Download 23. Oxidation and Reduction

NAME________________________________ PER ________ DATE DUE ___________________
ACTIVE LEARNING I N C HEMISTRY E DUCATION
CHAPTER 23
OXIDATION
AND
REDUCTION
(Part 1)
23-1
©1997, A.J. Girondi
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© 1997 A.J. Girondi, Ph.D.
505 Latshmere Drive
Harrisburg, PA 17109
[email protected]
Website: www.geocities.com/Athens/Oracle/2041
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©1997, A.J. Girondi
SECTION 23.1 Introduction To Oxidation–Reduction Reactions
In earlier chapters, we studied chemical reactions labeling them as combination, decomposition,
single replacement, and double replacement. In this chapter, we will study yet another group called
oxidation-reduction reactions. We can greatly simplify the classification of reactions by grouping them all
into two broad classes. These two classes are: (1) reactions in which there is no electron transfer from one
substance to another, and (2) reactions in which electrons transfer from one substance to another. All
chemical reactions fit into one or the other of these two categories.
In previous chapters we made use of oxidation numbers, but we did not define them. Now it is
time to do so.
An oxidation number is a signed number which is assigned to an atom or ion
according to a set of rules. It represents the "oxidation state" of the atom or ion.
The "oxidation state" of an atom or ion changes when it loses or gains electrons. You have previously
used oxidation numbers to write the correct formulas for compounds and polyatomic ions. You may recall
that there is a table of common oxidation numbers in the reference section of your ALICE materials. The
rules used to assign oxidation numbers to atoms and ions are listed below. You should know them well
enough to use them from memory during tests and quizzes.
Rules for Assigning Oxidation Numbers
1. The oxidation number of an atom of a free element is zero. Elements are free if they are not combined
with other elements. If atoms of an element are combined with themselves, they are still considered to be
free. For example, a free atom of Ag has an oxidation number of zero. In addition, atoms in molecules like
H2, Cl2, N2, O2, F2, Br2, I 2, P4, S8, etc., have oxidation numbers of zero.
2. The oxidation number of a monatomic ion is equal to its charge. A monatomic ion is one that formed
from only one atom. Ex: Ag1+
3. The algebraic sum of the oxidation numbers of the atoms in the formula of a compound is zero.
4. In compounds, the oxidation number of hydrogen is +1. (There is one exception. In compounds known
as hydrides, it can be -1. Sodium hydride is NaH.)
5. In compounds, the oxidation number of oxygen is -2. (Exceptions: oxygen is -1 in peroxide
compounds like H 2O2, and +2 when it combines with fluorine in OF2.)
6. In combinations of nonmetal atoms, the oxidation number of the less electronegative element is
positive and of the more electronegative element is negative. For example, in NO2, N = +4, and O = -2.
The element with the positive oxidation number is written first in the formula of a compound, such as in
NO2. An exception is ammonia, NH3, in which nitrogen (-3) is written before hydrogen (+1).
7. The algebraic sum of the oxidation numbers of the atoms in the formula of a polyatomic ion is equal to
the charge on the polyatomic ion. Example: In Cr2O72-, each chromium atom is +6, while each oxygen
atom is -2. Note that (+6 X 2) + (-2 X 7) = -2.
Because of your work in previous chapters, you are already somewhat familiar with rules 3 through 7.
Now, rules 1 and 2 will also be needed in order to understand redox reactions.
Reactions in which no electrons are transferred usually involve the separation and rejoining of
atoms or ions. For example, silver nitrate (AgNO3) and sodium chloride (NaCl) exist as ions in solution.
AgNO 3(s) ----> Ag1+(aq) + NO31-(aq)
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©1997, A.J. Girondi
NaCl(s) ----> Na1+(aq) + Cl1-(aq)
When added together, the positive Ag 1+ ions combine with the negative Cl1- ions to form a precipitate of
AgCl. The Na 1+ and NO31- ions are spectator ions which will combine only if the water is removed from the
system.
Formula Equation:
Ionic Equation:
AgNO 3(aq) + NaCl(aq) ----> AgCl(s) + NaNO3(aq)
Ag1+(aq) + NO31-(aq) + Na1+(aq) + Cl1-(aq) ----> AgCl(s) + NO31-(aq) + Na1+(aq)
Net Ionic Equation:
Ag 1+(aq) + Cl1-(aq) ----> AgCl(s)
Ag has an oxidation number of {1}______ when combined with NO31- and with Cl1-. The oxidation number
remains the same before and after the reaction. When the oxidation number does not change, there has
been no transfer of electrons. You can see from the equations above that there has been neither an
increase nor a decrease in oxidation number. This is true for the silver ions and for the other ions involved
in the reaction.
