Chem 112 Class Guide: BALANCING REDOX REACTIONS
Chapter 20, Sections 1 and 2 Learning Goals:
Upon completion of Chapter 19, Sections 1-4, you should be able to determine the following:
Determine the oxidizing agent and the reducing reagent in a redox reaction
Balance a redox reaction in acidic media
Balance a redox reaction in basic media
Chapter Reading Guide: Chapter 20, Sections 1 and 2
Section 1: Oxidation States and Oxidation-Reduction Reactions
NOTE: You may want to review Chapter 4, Section 4 (assigning oxidation numbers) before
continuing.
Read Chapter 20.1
The chemical substance that allows another substance to be oxidized is called the oxidizing
agent; one element in the oxidizing agent is generally reduced.
The chemical substance that allows another substance to be reduced is called the reducing
agent; one element in the reducing agent is generally oxidized.
Example:
Determine the oxidizing agent and the reducing agent in the following reaction:
PbS (s) + 4 H2O2 (aq) PbSO4 (s) + 4 H2O (l)
First, we need to assign oxidation numbers to all species in the reaction (Chapter 4!)
2 2
1
1
2 6 2
1 2
Pb S ( s )  4 H 2 O2 (aq)  Pb S O4 ( s )  4 H 2 O(l )
Now, we need to look at what changed:
S went from -2 as a reactant to +6 as a product; it was oxidized (LEO)
O went from -1 as a reactant to -2 as a product; it was reduced (GER)
PbS contains the element that was oxidized, it is the REDUCING AGENT
H2O2 contains the element that was reduced, it is the OXIDIZING AGENT
Try Practice exercise 20.1
Section 2: BALANCING CHEMICAL REACTIONS
Read Chapter 20.2
Redox equations cannot be balanced using conventional methods. We must take into
consideration that most of these equations are done in aqueous solution. This means that
water is present and may play a part in the reaction; if water is present that means that
protons (H+) and electrons are also present and must be accounted for in the balanced
reaction.
Follow this step-by-step process in order to balance a redox equation:
1) Divide the equation into one oxidation reaction and one reduction reaction
2) Balance each half-reaction
a. Balance elements other than H and O
b. Balance O by adding H2O as needed
c. Balance H by adding H+ ions as needed
d. Balance charge by adding e- as needed (the charge must be equal on both sides of
the arrow just as all the elements must equal)
3) Multiply the half reactions so that the number of e- are equal
4) Add the resulting half reactions together
Example:
Balance the following redox reaction in acidic media:
Cr2O72- (aq) + CH3OH (aq)  HCO2H (aq) + Cr3+ (aq)
1) Divide the equation into one oxidation reaction and one reduction reaction
To do this, we need to find out what was oxidized and what was reduced. So let’s assign
oxidation numbers:
6
2
2 1 2 1
1 2 2 1
3
Cr2 O72 (aq)  C H3 O H (aq)  H C O2 H (aq)  Cr 3 (aq)
So Cr went from +6 as a reactant to +3 as a product, it was reduced (Cr2O72- is the
oxidizing agent).
C went from -2 as a reactant to +2 as a product, it was oxidized (CH3OH is the reducing
agent).
The two half reactions will be:
Cr2O72- (aq)  Cr3+ (aq)
And:
CH3OH (aq)  HCO2H (aq)
2) Balance each half-reaction
Let’s start with the reduction reaction:
a) Balance elements other than O and H
Cr2O72- (aq) 2 Cr3+ (aq)
b) Balance O by adding H2O as needed
Cr2O72- (aq) 2 Cr3+ (aq) + 7 H2O (l)
c) Balance H by adding H+ ions as needed
14 H+ (aq) + Cr2O72- (aq) 2 Cr3+ (aq) + 7 H2O (l)
d) Balance charge by adding e- as needed
The reactants add up to a total charge of +12 (+14 from the H’s, -2 from the Cr2O7),
while the products equal +6 (all from the Cr3+). To balance that, we should add 6 eto the reactant side to get: +12 – 6 = +6
14 H+ (aq) + 6 e- + Cr2O72- (aq) 2 Cr3+ (aq) + 7 H2O (l)
Now, let’s do this for the oxidization equation:
a) Balance elements other than O and H
CH3OH (aq)  HCO2H (aq)
b) Balance O by adding H2O as needed
H2O (l) + CH3OH (aq)  HCO2H (aq)
c) Balance H by adding H+ ions as needed
H2O (l) + CH3OH (aq)  HCO2H (aq) + 4 H+ (aq)
d) Balance charge by adding e- as needed
The reactants add up to a total charge of 0, while the products equal +4 (all from the
H+). To balance that, we should add 4 e- to the product side to get: 0 = +4 - 4
H2O (l) + CH3OH (aq)  HCO2H (aq) + 4 H+ (aq) + 4 e3) Multiply the half reactions so that the number of e- are equal
We have 6 e- in the reduction reaction and 4 e- in the oxidation reaction. The lowest
common multiple between 6 and 4 is 12, so we need to multiply the reduction reaction
by 2 and the oxidation reaction by 3:
28 H+ (aq) + 12 e- + 2 Cr2O72- (aq) 4 Cr3+ (aq) + 14 H2O (l)
3 H2O (l) + 3 CH3OH (aq)  3 HCO2H (aq) + 12 H+ (aq) + 12 e4) Add the resulting half reactions together
2816 H  (aq)  12e   2Cr2O72 (aq)  4Cr 3 (aq)  14 11H 2O(l )
3H 2O(l )  3CH 3OH (aq)  3HCO2 H (aq)  12 H  (aq)  12e 
16 H  (aq)  2Cr2O72 (aq)  3CH 3OH (aq)  3HCO2 H (aq)  4Cr 3 (aq)  11H 2O(l )
Sometimes, our reaction doesn’t take place in acidic media, and instead it takes place in basic
media. If the solution is basic, the easiest way to balance the equation is to balance it like it is
acidic, but at the end of step 2), add the same number of OH- ions to each side of the equation
as you have H+ ions. Remember that OH- + H+  H2O. Let’s work through an example.
