic ion the left side and 2 atoms Al, 1 O atom, 2 H atoms, and 3
SO4 polyatomic ions on the right side.
Starting with A, the atoms of Al can be balanced by writing a 2 in front of the Al(OH)3
Looking at the SO4 ions, these are balance by writing a 3 in front of the H2SO4
Now, only the O atoms and H atoms remain unbalance. There are 6 atoms of
O and 12 atoms of H on the left-hand side of the equation and only 1 atom O and 2
atoms H on the right side. These can be balanced by writing a 6 in front of the H2O
The equation is now balance and it is interpreted to mean that 2 molecules of
Al(OH)2 react with 3 molecules of H2SO4 to form 1 molecule of Al2(SO4)3 and 6
molecules H2O.
QUIZ #2-2
Copy the following chemical equations on the answer sheet and below each
item, write the balanced equation.
Chapter 3
Stoichiometry
Introduction
A tremendous number of chemical compounds exists in nature and undergo
myriad reactions. By building and exploiting a systematic understanding of reactivity,
chemists also have produced an impressive array of man-made compounds. The
economics of any chemical process obviously depend on the amounts of each reactant
needed to produce a given amount of product. For processes carried out on an
industrial scale, even very small changes in efficiency can have enormous impact on
profitability. The quantitative relationships between the amounts of reactants and
products in a chemical reaction are referred to as stoichiometry.
Learning Outcomes
At the end of this chapter, you should be able to:
✓ Calculate the amount of product expected from a chemical reaction, given the
amount reactants used.
✓ Calculate the amounts of reactants needed in a chemical reaction to produce
a specified amount of product
✓ Identify a limiting reactant and calculate the amount of product formed from a
non-stochiometric mixture of reactants
✓ Calculate the percentage yield of a chemical reaction
Calculations Based on Balanced Equations
In doing these calculations, we shall use the mole method. The central part of
any calculation of this type is to determine the relative relationships between the moles
of reactants and products from the balanced equation. For example, the equation
representing the burning of methane (CH4) in oxygen to form carbon dioxide and water
is written as follow:
CH4 + 2 O2
CO2 + 2 H2O
The balanced equation gives us the relationships between moles of methane
and moles of oxygen, moles of methane and moles of carbon dioxide, moles of carbon
dioxide and moles of water, and so on. The following relationships can be obtained
from this balanced equation:
1 mol CH4
= 2 mol O2
1 mol CH4
= 1 mol CO2
1 mol CO2
= 2 mol H2O
and so on.
When considering mass relationships (or weight-weight calculations) from
balanced equation, there are four steps in calculations, although in some problems
some of these are not necessary.
1. Write a complete, balanced equation.
2. Convert from the given units to moles
3. Convert from moles of the given quantity to moles of the desired quantity – from
the balanced equation.
4. Convert from moles of the new quantity to the desired units, using formula
weight, density, Avogadro’s number, and so on.
PROBLEM EXAMPLE 3-1
Calculate the number of moles of NaOH that are necessary to produce moles
of Na2SO4 from the reaction
2 NaOH + H2SO4
Na2SO4 + 2 H2O
SOLUTION
Since the given quantity is in moles and the desired quantity is also in moles,
there is just one step involved – the conversion from moles of NaOH to moles of
Na2SO4, using the relationships from the balanced equation.
7.5 mol Na2SO4 x
2 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
1 𝑚𝑜𝑙 Na2SO4
= 15 mol NaOH
PROBLEM EXAMPLE 3-2
Calculate the number of grams of magnesium chloride that could be obtained
from 8.50 g of hydrochloric acid when the latter is reacted with an excess of
magnesium oxide (MgO).
SOLUTION
The problem involves all four steps: writing the balanced equation, converting
from grams of HCl (hydrochloric acid) to moles, converting from moles of HCl to moles
of MgCl2 (magnesium chloride) using the balanced equation, and converting from
moles of MgCl2 to grams.
