Oxidation and Reduction
My Goals for this Lesson:

Define oxidation and reduction.

Assign oxidation numbers to elements in a reaction.

Describe and explain what occurs during an oxidation-reduction reaction.
I’m preparing to understand and to be able to define, describe, explain and do calculations on oxidationreduction reactions.
Introduction
Answer the following questions from the Try It at Home lab (the
results are given in the interactive section).
Which element had a stronger attraction to the electrons?
How can this be used to make a battery?
Oxidation and Reduction
Answer these questions with your own words using the interactive section from the Lesson.
Definition
Examples
Where did the name
come from?
Oxidation
Reduction
Mnemonics
Use the lesson to explain the following 2 mnemonic devices to help you remember the difference
between oxidation and reduction.
OIL RIG LEO the lion says GER -
Real World Examples
Fill in the table using the Examples in the lesson.
Image
Reaction
Details
Electronegativity and Oxidation Numbers
Fill in the blanks using the lesson.
According to the concept of electronegativity, every atom
has a certain amount of
for the
electrons shared in a covalent bond. Some atoms have a
stronger
for these shared electrons
than others.
Picture a
bond as a sort of tug-of-war where both atoms are competing for the shared
electrons in the bond, much like two children fighting over the same toy.
Just as the
child can pull the toy closer to herself than a weaker child in a tug-of-war, the
atom with the
electronegativity has a stronger “pull” on the electrons than the atom with the
lower electronegativity.
When two nonmetals compete for the electrons that are “shared” in a covalent bond, the electrons spend
more of their time closer to the
electronegative element.
This uneven sharing of electrons in a covalent bond means that the
becomes more “negative” while the
electronegative element
electronegative element becomes more “positive.”
Even though the atoms involved in a covalent molecule do not have charges like an ion, we assign them
positive and negative
numbers to represent this “uneven sharing” that occurs. The
following rules for assigning
numbers will be useful when you need to determine if a
reaction involves oxidation and reduction.
Rules for Assigning Oxidation Numbers
Fill in the table using the Examples in the lesson.
Tab Title
Neutral Elements
Oxygen
Monatomic Ions
Covalent Compounds
Definition
Examples
Algebraic Sum
Compound
CaSO4
H2CO3
PO43MnO4-
Oxidation numbers
Ca =
S= O=
H=
C= O=
P=
O=
Mn = O =
Explain (or show) how each oxidation # is determined.
Fill in the oxidation numbers from the Example of the combustion of Methane in the lesson.
CH4 (g) + 2 O2 (g) →
CO2 (g)


 
(each O)
(each H)

(each H)
Explain the oxidation numbers for each compound:
CH4 O2 CO2 H2O Do the “Let’s Practice” for this section.
+ 2 H 2O
(g)
Oxidizing and Reducing Agents
Define these terms and fill in the blanks using the lesson.
Oxidizing Reagent Reducing Agent This may seem backward because the oxidizing agent is being reduced and the reducing agent is being
oxidized, but remember that if you are an agent of something, you cause that “something” to happen to
someone else. Define all of these terms in your notes and review them because it can be easy to mix
them up.