Example:
Balance the following redox reaction in basic media:
NO2- (aq) + Al (s)  NH4+ (aq) + AlO2- (aq)
1) Divide the equation into one oxidation reaction and one reduction reaction
To do this, we need to find out what was oxidized and what was reduced. So let’s assign
oxidation numbers:
3 2
0
3
1
1
2
N O2 (aq)  Al (s)  N H 4 (aq)  Al O2 (aq)
So N went from +3 as a reactant to -3 as a product, it was reduced (so NO2- is the
oxidizing agent).
Al went from 0 as a reactant to +1 as a product, it was oxidized (so Al is the reducing agent).
The two half reactions will be:
NO2- (aq)  NH4+ (aq)
And:
Al (s)  AlO2- (aq)
2) Balance each half-reaction
Let’s start with the reduction reaction:
a) Balance elements other than O and H
NO2- (aq)  NH4+ (aq)
b) Balance O by adding H2O as needed
NO2- (aq)  NH4+ (aq) + 2 H2O (l)
c) Balance H by adding H+ ions as needed
8 H+ (aq) + NO2- (aq)  NH4+ (aq) + 2 H2O (l)
d) Balance charge by adding e- as needed
The reactants add up to a total charge of +7 (8 from the hydrogen ion and -1 from
the nitrite), while the products equal +1 (all from the ammonium). To balance the
charge, we should add 6 e- to the reactant side of the equation:
8 H+ (aq) + 6 e- + NO2- (aq)  NH4+ (aq) + 2 H2O (l)
e) Add the same number of OH- ions to each side of the equation as you have H+ ions
(in this case, 8 OH- to each side):
8 H2O (l) + 6 e- + NO2- (aq)  NH4+ (aq) + 2 H2O (l) + 8 OH- (aq)
Now, let’s do this for the oxidation reaction:
a) Balance elements other than O and H
Al (s)  AlO2- (aq)
b) Balance O by adding H2O as needed
2 H2O (l) + Al (s)  AlO2- (aq)
c) Balance H by adding H+ ions as needed
2 H2O (l) + Al (s)  AlO2- (aq) + 4 H+ (aq)
d) Balance charge by adding e- as needed
The reactants add up to a total charge of 0, while the products equal +3 (-1 from the
aluminum compound and +4 from the hydrogen ion). To balance that, we should
add 3 e- to the product side to get: 0 = +3 -3
2 H2O (l) + Al (s)  AlO2- (aq) + 4 H+ (aq) + 3 ee) Add the same number of OH- ions to each side of the equation as you have H+ ions
(in this case 4 OH- to each side):
2 H2O (l) + Al (s) + 4 OH- (aq)  AlO2- (aq) + 4 H2O (l) + 3 e3) Multiply the half reactions so that the number of e- are equal
We have 6 e- in the reduction reaction and 3 e- in the oxidation reaction. The lowest
common multiple between 6 and 3 is 6, so the reduction reaction is fine the way it is
and we need to multiply the oxidation reaction by 2:
8 H2O (l) + 6 e- + NO2- (aq)  NH4+ (aq) + 2 H2O (l) + 8 OH- (aq)
4 H2O (l) + 2 Al (s) + 8 OH- (aq)  2 AlO2- (aq) + 8 H2O (l) + 6 e4) Add the resulting half reactions together
8 H 2O(l )  6e  NO2 (aq)  NH 4 (aq)  2 H 2O(l )  8OH  (aq)
4 2 H 2O(l )  2 Al ( s)  8OH  (aq)  2 AlO2 (aq)  8 H 2O(l )  6e
2 H 2O(l )  2 Al ( s)  NO2 (aq)  NH 4 (aq)  2 AlO2 ( aq)
Try Practice Exercises 20.2 and 20.3
Learning Resources
Chapter Learning Goals
Chapter 20, Sections 1 through 2 Learning Goals
Pre Class Assignment: This assignment must be completed prior to the next class. Check your syllabus
for the exact due date and time.
Complete to the pre class assignment
(http://berks.psu.edu/clt/chem112/BalancingRedoxReactions_HW.docx)
Submit a copy to the dropbox located in ANGEL called “Pre Class Assignment Submission:
Balancing Redox Reactions”
End of Chapter Problems:
Practice with these problems if you are having difficulty with any of the concepts covered in this
class guide AFTER we have met in class. If you cannot easily complete these problems, seek
help from your instructor, your mentor or the learning center
Chapter 20: 7, 13, 15, 17