The balanced equation is:
MgO + 2 HCl
H2O + MgCl2
The calculation is done as follows:
8.5 g HCl x
1 𝑚𝑜𝑙 𝐻𝐶𝑙
36.5 𝑔 𝐻𝐶𝑙
x
1 𝑚𝑜𝑙MgCl2
2 𝑚𝑜𝑙 𝐻𝐶𝑙
x
95.3 𝑔 MgCl2
1 𝑚𝑜𝑙 MgCl2
= 11.1 g MgCl2
QUIZ #3-1
For the reaction
4 FeS + 7 O2
2 Fe2O3 + 4 SO2
a) Calculate the number of moles of Fe2O3 that could be produced from 7.20
moles of FeS.
b) Calculate the number of grams of SO2 that could be produced from 3.25
moles of oxygen.
c) Calculate the number of grams of oxygen that would react with 0.125 g FeS.
d) How many moles of SO2 would be produced if 18.5 g Fe2O3 are produced in a
reaction.
Limiting Reagent
In general, when a chemical reaction is carried out, one of the reagents will be
used in excess of the amount needed. The reagent that is not present in excess is the
one that will determine how much product can be obtained as is thus referred to as the
limiting reagent. For example, in the reaction of hydrogen gas and oxygen gas to
form water: 2H2 + O2
2 H2O , 2 moles of hydrogen will combine with 1 mole
of oxygen to form 2 moles of water. If there is only 1 mole of oxygen present, and an
excess hydrogen, only 2 moles of water can be obtained. No matter how much
hydrogen is present, the oxygen will limit the amount of water that can be produced.
Thus, the oxygen is the limiting reagent.
The following steps can be used in doing calculations of this type:
1. Calculate the number of moles of product that could be obtained for each
reagent given.
2. The reagent that gives the least number of moles of product is the limiting
reagent and is the one that will determine the theoretical yield in the reaction;
that is, no matter how much of the excess reagent is present, no more product
can be obtained than that calculated from the limiting reagent.
3. Next, the moles of theoretical yield are converted to any other desired units,
such as grams.
4. To find the amount of excess reagent present, if desired, we first calculate the
amount of the excess reagent that will be used to produce the theoretical yield.
The difference between this amount and the amount present to start is the
amount of excess.
The amount of excess reagent that is used can be
calculated either from the theoretical amount of product attainable, as
calculated in step 2, or from the amount of limiting reagent present. In either
case, the mole relationships from the balanced equation are used to do the
calculation.
PROBLEM EXAMPLE 3-3
A 50.0 g sample of calcium carbonate is reacted with 35.0 g of phosphoric acid.
Calculate:
a) The number of grams of calcium phosphate that could be produced
b) The number of grams of excess reagent that will remain.
SOLUTION
The balanced equation is
3 CaCO3 + 2 H3PO4
Ca3(PO4)2 + 3 CO2 + 3 H2O
a) To calculate the theoretical yield, we first calculate the number of moles of
Ca3(PO4)2 that could be produced by each reagent, if it were used up,
50.0 g CaCO3 x
35.0 g H3PO4 x
1 𝑚𝑜𝑙 CaCO3
100 𝑔 CaCO3
1 𝑚𝑜𝑙 H3PO4
98.0 𝑔 H3PO4
x
x
1 𝑚𝑜𝑙 Ca3(PO4)2
3 𝑚𝑜𝑙 CaCO3
1 𝑚𝑜𝑙 Ca3(PO4)2
2 𝑚𝑜𝑙 H3PO4
= 0.167 mol Ca3(PO4)2
= 0.179 mol Ca3(PO4)2
This tells us that CaCO3 is the limiting reagent, and the maximum amount of
product that we can obtain is 0.167 mole of Ca3(PO4)2. To find the number of
grams that can be produced, we convert 0.167 mole of Ca3(PO4)2 to grams:
0.167 mol Ca3(PO4)2 x
310 𝑔𝑟𝑎𝑚𝑠Ca3(PO4)2
1 𝑚𝑜𝑙 Ca3(PO4)2
= 51.8 g Ca3(PO4)2
This is the theoretical yield.
b) To find the grams of excess reagent, we must first find how much of the reagent
that is in excess (H3PO4) is used. Then we subtract this from the amount of
H3PO4 present in the beginning, to find how much excess remains. We can
either start with the theoretical yield [0.167 mol Ca3(PO4)2] and calculate how
much H3PO4 it would take to produce this or start with the limiting reagent (50.0
g CaCO3) and find how much H3PO4 will react with that. Either way, we find
the same result. Let’s start with the 0.167 mol of Ca3(PO4)2.