In this chapter, however, we want to concentrate on the second type of reaction in which
electrons are transferred from one particle to another. Such oxidation-reduction equations are often
called "redox" equations for short. You have seen some redox reactions in previous chapters, although
you probably did not realize it. For example, single replacement reactions are also redox redox reactions.
We will now study some of the "whys" and "hows" of these reactions. Let's begin with some definitions.
Oxidation is defined as the LOSS of electrons by a substance.
Reduction is defined as the GAIN of electrons by a substance.
An example of a typical redox equation is:
2 Mg(s) + O2(g) ----> 2 MgO(s)
The reactants, Mg and O2, are neutral particles. The product, MgO, is made of equal numbers of Mg2+ and
O2- ions. These oppositely charged ions attract each other, and the result in an ionically bonded
compound. We can gain a better understanding of the reaction between Mg and O 2 by breaking it into two
separate parts, something that can be done to any redox equation. The separate parts are called halfequations or half-reactions. In the first half-equation below, magnesium is shown losing two electrons
(e- ). Remember that each electron has a charge of {2}_______. Magnesium is undergoing oxidation.
Oxygen, on the other hand, is shown gaining two electrons per atom (4 electrons total) and is, therefore,
undergoing reduction.
In oxidation half-equations, the electrons are always written on the product side (right side) to
show that they are being lost:
Oxidation half-equation:
Mg ----> Mg2+ + 2 e-
In the half-equation above, each magnesium atom is losing {3}________ electrons.
In reduction half-equations, the electrons are written on the reactant side to show that they are being
gained.
Reduction half-equation: O2 + 4 e- ----> 2 O2In the half-equation above, a total of {4}_______ electrons are being gained. Since two oxygen atoms are
involved, each oxygen atom is gaining {5}________ electrons.
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©1997, A.J. Girondi
In the overall equation, note that the oxidation number of the magnesium atom is changing from 0 to
{6}________. Also notice that the oxidation number of each of the two oxygen atoms is changing from 0
to {7}________.
0
0
+2
-2
2 Mg(s) + O2(g) ----> 2 MgO(s)
These changes in the oxidation numbers of magnesium and oxygen can be explained by the loss and
gain of electrons as shown in the half-equations.
You must learn to easily identify whether half-equations represent oxidations or reductions. One
way to remember the definitions is to think of electrons going O ut in O xidation, and electrons Returning
in Reduction.
Problem 1. Complete the exercises below by labeling each of the half-equations as oxidations or as
reductions.
Half-Equations
Oxidation or Reduction
a.
Cu1+ + e- ---> Cu
______________________
b.
F2 + 2e- ---> 2 F1-
______________________
c.
Cr3+ ---> Cr6+ + 3e-
______________________
d.
Bi3+ ---> Bi5+ + 2e-
______________________
e.
N5+ + 2e- ---> N3+
______________________
f.
2 Cl1- ---> Cl2 + 2e-
______________________
SECTION 23.2
Balancing Redox Half-equations
The total charge on each side of a redox equation must be equal. This is also true for halfequations. Notice that the sum of the oxidation numbers is zero on both sides of this equation taken
from section 23.1:
0
0
+2
-2
2 Mg(s) + O2(g) ----> 2 MgO(s)
The total charge on each side of a redox half-equation must also be equal. In each of the halfequations shown previously, this is also true. For example:
+3
+6
-3
Cr3+ ---->
Cr6+ + 3e-
You can see that the total charge on both sides of the half-reaction is {8}________.
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©1997, A.J. Girondi
This is an important rule about all redox equations:
In any redox equation or half-equation, the total charge of the reactants must
equal the total charge of the products.
This rule makes it possible for you to determine exactly how many electrons are gained or lost in any halfequation. If you were told that manganese changes from a charge of +7 to +4, you could write a halfequation for this change using the previous examples as your guide. For example, examine the halfequation below:
Mn7+ ----> Mn4+
The charges are not balanced (equal) on the reactant and product sides. By adding electrons (negative
charges) to the reactant (left) side, the charges can be balanced:
Mn7+ + 3 e- ----> Mn4+
Each side of the half-equation now has a total charge of {9}__________.
