Chapter #7
Reactions in Aqueous Media
Aqueous Ionic Compounds
Most ionic compounds dissolve in water to produce
solutions that conduct electricity. The degree to which
a solution will conduct electricity is used to determine
if a solute is a strong or weak electrolyte. Solutions
that do not conduct electricity are called
nonelectrolytes. Molecular compounds that dissolve in
water to that conduct electricity, are acids. Strong
acids are referred to as strong electrolytes since there
ae solutions are good conductors of electricity. Weak
acid solutions do not conduct electricity well and are
called weak electrolytes
Aqueous Ionic Compounds
Most ionic compounds dissolve in water to produce
solutions that conduct electricity. The degree to which
a solution will conduct electricity is used to determine
if a solute is a strong or weak electrolyte. Solutions
that do not conduct electricity are called
nonelectrolytes. Molecular compounds that dissolve in
water to that conduct electricity, are acids. Strong
acids are referred to as strong electrolytes since there
are solutions are good conductors of electricity. Weak
acid solutions do not conduct electricity well and are
called weak electrolytes.
How do we tell if a solution conducts electricity?
Solution Conductivity
Strong electrolyte
Weak electrolyte
Nonelectrolyte
The Solution Process
Earlier in the quarter we defined a solution as a
homogeneous mixture; a random combination of two or
more things. The part of the solution we have the most
of is the solvent and the minor components of a solution
are referred to as the solutes. Water is the most
common solvent and a good one for ionic solutes. Half
of a water molecule is slightly positive and the other half
is slightly negative and because of this is called polar.
When forming a solution between an ionic compound
and water, the positive region of the water attracts to an
anion in the solute crystal while the negative region of
another water molecule attracts to the cation in the
crystal lattice
Types of Aqueous Solutions
Solutions are homogeneous mixtures of a solute
and a solvent.
• The solute is the solution component in the smallest
amount while the solvent is the larger component of a
solution.
• Solutes whose solutions conduct electricity are called
electrolytes
• Solutes whose solutions do not conduct electricity are
called nonelectrolytes
• Electrolytes are solutes that form ions when they
dissolve. Ionic solutes or acids usually form solutions
that conduct electricity.
Solution Formation
Water is one of the best solvents known. It is able
to dissolve ionic solutes, such as sodium chloride,
to produce solutions that conduct electricity.
Molecules, containing a positive and negative
regions, are called polar. Water is an example of a
polar molecule and can dissolve ionic solutes by
the positive region of water attracting to the
negative ion of an ionic solute thus separating the
crystal lattice in to a solution of solvated ions.
Sodium Chloride Crystal Lattice
The Solvation Process
If the attractive force between the surface ion and the
solvent is greater than the forces between the ion and
the solid then the ion will enter the solution phase.
-
+
+
-
H2O The ion that has left the solid
and becomes completed
surrounded by water
molecules. It has become
solvated or hydrated.
-
+
+
K+
+
+
-
+
+
Solvated Ions
The process continues as new water molecules approach the crystal
until the crystal has been fully dissolved.
-
+
-
+
+
-
+
+
Note the different
orientation of water
molecules around the
oppositely charged
ions.
+
Reaction Driving Forces
Five Driving Forces Favor Chemical Change
1.
2.
3.
4.
5.
Formation of a solid
Formation of water
Transfer of electrons
Formation of a gas
Formation of a weak electrolyte
Precipitation
Precipitation is the formation of a solid when
two solutions are combined.
Solubility Rules
How do we determine if an ionic solute will
dissolve in water?
Solubility Rules
How do we determine if an ionic solute will
dissolve in water? Use the solubility rules
Writing Ionic Equations
Ionic compounds that are water soluble are indicated with
and (aq) after the symbol, and if insoluble then (s) goes
after the symbol.
Consider the following equation.
AgNO3 (aq) + NaCl (aq)
What are the products?
Writing Ionic Equations
Ionic compounds that are water soluble are indicated with
and (aq) after the symbol, and if insoluble then (s) goes
after the symbol.