0.167 mol Ca3(PO4)2 x
2 𝑚𝑜𝑙 H3PO4
1 𝑚𝑜𝑙Ca3(PO4)2
x
98 𝑔 H3PO4
1 𝑚𝑜𝑙 H3PO4
= 32.7 g H3PO4
This is the amount of H3PO4 used; therefore, the amount remaining is
35.0 g – 32.7 g = 2.30 g in excess.
QUIZ #3-2
From 12.5 g of CaCO3 and17.3 g of H3PO4, calculate the number of grams of Ca3(PO4)2
that could be produced from the following equation
3 CaCO3 + 2 H3PO4
Ca3(PO4)2 + 3 CO2 + 3 H2O
Percent Yield
All the calculations that have been done up to this point are theoretical yields.
This is the maximum amount of product that can be obtained in a particular reaction.
In general, however, when these reactions are carried out in the laboratory, we no not
actually obtain the theoretical amount of product. This is especially true in organic
reactions, where there are often side reactions taking place and where some of the
products are lost in the process of isolation and purifications. The amount of products
that is actually obtained is called the actual yield. The percent yield is the percent of
the theoretical yield that is actually obtained, calculated as follows:
percent yield =
𝑎𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
x 100
PROBLEM EXAMPLE 3-4
If 35.4 g Ca3(PO4)2 is actually obtained in problem example 3-3, what is the
percent yield?
SOLUTION
Actual yield = 35.4 g and theoretical yield = 51.8 g (from problem example 3-3);
therefore
Percent yield =
35.4 𝑔
51.8 𝑔
x 100 = 68.3%
PROBLEM EXAMPLE 3-4
If 15.0 g of MgO is treated with 18.5 g of H3PO4 and 17.6 g of Mg3(PO4)2 is
obtained, calculated the percent yield and the grams of excess reagent.
SOLUTION
The balanced equation is
3 MgO + 2 H3PO4
Mg3(PO4)2 + 3 H2O
a) Find the limiting reagent:
15.0 g MgO x
1 𝑚𝑜𝑙 𝑀𝑔𝑂
40.3 𝑔 𝑀𝑔𝑂
x
1 𝑚𝑜𝑙 Mg3(PO4)2
3 𝑚𝑜𝑙 𝑀𝑔𝑂
= 0.124 mol Mg3(PO4)2
18.5 g H3PO4 x
1 𝑚𝑜𝑙 H3PO4
98.0 𝑔 H3PO4
x
1 𝑚𝑜𝑙 Mg3(PO4)2
2 𝑚𝑜𝑙 H3PO4
= 0.0944 mol Mg3(PO4)2
Therefore, H3PO4 is the limiting reagent
b) Calculate the theoretical yield, using the limiting reagent
0.0944 mol Mg3(PO4)2 x
262.9 Mg3(PO4)2
1 𝑚𝑜𝑙 Mg3(PO4)2
= 24.8 g Mg3(PO4)2
c) Calculate the amount of excess reagent
using 0.0944 mole Mg3(PO4)2 gives
0.0944 mole Mg3(PO4)2 x
3 𝑚𝑜𝑙 𝑀𝑔𝑂
1 𝑚𝑜𝑙𝑒 Mg3(PO4)2
x
40.3 𝑔 𝑀𝑔𝑂
1 𝑚𝑜𝑙𝑒 𝑀𝑔𝑂
= 11.4 g
MgO
d) Calculate the percent yield
To do this we use the theoretical yield calculated in step (b) and the actual yield
given in the problem
Percent yield =
𝑎𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
=
17.6 𝑔
24.8 𝑔
x 100 = 71%
QUIZ #3-3
For the reaction
CaO + 2 HCl
H2O + CaCl2
1) If 1.23 g of CaO is reacted with excess hydrochloric acid (HCl) and 1.85 g of
CaCl2 is formed, what is the percent yield?
2) If 30.2 g of CaO is added to 34.5 g of HCl and 6.35 g of water is formed, what
is the percent yield?
Chapter 4
Electrochemistry
Introduction
Not only is matter, influenced by electricity but also many of the important
particles of matter are electrical in nature. All atomic nuclei are positively charged, and
all ions are charged either positively or negatively.