You have already learned that a chemical equation must be balanced according to mass, which
means that it must have the same number of atoms or ions on both sides. This is because of the Law of
Conservation of Mass which states that matter cannot be created or destroyed in ordinary chemical
reactions. Now you have also learned that charge cannot be created or destroyed, either. This is the Law
of Conservation of Charge.
A redox reaction or half-reaction must be balanced in two ways: first, according to
second, according to
{10}___________,
and,
{11}______________.
Problem 2. Complete the exercises below by adding the correct number of electrons needed to
balance the charges in each half-equation. Then, label each half-equation as an oxidation or reduction.
Review the first two rules for assigning oxidation numbers before you begin.
Half-Reactions
Oxidation or Reduction
a. Na ---> Na1+ + _____e-
______________________
b. Cl2 + _____e- ---> 2 Cl1-
______________________
c. 2 H1+ + _____e- ---> H2
______________________
d. 2 Al ---> 2 Al3+ + _____e-
______________________
e. 3 F2 + _____e- ---> 6 F1-
______________________
f. Fe2+ ---> Fe3+ + _____e-
______________________
g. Mn7+ + _____e- ---> Mn2+
______________________
h. 6 Cl1- ---> 3 Cl2 + _____e-
______________________
i. Cr6+ + _____e- ---> Cr3+
______________________
j. Cl1- ---> Cl5+ + _____e-
______________________
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©1997, A.J. Girondi
Problem 3. Write balanced half-equations for each situation presented below. You will need to decide
on which side to place the electrons. Make sure your half- equations are balanced according to mass and
charge. Then label each change as oxidation or reduction.
Half-Equation
Balanced Half-Equation
Ox or Red?
a. Cu2+ ---> Cu1+
__________________________
____________________
b. I 7+ ---> I1+
__________________________
____________________
c. 3 Br2 ---> 6 Br1-
__________________________
____________________
d. Sn2+ ---> Sn4+
__________________________
____________________
e. P5+ ---> P3+
__________________________
____________________
f. Cu2+ ---> Cu
__________________________
____________________
g. 2 Br2 ---> 4 Br1-
__________________________
____________________
h. 6 N3- ---> 3 N2
__________________________
____________________
SECTION 23.3
Adding Redox Half-Equations Together
Oxidation and reduction are processes that must occur together. This is true because it is
impossible to create or destroy electrons in any chemical change. Electrons can only be transferred from
one substance to another. The electrons that are lost in oxidation are gained in a reduction reaction. This
is why all oxidation-reduction reactions are the sum of two half-reactions. Let's consider the case in which
sodium reacts with fluorine:
Sodium atoms tend to lose {12}_______ electron(s) to become a stable ion:
Na ----> Na1+ + eFluorine atoms tend to gain one electron to become stable. Since fluorine is diatomic, F 2 would gain a
total of {13}________ electrons:
F2 + 2e- ----> 2 F1After the diatomic molecule gains electrons, is it still diatomic? {14}________ It has been changed into two
independent stable ions. Since F2 needs two electrons and sodium gives up only one electron per atom,
two sodium atoms will have to react with one F 2 molecule, so we multiply through the oxidation halfequation by two: 2 (Na ----> Na1+ + e- )
OR 2 Na ----> 2 Na1+ + 2 eand then we add the two half-equations together:
2 Na ----> 2 Na1+ + 2eF2 + 2e- ----> 2 F1-
<-- oxidation half-equation
<-- reduction half-equation
2 Na + F2 + 2e- ----> 2 Na1+ + 2 F1- + 2e-
<-- overall redox equation
Since there are two electrons on each side of the net equation, they can be dropped to yield the final
redox equation:
2 Na(s) + F2(g) ----> 2 NaF(s)
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©1997, A.J. Girondi
It's fairly simple to write redox equations from two half-equations. Let's try to write the redox
equation for the reaction between Cu atoms and N3+ ions. The two half-reactions are:
Cu ---> Cu2+ and 2 N3+ ---> N2
The 2 is needed in front of N 3+ in the second half-reaction because nitrogen is diatomic as a free element,
and the equation must be balanced by mass. While the two half-equations are balanced according to
mass, you can see that they are not balanced according to {15}_______________. To balance the
charge, we add electrons to the proper side of each half-reaction:
Cu ---> Cu2+ + 2e2 N3+ + 6e- ---> N2
<---- oxidation
<---- reduction
Now, since the number of electrons lost in the oxidation must equal the number of electrons gained in the
reduction, we multiply through the oxidation half-equation by {16}________:
3 Cu ----> 3 Cu2+ + 6 e(When trying to balance some redox equations, you may have to multiply through the reduction halfequation, or you may have to multiply through both half-equations.)