Consider the following equation.
AgNO3 (aq) + NaCl (aq)
NaNO3 (aq) + AgCl (s)
Called a formula equation
Ag+ (aq) + NO3-(aq) + Na+(aq) + Cl-(aq)
Na+(aq) + NO3-(aq) + AgCl (s)
Called an ionic equation
Writing Ionic Equations
Ionic compounds that are water soluble are indicated with
and (aq) after the symbol, and if insoluble then (s) goes
after the symbol.
Consider the following equation.
AgNO3 (aq) + NaCl (aq)
NaNO3 (aq) + AgCl (s)
Called a formula equation
Ag+ (aq) + NO3-(aq) + Na+(aq) + Cl-(aq)
Na+(aq) + NO3-(aq) + AgCl (s)
Called an ionic equation
Spectator Ions
Writing Ionic Equations
Ionic compounds that are water soluble are indicated with
and (aq) after the symbol, and if insoluble then (s) goes
after the symbol.
Consider the following equation.
AgNO3 (aq) + NaCl (aq)
NaNO3 (aq) + AgCl (s)
Called a formula equation
Ag+ (aq) + NO3-(aq) + Na+(aq) + Cl-(aq)
Na+(aq) + NO3-(aq) + AgCl (s)
Called an ionic equation
Spectator Ions
Ag+ (aq) + Cl-(aq)
AgCl (s)
Called the net ionic equation
Writing Ionic Equations
Ionic compounds that are water soluble are indicated with
and (aq) after the symbol, and if insoluble then (s) goes
after the symbol.
Consider the following equation.
AgNO3 (aq) + NaCl (aq)
NaNO3 (aq) + AgCl (s)
Called a formula equation
Ag+ (aq) + NO3-(aq) + Na+(aq) + Cl-(aq)
Na+(aq) + NO3-(aq) + AgCl (s)
Called an ionic equation
Spectator Ions
Ag+ (aq) + Cl-(aq)
AgCl (s)
Called the net ionic equation
This reaction goes to completion because the solid silver chloride is
formed. It is called a precipitate.
Ionic Equations
It is possible for all of the reactants and products
to be water soluble and thus produce all
spectator ions. If this is the case then all of the
ions cancel out and there is no net ionic
equation. When this occurs we say that there is
No Reaction, and give the label NR. This
makes sense, since in order for reactions to go
to completion a solid, water, gas, or electron
transfer must occur in order for a reaction to go
to completion.
Reaction Driving Forces
Five Driving Forces Favor Chemical Change
1.
2.
3.
4.
5.
Formation of a solid
Formation of water
Transfer of electrons
Formation of a gas
Formation of a weak electrolyte
Water Formation
Formation of water is a normal product between
acids and bases. Since acids and bases dissolve
in water to make solutions that are electrolytes,
then we conclude that acids and bases have
some ionic character. Since water does not
ionize, then when water is formed, we will also
have a net ionic equation and then a chemical
reaction.
Acids as Electrolytes
Strong acids and bases ionize 100%!
Memorized Strong acids and bases:
Acids
HCl (aq)
HI (aq)
HBr (aq)
HNO3
H2SO4
HClO4
Bases
Hydroxides of
group I and II
metals, except Be
and Mg
Acid-Base Reactions
Acids undergo characteristic double replacement
reactions with oxides, hydroxides, carbonates and
bicarbonates.
2HCl (aq) + CuO (s)  CuCl2 (aq) + H2O (l)
2HCl (aq) + Ca(OH)2 (aq)  CaCl2 (aq) + 2H2O (l)
2HCl (aq) + CaCO3 (aq)  CaCl2 (aq) + H2O (l) + CO2 (g)
2HC l (aq) + Sr(HCO3)2 (aq)  SrCl2 (aq) + 2H2O (l) + 2CO2 (g)
Acid-Base Reactions
Acids undergo characteristic double replacement
reactions with oxides, hydroxides, carbonates and
bicarbonates.