Learning Outcomes
After mastering this chapter, you should be able to:
✓ Convert electrical energy to chemical energy in electrolysis cells
✓ Convert chemical energy into electrical energy in galvanic or voltaic
cells
✓ Write anode, cathode and overall reaction equations that take place in
an electrolysis cell.
✓ Calculate equivalent weights and quantitative relationships in
electrolysis cells, in terms of coulombs, amperes, and faradays.
✓ Calculate the effect of concentration has on oxidation or reduction
potentials using the Nernst equation
LESSON 1
ELECTROLYTIC CELLS
In an electrolytic cell, electrical energy is
used to cause a chemical reaction to take place. An
example of an electrolytic cell is shown at the right.
The substance that conducts the current is molten
sodium chloride. The two electrodes (called the
anode and cathode) are placed into the molten
sodium chloride and an electric current is passed
through it. This electric current consists of a flow of
electrons. Thus, electrons are being force into the
cell. The cation (Na +) moves toward the electrode Figure 2- A simple electrolytic cell
called the cathode. At this electrode, Na + ions accept electrons to form sodium metal
according to the following reaction.
Na + + e -
Na
Therefore, reduction (gain of electrons or decrease in oxidation number takes
place at the cathode.
At the other, electrode, called the anode, the anion (Cl -) loses electrons to
form chlorine gas (Cl2) according to the equation
Cl2 + 2 e –
2 Cl -
Therefore, oxidation (loss of electrons or increase in oxidation number) takes
place at the anode.
In the cell, electrons move from the anode through the external circuit to the
cathode. The anions move toward the anode, where oxidation takes place. The
cations move toward the cathode where reduction takes place. It must be kept in mind
that these reactions only take place because of the electrical energy being put into the
system. Take note that both oxidation and reduction must take place in the cell. The
electrons lost by one substance must be gained by another.
To write the overall reaction, the cathode reaction must be doubled (or the
1
2
anode reaction could be written as Cl -
Cl2 + e – ), and then the two half-
reactions are added together.
Cathode:
2 Na + + 2 e 2 Cl -
Anode:
Overall:
2 Na + + 2 Cl -
Na
(reduction)
Cl2 + 2 e -
(oxidation)
2 Na + Cl2
Besides simple cells like the electrolysis of a cation to a free metal at the
cathode and an anion to a nonmetal at the anode, there are other types of reactions
that can take place at each electrode. Of course, in each case any reaction at the
cathode is reduction and at the anode is oxidation. Consider the following examples.
Anode Reactions (Oxidation)
1. Oxidation of an anion to an element, such as
2 Cl -
Cl2 + 2 e –
2. Oxidation of an anion or cation in solution to an ion of higher oxidation state,
such as
Fe 2+
Fe 3+ + e –
3. Oxidation of a metal anode, such as
Cu
Cu 2+ + 2e –
4. Oxidation of H2O to produce oxygen:
2 H2O
O2 + 4 H + + 4 e –
Cathode Reactions (Reduction)
1. Reduction of a cation to a metal, such as
Zn 2+ + 2e –
Zn
2. Reduction of an anion or cation in solution to an ion of lower oxidation state,
such as
Fe 3+ + e –
Fe 2+
3. Reduction of a non-metal to an anion, such as
Cl2 + 2e –
2 Cl –
4. Reduction of H2O to produce hydrogen gas:
2 H2O + 2e –
H2 + 2 OH –
In a particular electrolysis cell there are often several possible reactions that
could take place. For example, in any aqueous solution there is always the possibility
of the water being oxidized or reduced more readily than the cation or anion present.
The electrolysis of aqueous sodium chloride is one case where this takes place. It is
found that hydrogen gas is produced at the cathode and chlorine gas at the anode.
Thus, the overall reaction for this is
2 Na+ + 2 Cl- + 2H2O
H2 + Cl2 + 2 OH- + 2 Na+
This is one method that can be used for the commercial preparations for hydrogen and
chlorine, with an aqueous solution of sodium hydroxide as a third product.
As another example, it is found that, upon electrolysis of aqueous sodium
sulfate, hydrogen gas is produced at the cathode and oxygen gas at the anode.