Then, we add the two half-reactions together:
3 Cu ----> 3 Cu2+ + 6e2 N3+ + 6e- ----> N2
<-- oxidation half-equation
<-- reduction half-equation
2 N3+ + 3 Cu + 6e- ----> N2 + 3 Cu2+ + 6e-
<-- overall redox reaction
Since there are {17}________ electrons on both sides of the equation, they should be eliminated to give
the overall redox equation:
2 N3+ + 3 Cu ----> N2 + 3 Cu2+
How can you tell that the equation above is balanced according to mass?
{18}____________________________________________________________________________
Notice that in the equation above, {19}________ electrons are lost in the oxidation and {20}________
electrons are gained in the reduction. Electrons must always be balanced in this way in a redox equation.
Using the same procedure as above, write the overall redox equation for each of the half-equation
combinations given below. Identify each half-equation as oxidation or reduction. (Some elements in
these equations may have oxidation numbers that do not appear on your list of common oxidation
numbers.)
Problem 4. Li ----> Li1+ and O2 ----> 2 O2-
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©1997, A.J. Girondi
Problem 5. Fe ----> Fe2+ and Hg2+ ----> Hg
Problem 6. Cr3+ ----> Cr2+ and Al ----> Al3+
Problem 7. F2 ----> 2 F1- and Zn ----> Zn2+
SECTION 23.4
Changes In Oxidation Numbers
So far, the equations we have examined have been pretty simple. All of them involve either single
elements or single ions. In such cases, it is pretty easy to determine which elements or ions are losing
electrons and which are gaining electrons. Many redox equations are more complicated than this. Some
involve polyatomic ions like PO43- or NO 31-. Before we can attempt to understand redox equations that
include such ions, we must know how to determine the oxidation number of each atom in a polyatomic ion.
This is not new to you. You worked with this concept back in Chapter 14. For example, what is the
oxidation number of the Mn atom in the ion MnO41-? The total of the oxidation numbers of the atoms in
this ion must must equal the charge on the ion, which in this case is -1. Rule 5 in section 23.1 tells us that
the oxidation number of oxygen is -2. Since there are four of them, the total charge from oxygen is -8. If
the total is to be -1, the oxidation number of the one manganese atom present must be +7.
Let's try another one. What is the oxidation number of each chromium atom in the Cr2O72- ion?
The seven oxygen atoms have a total charge of -14. The total charge must add up to -2, since that is the
charge on the whole ion. The total positive charge from the chromium atoms must, therefore, be +12.
Since there are two Cr atoms present, each one must have an oxidation number of {21}_________.
You may want to reread the rules in section 23.1 and review the work you did with oxidation
numbers in Chapter 14 before completing problem 8.
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©1997, A.J. Girondi
Problem 8. Oxidation Numbers of Elements in Compounds or Polyatomic Ions
Compound or Ion
Determine Ox. No. of:
Oxidation No. is:
a.
HAsO 2
As
__________
b.
HBr
Br
__________
c.
KI
I
__________
d.
MnO 2
Mn
__________
e.
MnO 41-
Mn
__________
f.
H3PO 4
P
__________
g.
HCO31-
C
__________
h.
CO32-
C
__________
i.
ClO41-
Cl
__________
j.
ClO21-
Cl
__________
k.
SeO 42-
Se
__________
l.
Cr2O72-
Cr
__________
We can now use oxidation numbers to determine which atoms are being oxidized (losing
electrons) and which are being reduced (gaining electrons). In addition to losing and gaining electrons,
another definition of oxidation and reduction can be given in terms of changes in oxidation number:
When an atom or ion undergoes oxidation, its oxidation number goes up (becomes more
positive or less negative). When an atom or ion undergoes reduction, its oxidation
number goes down (becomes more negative or less positive.)
Oxidation involves an increase in oxidation number, while
in oxidation number.
{22}__________________
involves a decrease
According to our other definition, the half-equation below would be labeled as an oxidation
because electrons are being lost. Note that the oxidation number of Fe increases from 0 to +3, so it also
fits our new definition of oxidation:
Fe ----> Fe3+ + 3e-
oxidation
A similar relationship occurs when reduction occurs. In the half-equation shown below, Br2 gains 2
electrons to become 2 Br 1-. Electrons are gained, so this is reduction. The oxidation number of bromine
goes from 0 to -1. A decrease in oxidation number also indicates that a reduction has occurred:
Br2 + 2 e- ----> 2 Br1-
reduction
Use these additional definitions of oxidation and reduction to determine whether iron, Fe, is being
oxidized or reduced in each of the following processes.