2HCl (aq) + CuO (s)  CuCl2 (aq) + H2O (l)
2HCl (aq) + Ca(OH)2 (aq)  CaCl2 (aq) + 2H2O (l)
2HCl (aq) + CaCO3 (aq)  CaCl2 (aq) + H2O (l) + CO2 (g)
2HC l (aq) + Sr(HCO3)2 (aq)  SrCl2 (aq) + 2H2O (l) + 2CO2 (g)
Acid-Base Reactions
Bases undergo a double replacement reaction with
acids called neutralization:
NaOH (aq) + HCl (aq)  H2O (l) + NaC l (aq)
In words this well known reaction is often described as:
“acid plus base = salt plus water”
We previously discussed this reaction when describing
types of reactions.
Acid-Base Reactions
We have discussed the double replacement reactions
and ionic equations before. Since the acids and
bases undergo double replacement reactions called
neutralization reactions, then they can have ionic
equations too.
Formula equation:
HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l)
Total ionic equation:
H+ (aq) + Cl- (aq) + Na+ (aq) + OH- (aq)  Na+ (aq) + Cl- (aq) + H2O (l)
Net ionic equation:
H+ (aq) + OH- (aq)  H2O (l)
Acid-Base Reactions
Another property of acids is their reaction with certain metals to produce
hydrogen gas, H2 (g).
Zn (s) + 2HC l (aq)  H2 (g) + ZnCl2 (aq)
This is an example of a single replacement reaction and is a redox reaction.
Total ionic equation:
Zn (s) + 2H+ (aq) + 2Cl- (aq)  H2 (g) + Zn2+ (aq) + 2Cl- (aq)
Net ionic equation:
Zn (s) + 2H+ (aq)  H2 (g) + Zn2+ (aq)
Reaction Driving Forces
Five Driving Forces Favor Chemical Change
1.
2.
3.
4.
5.
Formation of a solid
Formation of water
Transfer of electrons
Formation of a gas
Formation of a weak electrolyte
Types of Chemical Reactions
REDOX reactions where the oxidation number
changes from reactants to products.
Oxidation is when the oxidation number increases,
by losing of electrons.
Reduction is when the oxidation number decreases
by gaining electrons.
Consider the following equation:
0
0
H2 + O2
H2O
What are the oxidation numbers of hydrogen and
oxygen?
Types of Chemical Reactions
REDOX reactions where the oxidation number
changes from reactants to products.
Oxidation is when the oxidation number increases,
by losing of electrons.
Reduction is when the oxidation number decreases
by gaining electrons.
Consider the following equation:
H2 + O2
H2O
What are the oxidation numbers of hydrogen and
oxygen?
REDOX REACTIONS
0
0
H2 + O2
2(1+) 2- = 0
H2O
How about hydrogen and oxygen in water?
REDOX REACTIONS
0
H2 + O2
oxidized
2(1+) 2- = 0
0
reduced
H2O
How about hydrogen and oxygen in water?
Oxidation is caused by the oxygen molecule, so
it is referred to as the oxidizing agent (OA)
Reduction is caused by the hydrogen molecule,
so it is referred to as the reducing agent (RA)
REDOX REACTIONS
Note:
• All of the previously discussed reactions are
REDOX except the double replacement
reactions.
• The number of electrons lost is equal to the
number of electrons gained in a reaction.
Why?
• Most elements have variable oxidation
numbers, except for hydrogen, oxygen, and
the memorized polyatomic ions.
REDOX REACTIONS
Oxidation numbers for a compound must add
up to equal zero, while the oxidation numbers
for a polyatomic ion must up to equal the
charge of that ion.
Consider the following chlorine compounds
1+
4(2-)=0
HClO4, HClO3, HClO2, HClO, Cl2, HCl
What is the oxidation number of chlorine in
each of these compounds, assuming
H 1+ and oxygen is 2-
REDOX REACTIONS
Oxidation numbers for a compound must add
up to equal zero, while the oxidation numbers
for a polyatomic ion must up to equal the
charge of that ion.