Therefore, the overall reaction here is simply
2 H2O
2H2 + O2
Electrolysis of water by this method can be accomplished with any solution as long as
the ions in the solution (which must be present in order for electrolysis to take place)
are not oxidized or reduce more readily than water.
Another application of electrolysis cells is in the final purification to obtain pure
copper. In this purification the impure copper acts as the anode and a thin plate of
pure copper acts as the cathode, with the electrolyte being copper (II) sulfate. The
reactions that take place in this electrolysis cell are
Anode:
Cu
(from the impure
sample)
Cu2+ + 2 e –
(formed
at the anode)
Cathode:
Cu2+ + 2 e –
Cu
(pure)
The copper that forms on the cathode is better than 99.9 % pure. Other metals found
in copper ores, such as silver, gold, and platinum, are deposited as a sludge at the
bottom of the cell and are further refined, and more active metals such as nickel remain
in solution and are later recovered, making the entire operation quite economical. This
process does require considerable amounts of electricity, however, and thus cheap
electrical power must be available near the electrolytic cell plant site.
LESSON 2
QUANTITATIVE RELATIONSHIPS IN ELECTROLYTIC CELLS
In electrolysis cells, one very important consideration is the amount of electrical
energy necessary for a given amount of material to react. One of the most commonly
used units in expressing amount of electrical energy is the coulomb (coul). Another
very important unit is the faraday (F), which is the amount of energy required for the
flow of 1 mole of electrons. To three significant digits, 1 faraday equals 96,500
coulombs. The unit commonly used to measure current flow is the ampere (A), which
is a rate such that 1 coulomb passes a given point in the circuit in 1 second. Therefore,
A = coul / s, or coulombs = (amperes)(seconds). Keep in mind that a coulomb is an
amount of electrical energy and an ampere is a rate at which it flows.
The equivalent weight, or gram-equivalent weight, of a substance is the mass
in grams that is equivalent to I mole of electrons; that is, it is the amount that reacts
with or is produced by 1 mole of electron. The faraday is therefore the amount of
electrical energy necessary to produce 1 equivalent of pure substance. To determine
the equivalent weight of a substance it is necessary to know how many moles of
electrons are transferred per mole of substance. Therefore, to calculate the equivalent
weight of a substance the formula weight is divided by the number of electrons
transferred.
PROBLEM EXAMPLE 4-1
Calculate the equivalent weight of each of the following:
a) Aluminum
Fe2+ + 2 e –
b) Fe, in the reaction Fe
c) Fe, in the reaction Fe3+ + 3 e –
d) Fe, in the reaction Fe3+ + e –
Fe
Fe2+
SOLUTION
a) Since Al will always lose three electrons to form Al3+, the equivalent weight is
27.0 / 3 = 9.00 g
b) In this case, the equivalent weight of Fe is, 55.8 / 2 = 27.9 g
c) 55.8 / 3 = 18.6 g
d) 55.8 / 1 = 58.8 g
Since 1 faraday is the amount of electrical energy necessary for the transfer of 1
mole of electrons, it is the amount of electrical energy necessary to produce 1
equivalent of any substance. For example, referring to sample problem 4-1, it
would take 1 F or 96,500 coul to produce 9.00 g of Al, 27.9 g of FeCl2, and 18.6 g
of Fe from FeCl3. We can also use relationships directly between equivalents and
moles, as follows:
1 mole Al
= 3 equivalents Al
1 mole Fe
= 2 equivalents Fe (from FeCl2)
1 mole Fe
= 3 equivalents Fe (from FeCl3)
PROBLEM EXAMPLE 4-2
How many coulombs would it take to deposit 1.00 g of silver from solution?
SOLUTION
Since Ag is always +1 in compounds, the equivalent weight of Ag is always 108/1 =
108.
1.0 g Ag x
1 𝑒𝑞𝑢𝑖𝑣 𝐴𝑔
108 𝑔 𝐴𝑔
x
96,500 𝑐𝑜𝑢𝑙
1 𝑒𝑞𝑢𝑖𝑣 𝐴𝑔
= 894 coul
PROBLEM EXAMPLE 4-3
How many grams of Zn could be deposited by 96.5 coul?
