Download OXIDATION-REDUCTION

OXIDATION is the process by which a substance loses
one or more electrons. Iron loses three electrons from
reactant to product.
0
+3
4 Fe + 3 O2  2 Fe2 O3
REDUCTION is the process by which a substance gains
one or more electrons. Iron gains three electrons from
reactant to product.
+3
0
2 Fe2 O3 + 3 C  4 Fe + 3 CO2
OXIDATION-REDUCTION REACTIONS
The oxidation number of an atom in a
substance is equal to the charge that the
atom would have if the electrons in each
bond belonged to the most electronegative
atom.
The more electronegative atom gains
electrons from the other atom because it is
treated as if it were reduced.
The less electronegative atom is oxidized.
OXIDATION-REDUCTION REACTIONS
You can use one of two mnemonic devices
to remember oxidation-reduction:
a. OIL RIG (Oxidation Is Loss of
electrons and Reduction Is Gain of
electrons.)
b. LEO GER (Lose Electrons - Oxidation;
Gain Electrons - Reduction.)
OXIDATION-REDUCTION REACTIONS
An OXIDIZING AGENT causes the oxidation of
another substance by accepting electrons from
that substance. The oxidizing agent contains the
atom that shows a decrease in oxidation number.
The oxidizing agent is itself reduced.
+3
0
0
+4
2 Fe2 O3 + 3 C  4 Fe + 3 CO2
C: 0  +4; C is oxidized, C is the reducing
agent
OXIDIZING AND REDUCING AGENTS
A REDUCING AGENT causes the reduction by providing
electrons to another substance. The reducing agent
contains the atom that shows an increase in
oxidation number. The reducing agent is oxidized.
+3
0
0
+4
2 Fe2 O3 + 3 C  4 Fe + 3 CO2
Fe: +3 to 0;
Fe is reduced, Fe2O3 is the oxidizing agent
OXIDIZING AND REDUCING AGENTS
0
0
+3 -2
4 Fe + 3 O2  2 Fe2 O3
Fe: 0 to +3; Fe is oxidized, Fe is the reducing agent
O: 0 to -2; O is reduced, O2 is the oxidizing agent
The substance that is oxidized or reduced is always a single
element.
The substance that is the oxidizing agent or the reducing agent,
is always whatever the reactant is – element or compound.
ANOTHER EXAMPLE
Okay, we will come back to oxidation and
reduction. First we have to learn how to
determine oxidation numbers. Get out your
periodic table.
Watch the video - Copy and paste this into the
URL window in Mozilla Firefox:
http://www.youtube.com/watch?v=8_CvNPuuhi
M&feature=related
A CLOSER LOOK
In order to identify a redox reaction, you must be able to write
the charges for each element. This is known as the oxidation
number. I have done these three for you. You write the
oxidation number above the compound and do the
bookkeeping below.
K2Cr2O7
+1 +6 -2
K2Cr2O7
+2 +12 -14 = 0
SO4-2
+6 -2
SO4-2
+6 -8 = -2
A CLOSER LOOK
Al(NO3)3
+3 +5 -2
Al(NO3)3
+3 +15 -18 = 0
1. The oxidation number for any uncombined
(free) element is zero (ex. O2, Na, Mg)
2. The oxidation number for a monatomic ion
equals its ionic charge. (ex. Ca+2, O-2)
3. The oxidation number of the more
electronegative atom is equal to the charge it
would have if it were an ion. Remember that
fluorine is the most electronegative element,
followed by oxygen and nitrogen.
OXIDATION NUMBER RULES
4. Some elements’ oxidation number corresponds to their
position on the periodic table:
 Elements in group 1A = +1
 Elements in group 2A = +2
 Aluminum is always +3
 Fluorine is always –1 (most electronegative)
 Hydrogen has an oxidation number of +1 when combined
with nonmetals. If bonded to a metal, it is a –1 because it is
more electronegative.
 Oxygen has an oxidation number of -2 in most compounds
and ions. Peroxides (O2-1) are the exception and if oxygen is
bonded to fluorine, the oxygen’s charge is a +2.