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©1997, A.J. Girondi
Problem 9. Determine the oxidation number of iron in each case. Write the oxidation number of iron
above each Fe shown, whether it is alone or combined with another element.
SECTION 23.5
Process
Ox or Red
a. Fe2O3 becomes Fe
__________________
b. FeO becomes Fe2O3
__________________
c. FeF2 becomes Fe
__________________
d. FeO becomes FeCl3
__________________
Oxidizing Agents And Reducing Agents
When discussing oxidation-reduction reactions, the terms oxidizing agent and reducing agent are
often used. The substance that undergoes a decrease in its oxidation number is called the oxidizing
agent. The substance whose oxidation number increases is called the reducing agent. The oxidizing
agent does the oxidizing, and, as a result, gets reduced. The reducing agent does the reducing, and, as
a result, gets oxidized. This can be a little confusing at first, so let's organize the data in another way.
Oxidizing Agent
Reducing Agent
1. its oxidation number decreases
2. it does the oxidizing
3. it gets reduced (gains e- )
1. its oxidation number increases
2. it does the reducing
3. it gets oxidized (loses e- )
It makes sense if you think about it. The oxidizing agent does the oxidizing, meaning that it causes
something else to lose electrons. In the process the oxidizing agent gains those electrons; thus, the
oxidizing agent gets reduced. On the other hand, the reducing agent has to give away electrons in order
to do the reducing. Therefore, the reducing agent ends up losing electrons and being oxidized.
Complete problem 10 by identifying the reducing agent and oxidizing agent in each equation. As you
complete the table, be sure to indicate whether you are answering with the atom of an element or the ion
(the first two are done for you). Be sure to include the charge when answering with ions.
Problem 10. Identifying Oxidizing and Reducing Agents
Reducing Agent
Oxidizing Agent
a. 2 Na + Ni2+ ----> 2 Na1+ + Ni
Na
Ni2+
b. Hg2+ + 2 Fe2+ ----> Hg + 2 Fe3+
Fe 2+
Hg2+
c. 2 Al + 3 Pb2+ ----> 2 Al3+ + 3 Pb
______________
_______________
d. Mg2+ + 2 Li ----> Mg + 2 Li1+
______________
_______________
e. O2 + 4 K ----> 2 K2O
______________
_______________
f. Fe + Ni2+ ----> Fe2+ + Ni
______________
_______________
______________
_______________
______________
_______________
______________
_______________
g. Cu + 2
h.
Ag1+ ----> Cu2+
+ 2 Ag
Cd + 2 H1+ ----> Cd2+ + H2
i. 3 Co2+ + 2 Al ----> 3 Co + 2 Al3+
23-11
©1997, A.J. Girondi
ACTIVITY 23.6
Observing Redox Reactions
So far you have worked with redox equations. Now it is time to actually carry out several redox
reactions in the laboratory. You will be able to see changes in substances being oxidized and reduced,
and you will be attempting to identify the substances being oxidized and reduced. Be sure to wear safety
glasses and an apron.
Procedure - Part A:
1. Place 20 mL of 0.2 M CuSO4 solution into a clean 125 mL Erlenmeyer flask. This solution contains
Cu2+ ions. We can ignore the SO42- ions, because they do not get involved in the reaction.
2. Put a small amount of powdered zinc metal into the Cu 2+ solution. Allow the reaction to proceed for at
least five minutes before recording any observations. Swirl the mixture in the flask every minute or so.
What happens to the zinc metal after 5 minutes? {23}________________________________________
______________________________________________________________________________
What happens to the solution's color? {24}________________________________________________
______________________________________________________________________________
3. Atoms of zinc metal, Zn, have been converted into colorless Zn2+ ions which go into solution. The blue
Cu2+ ions present in the CuSO4 solution have been converted into solid copper atoms, Cu, which appear
on the bottom of the flask. Write a redox equation that describes the reaction between Zn and Cu2+:
{25}____________________________________________________________________________
4. Complete the items below. Be sure to indicate whether your answer refers to an atom or an ion of an
element.