Consider the following chlorine compounds
1+ 7+ 4(2-)=0
HClO4, HClO3, HClO2, HClO, Cl2, HCl
What is the oxidation number of chlorine in
each of these compounds, assuming H is 1+
and oxygen is 2-
REDOX REACTIONS
Oxidation numbers for a compound must add
up to equal zero, while the oxidation numbers
for a polyatomic ion must up to equal the
charge of that ion.
Consider the following chlorine compounds
1+ 7+ 4(2-)=0
5+
HClO4, HClO3, HClO2, HClO, Cl2, HCl
What is the oxidation number of chlorine in
each of these compounds, assuming H is 1+
and oxygen is 2-
REDOX REACTIONS
Oxidation numbers for a compound must add
up to equal zero, while the oxidation numbers
for a polyatomic ion must up to equal the
charge of that ion.
Consider the following chlorine compounds
1+ 7+ 4(2-)=0
5+
3+
HClO4, HClO3, HClO2, HClO, Cl2, HCl
What is the oxidation number of chlorine in
each of these compounds, assuming H is 1+
and oxygen is 2-
REDOX REACTIONS
Oxidation numbers for a compound must add
up to equal zero, while the oxidation numbers
for a polyatomic ion must up to equal the
charge of that ion.
Consider the following chlorine compounds
1+ 7+ 4(2-)=0
5+
3+
1+
HClO4, HClO3, HClO2, HClO, Cl2, HCl
What is the oxidation number of chlorine in
each of these compounds, assuming H is 1+
and oxygen is 2-
REDOX REACTIONS
Oxidation numbers for a compound must add
up to equal zero, while the oxidation numbers
for a polyatomic ion must up to equal the
charge of that ion.
Consider the following chlorine compounds
1+ 7+ 4(2-)=0
5+
3+
1+
0
HClO4, HClO3, HClO2, HClO, Cl2, HCl
What is the oxidation number of chlorine in
each of these compounds, assuming H is 1+
and oxygen is 2-
REDOX REACTIONS
Oxidation numbers for a compound must add
up to equal zero, while the oxidation numbers
for a polyatomic ion must up to equal the
charge of that ion.
Consider the following chlorine compounds
1+ 7+ 4(2-)=0
5+
3+
1+
0
1HClO4, HClO3, HClO2, HClO, Cl2, HCl
What is the oxidation number of chlorine in
each of these compounds, assuming H is 1+
and oxygen is 2-
REDOX REACTIONS
3(2-)=2-
How about sulfur in SO3 2-
REDOX REACTIONS
4+ 3(2-)=2-
How about sulfur in SO3 212(1+) +6(2-)=0
How about carbon in C6H12O6
REDOX REACTIONS
4+ 3(2-)=2-
How about sulfur in SO3 20 + 12(1+) +6(2-)=0
How about carbon in C6H12O6
Types of Chemical Reactions
REDOX reactions where the oxidation number
changes from reactants to products.
Oxidation is when the oxidation number increases,
by losing of electrons.
Reduction is when the oxidation number decreases
by gaining electrons.
Consider the following equation:
H2 + O2
H2O
What are the oxidation numbers of hydrogen and
oxygen?
Types of Chemical Reactions
REDOX reactions where the oxidation number
changes from reactants to products.
Oxidation is when the oxidation number increases,
by losing of electrons.
Reduction is when the oxidation number decreases
by gaining electrons.
Consider the following equation:
0
0
H2 + O2
H2O
What are the oxidation numbers of hydrogen and
oxygen?
REDOX REACTIONS
0
0
H2 + O2
2(1-) 2- = 0
H2O
How about hydrogen and oxygen in water?
REDOX REACTIONS
0
2(1+) 2- = 0
0
H2 + O2
oxidized
H2O
reduced
How about hydrogen and oxygen in water?