OXIDATION NUMBER
Here are some
common oxidation
numbers. Notice
that nonmetals
tend to have
multiple oxidation
numbers.
4. The sum of the oxidation numbers of all the
atoms in a particle must equal the charge of
that ion. (ex. SO4-2)
5. The sum of all oxidation numbers for all atoms
in a neutral compound is zero.
OXIDATION NUMBER
Write the oxidation numbers for the following
compounds. Write the numbers above each element.
1. HNO3
8. NO
2. H3PO4
9. CrI3
3. AgNO3
10. SO2
4. Cu(NO3)2
11. K2SO4
5. 2 Fe
12. H2S2O7
6. H2SO4
13. P4O6
7. KH2PO4
14. AsO4-3
PRACTICE
+1 +5 -2
1. HNO3
+1+5 -6=0
+1 +5 -2
2. H3PO4
3.
+2 +5 -2
4. Cu(NO3)2
+2 +10 -12=0
0
5. 2 Fe
+3 +5 -8 =0
+1 +5 -2
3. AgNO3
+1+5 -6=0
ANSWERS
+1 +1 +5 -2
7. KH2PO4
+1+2 +5 -8=0
+2 -2
8. NO
+2 -2 =0
+1 +6 -2
+3 -1
6. H2SO4
9. CrI3
+2 +6 -8 =0
+3 -3 =0
+4 -2
10. SO2
+4 -4 =0
+1 +6 -2
11. K2SO4
+2+6 -8=0
+1 +6 -2
12. H2S2O7
+2 +12 -14=0
ANSWERS
+3 -2
13. P4O6
+12 -12=0
+5 -2
14. AsO4-3
+5 -8 = -3
Now, put it all back together – see how to
determine what is oxidized and what is
reduced.
Watch the video - Copy and paste this into the
URL window in Mozilla Firefox:
http://www.youtube.com/watch?v=-vKOPD3K6g
REVIEW
A. Determine the oxidation number for each atom.
B. State what is being oxidized and what is being reduced.
C. State the oxidizing agent and the reducing agent.
1. Cu + 2 AgNO3  Cu(NO3) 2 + 2 Ag
2. 3 H2S + 2 HNO3  3 S + 2 NO + 4 H2O
**REMEMBER: you must have something oxidized and
something reduced in all redox equations
NOW, PRACTICE THIS
Cu + 2 AgNO3  Cu(NO3) 2 + 2 Ag
PRACTICE
3 H2S + 2 HNO3  3 S + 2 NO + 4 H2O
PRACTICE
0
+1 +5 -2
+2
+5 -2
0
A. Cu + 2 AgNO3  Cu(NO3) 2 + 2 Ag
B. Cu 0  +2 lost electrons, oxidized
Ag +1 0 gained electrons, reduced
C. Cu is oxidized; Cu is the reducing agent
Ag is reduced; AgNO3 is the oxidizing
agent
ANSWERS
+1 -2
+1 +5 -2
0
+2 -2
+1 -2
A. H2S + 2 HNO3  3 S + 2 NO + 4 H2O
B. S -2  0 lost electrons, oxidized
N +5 +2 gained electrons, reduced
C. S is oxidized; H2S is the reducing agent
N is reduced; HNO3 is the oxidizing agent
ANSWERS
Write the oxidation numbers for the following
compounds. Write the charges above the elements:
1. H2SO4
2. Ca2(OH)PO4
3. ClO4-1
4. NaHCO3
State the charge changes, which compound lost or
gained electrons, which is oxidized, which is reduced,
and name the oxidizing and reducing agents:
1. Cu + AgNO3  Cu(NO3)2 + Ag
2. H2S + HNO3  S + NO + H2O
DO NOW
Oxidation Numbers
+1 +6 -2
+7 -2
1. H2SO4
3. ClO4-1
+2 +6 -8 = 0
+7 -8 = -1
+2 -2 +1 +5 -2
+1 +1+4-2
2. Ca2(OH)PO4
4. NaHCO3
+4 -2 +1 +5 -8 = 0
+1 +1+4-6 = 0
DO NOW ANSWERS
0
+1+5-2
+2 +5 -2
0
1.Cu + AgNO3  Cu(NO3)2 + Ag
Cu 0  +2 lose e-, oxidized Cu = reducing agent
Ag +1  0 gain e-, reduced AgNO3 = oxidizing
agent
DO NOW ANSWERS
+1 -2
+1+5-2
0
+2-2
+1 -2
2.H2S + HNO3  S + NO + H2O
S -2  0 lose e-, oxidized H2S = reducing
agent
N +5  +2 gain e-, reduced HNO3 = oxidizing
agent
DO NOW ANSWERS
1. Corrosion
This is the oxidation of a metal
caused by a reaction between the
metal and some substance in the
environment.