SOLUTION
Equivalent weight of Zn = atomic weight / 2 = 65.4 / 2 = 32.7
96.5 coul x
1 𝑒𝑞𝑖𝑣 𝑍𝑛
96,500 𝑐𝑜𝑢𝑙
x
32.7 𝑔 𝑍𝑛
1 𝑒𝑞𝑢𝑖𝑣 𝑍𝑛
= 3.27 x 10-2 g Zn
PROBLEM EXAMPLE 4-4
How long would it take a current of 100 A to deposit 10.0 g of Fe from a
solution of FeCl2?
SOLUTION
100 A = 100 coul / s
Equivalent weight of Fe in FeCl2 = 55.8 / 2 = 27.9
10.0 g Fe x
1 𝑒𝑞𝑢𝑖𝑣 𝐹𝑒
27.9 𝑔 𝐹𝑒
x
96,500 𝑐𝑜𝑢𝑙
1 𝑒𝑞𝑢𝑖𝑣 𝐹𝑒
x
1𝑠
100 𝑐𝑜𝑢𝑙
= 346 s
PROBLEM EXAMPLE 4-5
How many grams of Al could be obtained by a current of 20 A flowing for 1
hr?
SOLUTION
1 hr x
60 𝑚𝑖𝑛
1 ℎ𝑟
x
60 𝑠
1 𝑚𝑖𝑛
= 3600 s
Since coul = A x s, the number of coulombs is
(20 A)(3600 s) = 72,000 A-s = 72,000 coul
72,000 coul x
1 𝑒𝑞𝑢𝑖𝑣 𝐴𝑙
96,000 𝑐𝑜𝑢𝑙
x
9.00 𝑔 𝐴𝑙
1 𝑒𝑞𝑢𝑖𝑣 𝐴𝑙
= 6.7 g Al
QUIZ #4-1
1. Calculate the equivalent weight of Sn for the reaction
Sn
Sn 2+ + 2 e –
2. Calculate the equivalent weight of I2 for the reaction
2I-
I2 + 2 e –
3. Calculate the number of grams of aluminum that could be deposited by
a) 5,000 coul
b) a current of 100 A flowing for 10 minutes
4. Calculate the time, in seconds, that it would take
a) for a current of 50.0 A to deposit 18.5 g of zinc
b) for a current of 18.5 A to form 42.0 g of iron from a solution of iron (III)
chloride (Fe3Cl)
5. Calculate the number of amperes necessary to
a) deposit 125 g of silver in 3.50 hours
b) deposit 84.5 g of nickel from a solution of NiCl3 in 45.0 minutes
6. Calculate the number of coulombs necessary to
a) Plate out 18.5 g of chromium from a solution of Cr2(SO4)3
b) Plate out 245 mg of tin from a solution of Sn(NO3)2
LESSON 3
VOLTAIC OR GALVANIC CELLS
In a voltaic or galvanic cell, a chemical reaction produces electrical energy. In
a cell of this type, the oxidation-reduction reaction is carried out in such a way that the
electrons being transferred from the reducing agent to the oxidizing agent travel
through a wire and thus produce an electric current.
One example of a voltaic cell involves zinc metal in a solution containing Cu2+
ions. The overall reaction that takes place is
Zn + Cu2+
Zn2+ + Cu
If the reaction is carried out in such a way that the zinc metal does not come in
contact with the Cu2+ ions, the energy given off may be obtained as electricity. This is
Figure 3- A simple voltaic cell using (a) a porous partition and (b) a salt bridge
done in a cell like that shown in the Figure 3. In this cell, the solution containing the
Cu2+ ions are separated from the solution in which the zinc metal is placed. The two
solutions are separated by a porous plate.as shown if Figure 3a. They can also be
separated by a salt bridge, which contains an ionic substance and allows ions to pass
from a solution in one container to a solution in another container, as shown in Figure
3b. When switch in the external circuit is closed, the reaction will begin to take place
since electrons can then pass through from the anode to the cathode. The two halfreaction that take place are
Anode:
Zn
Cathode:
Cu2+ + 2 e –
Zn2+ + 2 e –
Cu
(oxidation)
(reduction)
LESSON 4
STANDARD ELECTRODE POTENTIALS
The electromotive series of metals, or sometimes the activity series, tells us
that any metal will displace a lower metal from a solution of its ions. The basis of the
series is the standard electrode potentials (Eo) as shown in Table 4-1. These give the
potential, in volts, for a particular half-reaction to take place. The potentials given in
Table 4-1 are written as reduction potentials, where each half-reaction is written as
a reduction reaction. If these were written as oxidation potentials, the reaction would
be the reverse of that given, and the sign would be opposite. For example, for the
oxidation half-reaction, Li
Li+ + e –, the potential (Eo) is + 3.04 volts (V).