substance oxidized:{26}______________
oxidizing agent:
{27}______________
substance reduced: {28}______________
reducing agent: {29}______________
Procedure - Part B:
1. Obtain a solution of 3.0 M HCl. Place 20 mL of the HCl into a 125 mL Erlenmeyer flask. This solution
contains H 1+ ions. Place the beaker under the fume hood and add a small (2 cm) piece of magnesium
metal to the solution. Observe what happens. What evidence was there that a chemical reaction
occurred? {30}_____________________________________________________________________
2. Magnesium atoms, Mg(s) , react with the hydrogen ions, H1+, in the HCl solution to form hydrogen gas,
H2(g), and magnesium ions, Mg2+(aq), which go into solution. Write a redox equation for the reaction that
occurred. {31}_____________________________________________________________________
3. Complete the items below. Be sure to indicate whether your answer refers to an atom or an ion of an
element.
substance oxidized: {32}______________
oxidizing agent:
substance reduced: {34}______________
reducing agent: {35}______________
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{33}______________
©1997, A.J. Girondi
Procedure - Part C:
1. Add a small piece of copper metal, Cu, to 20 mL of 0.1 M HCl. Do you observe any reaction?_______
2. Based on your observation, is the following statement true or false {36}_____________?
"H1+ ions will oxidize Mg metal much more readily than they oxidize Cu metal."
From these results you see that oxidizing and reducing agents are not always strong enough for a redox
reaction to occur spontaneously. H1+ was able to oxidize Mg, but it was not strong enough to oxidize the
copper.
Cu + HCl ----> no reaction
OR
Cu + H1+ ----> no reaction
If this reaction had occurred, it would have fallen into the category of single replacement. Single
replacement reactions are also redox reactions. Remember when you used the activity series in chapter
6? (A copy of the activity series can also be found in your reference notebook.) If you check it, you will
find that copper is below hydrogen on the activity series, meaning no reaction will occur. You will also note
in the series that Mg is located above both copper and hydrogen, which explains why the reactions in
parts A and B of this activity did occur.
SECTION 23.7
Balancing Redox Equations
In Chapters 6 and 7 you learned how to balance chemical equations. The equations you worked
with were classified into four groups - combination, decomposition, single replacement, and double
replacement. You learned how to balance them by inspection. This means that you looked from side to
side in each equation changing the coefficients, in your attempt to make the number of atoms of each
element on the left side equal to the number of atoms of each element on the right side.
Many redox equations are difficult to balance by inspection. For example:
?? H2S + ?? HNO3 ----> ?? S + ?? NO + ?? H2O
Balancing such equations can be made easier by using oxidation numbers. This fact is the basis for a
method of balancing redox equations known as the electron-transfer method. It is a much more
systematic method of balancing equations than is balancing by inspection. The steps involved in the
electron-transfer method will be illustrated using the equation above. (For simplicity's sake, sulfur is
represented as S in this example, not S8)
Step 1: Assign oxidation numbers to each element in the equation. (Follow the rules in section 23.1)
+1
-2
+1
-2
0
\ /
\
/
H2S + HNO3 ---->
|
+2
-2
+1
-2
|
\ /
\ /
S + NO + H2O
+5
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©1997, A.J. Girondi
Step 2: Identify the elements that have been oxidized and reduced, and then write the half-equations
(balanced according to mass and charge).
Sulfur's oxidation number increases from -2 to 0, so it has been oxidized. Nitrogen's oxidation
number decreases from +5 to +2, so it has been reduced. Make sure both half-equations are balanced
according to mass and according to charge.
oxidation:
S 2- ----> S + 2e-
reduction:
N5+ + 3e- ----> N2+
Step 3: Conserve electrons by multiplying each half-equation by a coefficient that makes the number of
electrons lost in the oxidation equal to the number of electrons gained in the reduction.
3 (S2- ----> S + 2e- ) = 3 S2- ----> 3 S + 6e2 (N5+ + 3e- ----> N2+)
=
2 N5+ + 6e- ----> 2 N2+
Step 4: Add the two half-reactions. (The electrons should cancel out.)
oxidation:
3 S2- ----> 3 S + 6e-
reduction:
2 N5+ + 6e- ----> 2 N2+
overall:
3 S2- + 2 N5+ ----> 3 S + 2 N2+
Step 5: Place the coefficients in front the proper substances in the original equation.
3 H2S + 2 HNO3 ----> 3 S + 2 NO + ?? H2O
Step 6: Add any other coefficients or make any changes needed to balance the equation. (In our
example, you need a coefficient in front of the H2O. A 4 will balance it.)
3 H2S + 2 HNO3 ----> 3 S + 2 NO + 4 H2O
DONE!!