The best example is rusting.
Corrosion is a problem for it results in
the loss of structural strength of
the metal.
Can be prevented by coating the
metal with paint, plastic or another
metal.
APPLICATIONS
2. Bleaching
A chemical substance that is used to
eliminate unwanted color from fabrics
and other materials.
Bleaches are oxidizing reagents because
they remove electrons from the
pigments that cause color.
Common bleaches are chlorine (Cl2),
hydrogen peroxide (H2O2), and
hypochlorite ion (ClO-1).
APPLICATIONS
3. Fuels and Explosives
Fuels release energy as they are oxidized.
Common fuels (gasoline, natural gas) are
composed largely of carbon and
hydrogen.
Once ignited, they are oxidized by oxygen,
forming water and carbon dioxide.
Nitroglycerin is both an oxidizer (hydrogen
and oxygen) and a reducer (nitrogen).
APPLICATIONS
4. Photography
Photography involves the capturing
of a light image on a lightsensitive medium and the
processing of the image to make
a permanent record.
The process is based on the redox
of silver halides, such as AgBr.
APPLICATIONS
5. Electrochemical cells (page 664)
These involve a transfer of electrons from an oxidized substance to a
reduced substance.
If a zinc strip is in contact with copper(II) sulfate solution, the zinc
strip (the anode) loses electrons to the copper(II) ions (the
cathode) in solution.
The copper(II) ions accept the electrons and drop out of solution as
copper atoms (now neutral).
APPLICATIONS
As the electrons are transferred, energy is released in the form of heat
and the temperature rises.
If the two metals cannot touch and are separated, a transfer of
electrical energy instead of heat accompanies the electron transfer.
To complete the circuit, electrons flow in one direction from metal to
metal.
Then a salt bridge, or some other set-up, allows the ions to pass from
one side to another, completing the electrical circuit.
APPLICATIONS
An electrochemical
cell converts
chemical energy to
electrical energy by a
spontaneous redox
reaction. It was
invented by
Allesandro Volta and
is also called a
voltaic cell.
APPLICATIONS
Many redox reactions are easy to balance by the
usual methods.
KClO3  KCl + O2
Some are harder.
Cu + HNO3  Cu(NO3)2 + NO2 + H2O
BALANCING REDOX EQUATIONS
The OXIDATION NUMBER METHOD allows you to
apply your knowledge of oxidation numbers in
order to balance equations.
The fundamental principle in balancing redox
equations is that the number of electrons lost in
an oxidation process (increase in oxidation
number) must equal the number of electrons
gained in the reduction process (decrease in
oxidation number).
BALANCING REDOX EQUATIONS
Two methods:
1. Oxidation Number/Bracket Method
This will not always work if the reaction is
complicated.
2. Half Reaction Method
This method always works but takes longer.
BALANCING REDOX EQUATIONS
STEP ONE:
Assign oxidation numbers to all atoms in the
equations. Write them above the element.
S + HNO3  SO2 + NO + H2O
OXIDATION NUMBER METHOD
STEP TWO:
Identify the element oxidized and the element
reduced. Determine the change in oxidation
number of each oxidized and reduced element.
STEP THREE:
Connect the atoms that change oxidation number
using a bracket. Write the change in oxidation
number at the midpoint.
OXIDATION NUMBER METHOD
STEP FOUR:
Choose coefficients that make the total increase
in oxidation number equal the total decrease.