Oxidation is caused by the oxygen molecule, so
it is referred to as the oxidizing agent (OA)
Reduction is caused by the hydrogen molecule,
so it is referred to as the reducing agent (RA)
REDOX REACTIONS
Note:
• All of the previously discussed reactions are
REDOX except the double replacement
reactions.
• The number of electrons lost is equal to the
number of electrons gained in a reaction.
Why?
• Most elements have variable oxidation
numbers, except for hydrogen, oxygen, and
the memorized polyatomic ions.
Balancing Redox Reactions
I. Oxidation Number Method
a. Assign oxidation numbers to each element
b. Determine the elements oxidized and reduced
c. Balance the atoms that are oxidized and reduced
d. Balance the electrons lost or gained, to conform to the Law of
Conservation of Matter, by placing coefficients in front of the
formulas containing the atoms oxidized and reduced to both
sides of the equation.
e. The remaining atoms are balanced by inspection
f. Balance oxygen, or hydrogen by adding H2O
g. Balance remaining hydrogen atoms by adding H+
h. Simplify
i. For basic reactions add the same number of OH- ions to both sides
of the equation as there are H+ ions.
j. Combine H+ and OH- ions to make water
k. Simplify again if necessary.
HNO3
+
Cu2O
→
Cu(NO3)2
+
NO
+
H 2O
Balancing REDOX Reactions
I. Oxidation Number Method
a.
b.
c.
d.
Assign oxidation numbers to each element
Determine the elements oxidized and reduced
Balance the atoms that are oxidized and reduced
Balance the electrons lost or gained, to conform to the Law of
Conservation of Matter, by placing coefficients in front of the
formulas containing the atoms oxidized and reduced to both sides
of the equation.
e. The remaining atoms are balanced by inspection
f. Balance oxygen, or hydrogen by adding H2O
g. Balance remaining hydrogen atoms by adding H+
h. Simplify
i. For basic reactions add the same number of OH- ions to both sides of
the equation as there are H+ ions.
j. Combine H+ and OH- ions to make water
k. Simplify again if necessary.
1+ ? 3(2-)=0
HNO3 + Cu2O
→
Cu(NO3)2
+
NO
+
H2O
Balancing REDOX Reactions
I. Oxidation Number Method
a.
b.
c.
d.
Assign oxidation numbers to each element
Determine the elements oxidized and reduced
Balance the atoms that are oxidized and reduced
Balance the electrons lost or gained, to conform to the Law of Conservation of
Matter, by placing coefficients in front of the formulas containing the atoms
oxidized and reduced to both sides of the equation.
e. The remaining atoms are balanced by inspection
f. Balance oxygen, or hydrogen by adding H2O
g.
Balance remaining hydrogen atoms by adding H+
h.
Simplify
i. For basic reactions add the same number of OH- ions to both sides of the
equation as there are H+ ions.
j. Combine H+ and OH- ions to make water
k. Simplify again if necessary.
1+ 5+ 3(2-)=0 2(?)+2-=0
HNO3 + Cu2O →
Cu(NO3)2
+
NO
+
H2O
Balancing REDOX Reactions
I. Oxidation Number Method
a.
b.
c.
d.
Assign oxidation numbers to each element
Determine the elements oxidized and reduced
Balance the atoms that are oxidized and reduced
Balance the electrons lost or gained, to conform to the Law of Conservation of
Matter, by placing coefficients in front of the formulas containing the atoms
oxidized and reduced to both sides of the equation.
e. The remaining atoms are balanced by inspection
f. Balance oxygen, or hydrogen by adding H2O
g. Balance remaining hydrogen atoms by adding H+
h. Simplify
i. For basic reactions add the same number of OH- ions to both sides of the
equation as there are H+ ions.
j. Combine H+ and OH- ions to make water
k. Simplify again if necessary.
1+ 5+ 3(2-)=0 2(1+)+2-=0
HNO3 + Cu2O →
Cu(NO3)2
+
NO
+
H2O
Balancing Redox Reactions
I. Oxidation Number Method
a.
b.
c.
d.