Here's another example. Let's follow the six steps to balance this equation:
I 2 + HNO3 ----> HIO3 + NO2 + H2O
Step 1:
0
|
I2
+1
\
+
-2
/
HNO3
+1
\
---->
-2
/
HIO3
|
|
+5
+5
23-14
+4
\
+
-2
/
NO2
+1
\
+
-2
/
H2O
©1997, A.J. Girondi
Step 2:
oxidation:
I 2 ----> 2 I5+ + 10e-
reduction:
N5+ + e- ----> N4+
Notice that if gases are diatomic in the equation to be balanced, they are also written in diatomic form in the
half-reactions. This is why the oxidation half-reaction above is I 2 ---> 2 I5+ + 10e- rather than
I ---> I5+ + 5e- . A coefficient 2 is needed in front of I5+ to balance mass, and then 10e- are needed to
balance charge.
Step 3:
1 (I 2 ----> 2 I5+ + 10e- ) = I2 ----> 2 I 5+ + 10e1 0 (N5+ + e- ----> N4+) = 10 N5+ + 10e- ----> 10 N4+
Step 4:
oxidation:
I 2 ----> 2 I 5+ + 10e-
reduction:
10 N5+ + 10e- ----> 10 N4+
overall:
I 2 + 10 N5+ ----> 2 I5+ + 10 N4+
Step 5:
I 2 + 10 HNO3 ----> 2 HIO3 + 10 NO2 + ?? H2O
Step 6:
I 2 + 10 HNO3 ----> 2 HIO3 + 10 NO2 + 4 H2O
DONE!!
Problem 11. Use the six steps of the electron-transfer method to balance the redox equations below.
Again, for simplicity, sulfur is represented as S instead of S8. (Some elements in these equations may
have oxidation numbers that do not appear on your list of common oxidation numbers.)
a. HNO3 + H2S -----> H2O + NO + S
b. S + HNO3 -----> SO2 + NO + H2O
c. HIO3 + NO2 + H2O ----> HNO3 + I 2
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©1997, A.J. Girondi
d. Al + H1+ + SO42- ----> Al3+ + SO2 + H2O
(In order to balance more complex equations, you may need to change one or more coefficients even
after you have completed the 6 steps. This will be necessary in equation e below.
e. HCl + KMnO4 ----> KCl + MnCl2 + Cl2 + H2O
The equations below are optional. You may balance them on a separate page if you feel that you need
more practice.
f. HNO3 + KI ----> KNO3 + I 2 + NO + H2O
g. Fe2+ + MnO41- + H1+ ----> Mn2+ + Fe3+ + H2O
h. Sn2+ + Ce4+ ----> Sn4+ + Ce3+
j. Sn + H1+ + NO31- ----> SnO2 + NO2 + H2O
j. NaI + H2SO 4 ----> H2S + I 2 + Na2SO 4 + H2O (This one can be fun!)
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©1997, A.J. Girondi
SECTION 23.8
Learning Outcomes
This is the end of Chapter 23. The subject of oxidation-reduction is continued in Chapter 24.
Review the learning outcomes below. Arrange to take any quizzes or exams on Chapter 23, and then
move on to Chapter 24.
_____1. Distinguish between oxidation and reduction.
_____2. Identify the oxidizing and reducing agents in a redox equation.
_____3. Assign oxidation numbers to all elements in any compound or polyatomic ion in a redox
equation using the rules for assigning oxidation numbers.