STEP FIVE:
Balance the remaining elements by inspection
using the conventional method and check the
final equation.
OXIDATION NUMBER METHOD
1. Cu + HNO3  Cu(NO3)2 + NO2 + H2O
PRACTICE
2. Cl2 + KBr  KCl + Br2
PRACTICE
HALF REACTIONS
These reactions are used to balance redox
equations. The reaction is broken into two
parts – the oxidation part and the reduction
part. The half –reaction method allows you to
apply your knowledge of oxidation numbers in
order to balance equations.
BALANCING REDOX REACTIONS
STEPS TO FOLLOW:
1. Write out the equation. Then change it to a
net ionic equation if it is not one already, omit
the spectator ions, and assign oxidation
numbers to all atoms in the equation. Write
them above the element.
HS-1 + IO3-1  I-1 + S + H2O
HALF REACTION METHOD
2. Write the separate half-reactions.
HS-1  S
IO3-1  I-1
3. Balance all the elements except O and H
(already balanced in this one).
HS-1  S
IO3-1  I-1
HALF REACTION METHOD
4. If the oxygen is unbalanced, add enough water
(H2O) to the side deficient in oxygen.
HS-1  S
IO3-1  I-1 + 3H2O
5. Add sufficient hydrogen ions (H+) to the side
deficient in hydrogen to balance the hydrogen.
HS-1  S + H+1
6H+1 + IO3-1  I-1 + 3H2O
HALF REACTION METHOD
6. Write the electrons in each half reaction.
-2
0
HS-1  S + H+1 + 2e-
6e- + 6H+1
+5
-1
+ IO3-1  I-1 + 3H2O
HALF REACTION METHOD
7. Determine the least common multiple and
multiply each to get it so that the number of
electrons gained equals the number of
electrons lost.
(x3) 3HS-1  3S + 3H+1 + 6e(x1) 6e- + 6H+1 + IO3-1  I-1 + 3H2O
HALF REACTION METHOD
8. Add the two half reactions together and return the spectator
atoms. Delete anything that exactly occurs on both sides.
(Notice how the water showed back up!)
3HS-1 + 6e- + 6H+1 + IO3-1  3S + 3H+1 + 6e+ I-1 + 3H2O
becomes 3H+1
3HS-1 + 3H+1 + IO3-1  3S + I-1 + 3H2O
HALF REACTION METHOD
PRACTICE (ACIDIC):
MnO4-1 + H2SO3  Mn+2 + HSO4-1 + H2O
HALF REACTION METHOD
BASIC: Basic solutions add OH-1 to get rid of the
H+1. Balance them the same way that you
balance acidic solutions (H2O and then H+1)
only add one more step.
Anywhere that you added an H+1, you need to
add an OH-1 to both sides. Combine any OH-1
with any H+1 on the same side to made water.
BALANCING REDOX REACTIONS
PRACTICE (Basic):
I-1 + OCl-1  I2 + Cl-1 + H2O
HALF REACTION METHOD
PRACTICE (BASIC)
-1
-2 +1
0
-1
+1 -2
I-1 + OCl-1  I2 + Cl-1 + H2O
+1
0
-1
0
2e- + 2H+ + OCl-1  Cl-1 + H2O 2I-1  I2 + 2e-
2e- + 2H+ + OCl-1 + 2I-1  Cl-1 + H2O +  I2 + 2e-
2H+ + OCl-1 + 2I-1  Cl-1 + H2O + I2
-1
-2 +1
0
-1
+1 -2
I-1 + OCl-1  I2 + Cl-1 + H2O
+1
0
2e- + 2H+ + OCl-1  Cl-1 + H2O
-1
0
2I-1  I2 + 2e-
2e- + 2H+ + OCl-1 + 2I-1  Cl-1 + H2O +  I2 + 2e-
2H+ + OCl-1 + 2I-1  Cl-1 + H2O + I2
BASIC 2OH- + 2H+ + OCl-1 + 2I-1  Cl-1 + H2O + I2 + 2OH2H2O
H2O + OCl-1 + 2I-1  Cl-1 + I2 + 2OH-
HALF REACTION METHOD