Assign oxidation numbers to each element
Determine the elements oxidized and reduced
Balance the atoms that are oxidized and reduced
Balance the electrons lost or gained, to conform to the Law of Conservation of
Matter, by placing coefficients in front of the formulas containing the atoms
oxidized and reduced to both sides of the equation.
e. The remaining atoms are balanced by inspection
f. Balance oxygen, or hydrogen by adding H2O
g. Balance remaining hydrogen atoms by adding H+
h. Simplify
i. For basic reactions add the same number of OH- ions to both sides of the
equation as there are H+ ions.
j. Combine H+ and OH- ions to make water
k. Simplify again if necessary.
1+ 5+ 3(2-)=0 2(1+)+2-=0
HNO3 + Cu2O →
? + 2(1-)=0
Cu(NO3)2 +
NO
+
H2O
Balancing REDOX Reactions
I. Oxidation Number Method
a.
b.
c.
d.
Assign oxidation numbers to each element
Determine the elements oxidized and reduced
Balance the atoms that are oxidized and reduced
Balance the electrons lost or gained, to conform to the Law of Conservation of
Matter, by placing coefficients in front of the formulas containing the atoms
oxidized and reduced to both sides of the equation.
e. The remaining atoms are balanced by inspection
f. Balance oxygen, or hydrogen by adding H2O
g. Balance remaining hydrogen atoms by adding H+
h. Simplify
i. For basic reactions add the same number of OH- ions to both sides of the
equation as there are H+ ions.
j. Combine H+ and OH- ions to make water
k. Simplify again if necessary.
1+ 5+ 3(2-)=0 2(1+)+2-=0
HNO3 + Cu2O →
2+ + 2(1-)=0
Cu(NO3)2 +
NO
+
H2O
Balancing REDOX Reactions
I. Oxidation Number Method
a.
b.
c.
d.
Assign oxidation numbers to each element
Determine the elements oxidized and reduced
Balance the atoms that are oxidized and reduced
Balance the electrons lost or gained, to conform to the Law of Conservation of
Matter, by placing coefficients in front of the formulas containing the atoms
oxidized and reduced to both sides of the equation.
e. The remaining atoms are balanced by inspection
f. Balance oxygen, or hydrogen by adding H2O
g. Balance remaining hydrogen atoms by adding H+
h. Simplify
i. For basic reactions add the same number of OH- ions to both sides of the
equation as there are H+ ions.
j. Combine H+ and OH- ions to make water
k. Simplify again if necessary.
1+ 5+ 3(2-)=0 2(1+)+2-=0
HNO3 + Cu2O →
2(1-)=0
? + 2- =0
Cu(NO3)2 + NO + H2O
Balancing REDOX Reactions
I. Oxidation Number Method
a.
b.
c.
d.
Assign oxidation numbers to each element
Determine the elements oxidized and reduced
Balance the atoms that are oxidized and reduced
Balance the electrons lost or gained, to conform to the Law of Conservation of
Matter, by placing coefficients in front of the formulas containing the atoms
oxidized and reduced to both sides of the equation.
e. The remaining atoms are balanced by inspection
f. Balance oxygen, or hydrogen by adding H2O
g. Balance remaining hydrogen atoms by adding H+
h. Simplify
i. For basic reactions add the same number of OH- ions to both sides of the
equation as there are H+ ions.
j. Combine H+ and OH- ions to make water
k. Simplify again if necessary.
1+ 5+ 3(2-)=0 2(1+)+2-=0
HNO3 + Cu2O →
2(1-)=0
2 + 2- =0
Cu(NO3)2 + NO + H2O
Balancing REDOX Reactions
I. Oxidation Number Method
a.
b.
c.
d.