An oxidizing agent causes another substance to be

The only way to cause oxidation is to remove
oxidation is loss of

.
from that other substance (because
).
Therefore, the oxidizing agent is
as it gains electrons from the other substance.
Be sure to do the “Let’s Practice” section on the bottom of the Lesson page.
*****HONORS STUDENTS ONLY – COMPLETE THE REST OF THIS FILE*****
Half Reactions
Some oxidation-reduction reactions can be balanced using the methods we have used throughout this
course, while others require a more involved balancing method. There are several ways to approach the
balancing of redox reactions, but we will focus on one approach that involves half reactions.
To be able to keep track of the moles of
as
involved in a redox reaction, they are often written
reactions. This means that we write one reaction for the
reaction for the
Because oxidation is the
process and another
process.
of electrons, you will see electrons given off as a
for the
of electrons, you will see electrons written as a
for the
oxidation half reaction.
Because reduction is the
reduction half reaction.
From the “Did You Know?” box, explain how batteries work and relate to half reactions.
Balancing Redox Equations Using the Half Reaction Method
Steps
Use the following steps to balance oxidation-reduction reactions using the half-reaction method.
1) Identify the element that is oxidized and the element that is reduced in the equation, using
oxidation numbers if needed.
Cr2O72-(aq) + HNO2 (aq) → Cr3+ (aq) + NO3- (aq)
Cr2O72- is reduced.
HNO2 is oxidized.
2) Split the reaction into two half reactions, one representing the oxidation and all other elements
involved, one representing the reduction and all other elements involved.
Cr2O72- → Cr3+ Chromium went from +6 to +3. It was reduced. To go from +6 to +3 it gained 3 e-.
HNO2 → NO3- Nitrogen went from +3 to +5. It was oxidized. To go from +3 to +5 it lost 2 e-.
Balance each half reaction separately using the following steps (This is different than the balancing that
you are used to, so pay close attention).
3) Balance all elements, except hydrogen and oxygen, first using coefficients.
Cr2O72- → 2 Cr3+ (balanced Cr ONLY)
HNO2 → NO3- (N is already balanced)
4) Balance any oxygen atoms by adding enough H2O to the other side. (You are allowed to do this
because there is plenty of water available in the aqueous solution.)
Cr2O72- → 2 Cr3+ + 7 H2O
This half reaction needs seven oxygen atoms on the right, so we add seven H2O molecules.
HNO2 + H2O → NO3This half reaction needs one more oxygen atom on the left, so we add one H2O molecule.
5) Balance any H’s by adding enough H+ ions to the other side (you are allowed to do this because
the solution is acidic, so there are H+ ions available).
Cr2O72- + 14 H+ → 2 Cr3+ + 7 H2O
This half reaction needs 14 hydrogen atoms on the left to balance the 14 hydrogen atoms in the
seven H2O molecules, so we add 14 H+ ions to the left.
HNO2 + H2O → NO3- + 3H+
This half reaction needs three hydrogen atoms on the right to balance the three hydrogen atoms
on the left, so we add three H+ ions to the right.
6) Balance the total charges of the ions and compounds on each side (charges, not oxidation
numbers) last by adding enough electrons to the side with the higher overall charge to bring it
down to equal the charge of the other side.
6 e− + Cr2O72- + 14 H+ → 2 Cr3+ + 7 H2O
The sum of the charges on the left side of the chromium half reaction is +12 (-2 for the Cr2O72plus +14 for the 14 H+ and the sum on the right side is +6 (for the 2 Cr3+). If we add six electrons
to the left side, the sum of the charges on each side of the equation becomes +6.
HNO2 + H2O → NO3- + 3 H+ + 2 e−
The sum of the charges on the left side of the nitrogen half reaction is zero. The sum of the
charges on the right side of the nitrogen half-reaction is +2 (−1 for the nitrate plus +3 for the
three H+). If we add two electrons to the right side, the sum of the charges on each side of the
equation becomes zero.
We can combine the two half reactions and get a whole redox equation.

The most important thing to remember when doing this is that the number of moles of
electrons (represented by the coefficient) must be equal for both half reactions before
they can combine.

This is because it is the same electrons being lost in the oxidation process that are gained
in the reduction process.
7) If the number of electrons lost in the oxidation half reaction is not equal to the number of
electrons gained in the reduction half reaction, multiply one or both of the half reactions by a
number that will make the number of electrons gained equal to the number lost.
When multiplying a half reaction by a number, that number gets multiplied by every coefficient in
that half reaction!
6 e− + Cr2O72- + 14 H+ → 2 Cr3+ + 7 H2O
3(HNO2 + H2O → NO3- + 3 H+ + 2 e−) equals 3 HNO2 + 3 H2O → 3 NO3- + 9 H+ + 6 e−
8) Once the number of electrons is equal, add the two half reactions together to make one balanced
redox equation.
Cr2O72- + 3 HNO2 + 3 H2O + 14 H+ → 2 Cr3+ + 3 NO3- + 9 H+ + 7 H2O
The three H2O in the second half reaction cancel three of the seven H2O in the first half
reaction to yield four H2O on the right of the final equation.
The nine H+ on the right of the second half reaction cancel nine of the 14 H+ on the left of the
first half reaction, leaving five H+ on the left of the final equation.
Tips:

If you have equal amounts of the same substance on both sides of the balanced equation (like you
will for the electrons), they cancel out and do not get written in the final equation.

If the same substance is on both sides of the balanced equation with different coefficients, you
can reduce them by subtracting the lower number from both (one side will cancel, the other will
reduce).

If the same substance appears more than once on the same side of the equation, you can combine
them by adding the coefficients together.
Example
Try to complete this yourself but use the lesson interactive walk through if you have trouble.
1) Identify the element that is oxidized and the element that is reduced in the equation, using
oxidation numbers if needed.
MnO4- + Fe 2+ → Mn2+ + Fe3+
The
is reduced.
The
is oxidized.
2) Split the reaction into two half reactions, one representing the oxidation and all other elements
involved, one representing the reduction and all other elements involved.
Balance each half reaction separately using the following steps (This is different than the balancing that
you are used to, so pay close attention).
3) Balance all elements, except hydrogen and oxygen, first using coefficients.
4) Balance any oxygen atoms by adding enough H2O to the other side (you are allowed to do this
because there is plenty of water available in the aqueous solution).
5) Balance any H's by adding enough H+ ions to the other side. (You are allowed to do this because
the solution is acidic, so there are H+ ions available.)
6) Balance the total charges of the ions and compounds on each side (charges, not oxidation
numbers) last by adding enough electrons to the side with the higher overall charge to bring it
down to equal the charge of the other side.
7) If the number of electrons lost in the oxidation half reaction is not equal to the number of
electrons gained in the reduction half reaction, multiply one or both of the half reactions by a
number that will make the number of electrons gained equal to the number lost.
When multiplying a half reaction by a number, that number gets multiplied by every coefficient in
that half reaction!
8) Add the two half reactions together, and do not forget to cancel or reduce anything that appears
on both sides of the arrow. Remember that the electrons never appear in the final balanced
equation.