_____4. Determine the number of electrons transferred in redox half- equations and overall equations.
_____5. Write balanced net redox equations given two half-equations.
_____6. Balance redox equations by the electron-transfer method.
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SECTION 23.9 Answers to Questions and Problems
Questions:
{1} +1; {2} -1; {3} 2; {4} 4; {5} 2; {6} +2; {7} -2; {8} +3; {9} +4; {10} mass; {11} charge; {12} one; {13} 2;
{14} no; {15} charge; {16} three; {17} six;
{18} It has the same number of atoms of each element on both sides of the equation;
{19} six; {20} six; {21} +6; {22} reduction; {23} It gets oxidized (Zn(s) ---> Zn2+(aq) + 2e- ;
{24} It fades as the blue Cu2+ ions are reduced to Cu atoms; {25} Zn(s) + Cu2+(aq) ---> Zn2+(aq) + Cu(s)
{26} Zn; {27} Cu2+; {28} Cu2+; {29} Zn; {30} A gas is evolved (given off);
{31} Mg(s) + 2 H1+(aq) ---> Mg2+(aq) + H2(g); {32} Mg; {33} H1+; {34} H1+; {35} Mg; {36} true
Problems:
1. a. red; b. red; c. ox; d. ox; e. red; f. ox
2. a. Na ---> Na1+ + __1___eb. Cl2 + ___2__e- ---> 2 Cl1c. 2 H1+ + __2___e- ---> H2
d. 2 Al ---> 2 Al3+ + __6___ee. 3 F2 + __6___e- ---> 6 F1f. Fe2+ ---> Fe3+ + ___1__eg. Mn7+ + __5___e- ---> Mn2+
h. 6 Cl1- ---> 3 Cl2 + __6___ei. Cr6+ + __3___e- ---> Cr3+
j. Cl1- ---> Cl5+ + ___6__e3. a. Cu2+ ---> Cu1+
b. I 7+ ---> I1+
c. 3 Br2 ---> 6 Br1d. Sn2+ ---> Sn4+
e. P5+ ---> P3+
f. Cu2+ ---> Cu
g. 2 Br2 ---> 4 Br1h. 6 N3- ---> 3 N2
ox
red
red
ox
red
ox
red
ox
red
ox
Cu2+ + 1e- ---> Cu1+
I 7+ + 6e- ---> I1+
3 Br2 + 6e- ---> 6 Br1Sn 2+ ---> Sn4+ + 2eP 5+ + 2e- ---> P3+
Cu2+ + 2e- ---> Cu
2 Br2 + 4e- ---> 4 Br16 N3- ---> 3 N2 + 18 e-
red
red
red
ox
red
red
red
ox
4. ox:
red:
overall:
4(Li ---> Li1+ + e- )
O2 + 4e- ---> 2 O24 Li + O2 ---> 4 Li1+ + 2 O2-
5. ox:
Fe ---> Fe2+ + 2ered:
Hg2+ + 2e- ---> Hg
overall: Fe + Hg2+ ---> Fe2+ + Hg
6. ox:
red:
overall:
Al ---> Al3+ + 3e3(Cr3+ + e- ---> Cr2+)
3 Cr3+ + Al ---> 3 Cr2+ + Al3+
7. ox:
Zn ---> Zn2+ + 2ered:
F2 + 2e- ---> 2 F1overall: F2 + Zn ---> 2 F1- + Zn2+
8. a. 3+; b. 1-; c. 1-; d. 4+; e. 7+; f. 5+; g. 4+; h. 4+; i. 7+; j. 3+; k. 6+; l. 6+
9. a. red; b. ox; c. red; d. ox
10. c. Al, Pb2+; d. Li, Mg2+; e. K, O2; f. Fe, Ni2+; g. Cu, Ag1+; h. Cd, H1+; i. Al, Co2+
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©1997, A.J. Girondi
11. a.
ox:
red:
overall:
final:
3 (S2- ---> S + 2e- )
2 (3e- + N5+ ---> N2+)
3 S2- + 2 N5+ ---> 3 S + 2 N2+
2 HNO3 + 3 H2S ---> 4 H2O + 2 NO + 3 S
11. b.
ox:
red:
overall:
final:
3 (S ---> S4+ + 4e- )
4(3e - + N5+ ---> N2+)
3 S + 4 N5+ ---> 3 S4+ + 4 N2+
3 S + 4 HNO3 ---> 3 SO2 + 4 NO + 2 H2O
11. c.
ox:
red:
overall:
final:
10(N 4+
1(10e10 N4+
2 HIO3
11. d.
ox:
red:
overall:
final:
2(Al ---> Al3+ + 3e- )
3(2e - + S6+ ---> S4+)
2 Al + 3 S6+ ---> 2 Al3+ + 3 S4+
2 Al + 12 H1+ + 3 SO42- ---> 2 Al3+ + 3 SO2 + 6 H2O
11. e.
ox:
red:
overall:
final:
5(2 Cl1- ---> Cl2 + 2e-)
2(5e - + Mn7+ ---> Mn2+)
10 Cl1- + 2 Mn7+ ---> 5 Cl2 + 2 Mn2+
16 HCl + 2 KMnO4 ---> 2 KCl + 2 MnCl2 + 5 Cl2 + 8 H2O
---> N5+ + e- )
+ 2 I 5+ ---> I 2)
+ 2 I5+ ---> 10 N5+ + I2
+ 10 NO2 + 4 H2O ---> 10 HNO3 + I 2
11. "f" through "j" - answers not given to the optional ones. You figure them out!
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SECTION 23.10 Student Notes
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©1997, A.J. Girondi