Table 4-1. Standard reduction potentials for some common metals - for aqueous solutions
of ions at 1.0 M concentration at 25oC and H2 gas at 1.0 atm pressure and 25oC
It is not possible to measure the potential for any half-reaction by itself, it can
only be measured as compared to something else. Thus, each of these is compared
to the 2 H+ + 2 e –
H2 half reaction; that is, this half-reaction is considered
to have a standard potential of 0.00 (at 25oC and 1 atm pressure). Then all the other
half-reactions are compared to this. The cell used to carry this out is shown in Figure
4.Figure 4 - Cell used to measure standard potentials
The metal in a solution containing its ions is used as one electrode and the
hydrogen gas being passed over platinum metal is the other electrode. Either the
reaction of metal ion going to metal ion (M n+ + n e –
metal ion (M
Ms
n+
For example, if the Zn – Zn
2+
M) or metal going to
–
+ n e ) will take place, and the voltage is measured.
half-cell is used for the electrode, the Zn goes to Zn
and the potential is 0.76 V; therefore, the potential for the reverse reaction (Zn
e–
2+
2+
+ 2
Zn) is -0.76 V, (as compared to the hydrogen electrode) as given in the
Table. If Cu and Cu
2+
ion are used as the other electrode, along with the hydrogen
electrode, the reduction half-reaction ( Cu
2+
+ 2e
–
Cu) takes place and
produces a potential of 0.34 V (as compared to the hydrogen electrode). In all these
cells the concentration of the ions is 1.0 M, which is taken as the standard. Thus, the
potentials given here are called standard electrode potentials or, for the particular
values given in the table, standard reduction potentials.
To determine if an oxidation-reduction reaction will take place spontaneously,
the two half-reactions are written and the potentials are added. If the potential is
positive the reaction will go spontaneously; if it is negative, it will not. Consider the
following examples (all concentrations 1.00 M):
1. Mg 2+
+
Zn 2+ + Mg
Zn
Mg 2+ + 2 e –
Zn
Mg 2+ + Zn
Mg
Eo = -2.36 V
Zn 2+ + 2 e –
Eo = +0.76 V
Mg + Zn 2+
Eo = -1.60 V
Therefore, this reaction will not go spontaneously as written (the reverse reaction
would since the potential would be + 1.60 V)
2. 3 Mg +
2 Al 3+
3 Mg 2+ + 2Al
3 (Mg
2(Al
3+
+
3e
–
3 Mg + 2 Al 3+
Mg 2+ + 2 e –)
Eo = +2.36 V
Al)
Eo = -1.66 V
3 Mg 2+ + 2 Al
Eo = +0.70
This reaction will go spontaneously as written. (The potential is not multiplied by
3 or 2 in the half-reactions)
Calculations of this type not only tell what reactions will go spontaneously but
also tell us something about the reverse reaction; that is, in the first example the given
reaction will not go spontaneously but the reverse reaction would.
QUIZ #4-2
1. Calculate the voltage that would be developed by cells using the following
reactions (all concentrations 1M). Use Table 4.1
a) Zn + Ni 2+
Zn 2+ + Ni
b) Pb + Zn 2+
Pb 2+ + Zn
2. From the standard reduction potentials for metals (Table 4.1), determine
whether the following reactions will occur
a) Zn + Ni 2+
Zn 2+ + Ni
b) Pb + Zn 2+
Pb 2+ + Zn
LESSON 5
THE NERNST EQUATION
All the potential used in the previous lesson involved solutions in which the
concentration of the ions was 1 M. This is known as the standard state. As the
concentration changes, the potential for the half-reaction also changes according to
the Nernst equation, which states that
E = Eo -
where
0.059
𝑛
log
[𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠] 𝑥
[𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠] 𝑦
Eo = the standard oxidation or reduction potential
N = the number of electrons transferred
x and y = the coefficients in the balance equation
0.059 = a constant at 25 oC
In using this equation it is very important always to write the log factor as the
concentration of products over reactants, just as in equilibrium constants, for the
reaction as written; that is, if Eo is an oxidation potential (from the reduced to oxidized
state, often written as Eored, ox) it is [oxidized state]/[reduced state]. And if Eo is a
reduction potential (Eored, ox) it is [reduced state/oxidized state].
For anything in the free state, such as a free metal (Mo), the concentration is
taken as one.
It is actually the activity that is one but most purposed molar
concentration of one is practically the same.
PROBLEM EXAMPLE 4-6
Calculate the half-cell potential for the reaction
Cu 2+ + 2 e-
Cuo
When the [Cu 2+] is 1 x 10 -4
SOLUTION
Since the [Cu 2+] is less than one, we would expect the potential to decrease (whether
it is a positive or a negative value, it should become smaller). The value of Eo for this
half-reaction is +0.34 V (from Table 4-1). The Nernst equation is
o
E=E -
0.059
log
2
[𝐶𝑢𝑜 ]
[𝐶𝑢2+ ]
Putting the known values into this (recall that [Cuo] = 1) gives
E = 0.34 -
0.059
2
E = 0.34 - (
log
0.059
2
E = 0.22 V
1
1 𝑥 10−4
) (4)
This value is less than 0.34, as predicted
PROBLEM EXAMPLE 4-7
For the reaction 2 Al + 3 Cu 2+
2 Al 3+ + 3 Cu, calculate the potential
if the [Cu 2+] is 1 x 10 -4 and the [Al 3+] is 1 x 10 -3
SOLUTION
This can be solved either as two separate steps or as the overall reaction.
a) EAl, Al3+ = 1.66 –
ECu2+,Cu = 0.34 -
0.059
3
0.059
2
log
log
1
1 𝑥 10−3
1
1 𝑥 10−4
Eoverall = 1.72 V + 0.22 V = 1.94 V
= 1.66 + 0.059 = 1.72 V
= 0.34 – 0.118 = 0.22 V
In this case, it is not necessary to multiply each of the half-reactions by 2 and
3 since we are using the individual cases and simply adding the two new potentials. It
should be obvious, from your knowledge of logarithms, that even if the Al, Al3+ halfreactions were multiplied through by 2, for example, the answer would have come out
the same since the 2 would then show up both in the n term and as a second power in
the log term, as
1.66 -
(0.059)
(3)(2)
log (1 x 10-3)2 = 1.66 -
0.059
(3)(2)
(-3)(2)
b) Using the overall reaction, from the sum of the two half-reactions, it can be
seen that there are six electrons involved:
2 Al3+ + 6 e -
2 Al
3 Cu 2+ + 6 e -
3 Cu
___________________________________
2 Al + 3 Cu 2+
2 Al3+ + 3 Cu
Eooverall = 1.66 + 0.34 = 2.00 V (from Table 4-1)
Since the overall reaction involves six electrons
Eoverall = 2.00 -
= 2.0 -
= 2.0 -
0.059
6
0.059
6
0.059
6
log
log
[𝐴𝑙3+ ] 2
[𝐶𝑢2+ ] 3
10−6
= 2.00 10−12
0.059
6
log (1 x 106)
(6) = 2.00 - 0.059 = 1.94 V
QUIZ #4-3
1. Calculate the potential of each of the following half-reaction with the given
concentration (use Table 4.1)
a) Cu 2+ + 2 e -
Cu
[Cu 2+] = 1.0 x 10
-3
Cu 2+ +
b) Cu
2e–
[Cu 2+] = 1.0 x 10 -4
2. Calculate the potential for each of the following reactions, at the given
concentrations (use Table 4.1)
a) Cu 2+ + Zn
Zn 2+ + Cu
[Cu 2+] = 1.0 x 10 -3
[Zn 2+] = 1.0 x 10 -4
b) 2 Na + Fe 2+
[Na +] = 1.0 x 10 -2
[Fe 2+] = 1.0 x 10 -3
2 Na + + Fe
THE PERIODIC TABLE