Assign oxidation numbers to each element
Determine the elements oxidized and reduced
Balance the atoms that are oxidized and reduced
Balance the electrons lost or gained, to conform to the Law of Conservation of
Matter, by placing coefficients in front of the formulas containing the atoms
oxidized and reduced to both sides of the equation.
e. The remaining atoms are balanced by inspection
f. Balance oxygen, or hydrogen by adding H2O
g. Balance remaining hydrogen atoms by adding H+
h. Simplify
i. For basic reactions add the same number of OH- ions to both sides of the
equation as there are H+ ions.
j. Combine H+ and OH- ions to make water
k. Simplify again if necessary.
1+ 5+ 3(2-)=0 2(1+)+2-=0
HNO3 + Cu2O →
oxidized
reduced
2+ 2(1-)=0
2 + 2- =0
Cu(NO3)2 + NO + H2O
Balancing REDOX Reactions
I. Oxidation Number Method
a.
b.
c.
d.
Assign oxidation numbers to each element
Determine the elements oxidized and reduced
Balance the atoms that are oxidized and reduced
Balance the electrons lost or gained, to conform to the Law of Conservation of
Matter, by placing coefficients in front of the formulas containing the atoms
oxidized and reduced to both sides of the equation.
e. The remaining atoms are balanced by inspection
f. Balance oxygen, or hydrogen by adding H2O
g. Balance remaining hydrogen atoms by adding H+
h. Simplify
i. For basic reactions add the same number of OH- ions to both sides of the
equation as there are H+ ions.
j. Combine H+ and OH- ions to make water
k. Simplify again if necessary.
+
2 + 2- =0
1+ 5+ 3(2-)=0 2(1+)+2-=0 2 2(1 )=0
HNO3 + Cu2O → 2 Cu(NO3)2 + NO +
oxidized
reduced
H2O
Balancing REDOX Reactions
I. Oxidation Number Method
a.
b.
c.
d.
Assign oxidation numbers to each element
Determine the elements oxidized and reduced
Balance the atoms that are oxidized and reduced
Balance the electrons lost or gained, to conform to the Law of Conservation of
Matter, by placing coefficients in front of the formulas containing the atoms
oxidized and reduced to both sides of the equation.
e. The remaining atoms are balanced by inspection
f. Balance oxygen, or hydrogen by adding H2O
g. Balance remaining hydrogen atoms by adding H+
h. Simplify
i. For basic reactions add the same number of OH- ions to both sides of the
equation as there are H+ ions.
j. Combine H+ and OH- ions to make water
k. Simplify again if necessary.
+
1+ 5+ 3(2-)=0 2(1+)+2-=0 2 2(1 )=0
HNO3 + 3 Cu2O → 3 (2) Cu(NO3)2
Oxidized3( -2e)
Reduced 2(+3)e
2 + 2- =0
+ NO + H2O
Balancing REDOX Reactions
I. Oxidation Number Method
a.
b.
c.
d.
Assign oxidation numbers to each element
Determine the elements oxidized and reduced
Balance the atoms that are oxidized and reduced
Balance the electrons lost or gained, to conform to the Law of Conservation of
Matter, by placing coefficients in front of the formulas containing the atoms
oxidized and reduced to both sides of the equation.
e. The remaining atoms are balanced by inspection
f. Balance oxygen, or hydrogen by adding H2O
g. Balance remaining hydrogen atoms by adding H+
h. Simplify
i. For basic reactions add the same number of OH- ions to both sides of the
equation as there are H+ ions.
j. Combine H+ and OH- ions to make water
k. Simplify again if necessary.
+
1+ 5+ 3(2-)=0 2(1+)+2-=0 2 2(1 )=0
2HNO3 + 3 Cu2O → 3 (2) Cu(NO3)2
Oxidized3( -2e)
Reduced 2(+3)e
2 + 2- =0
+ 2NO +
H2 O
Balancing REDOX Reactions
I. Oxidation Number Method
a.
b.
c.
d.
Assign oxidation numbers to each element
Determine the elements oxidized and reduced
Balance the atoms that are oxidized and reduced
Balance the electrons lost or gained, to conform to the Law of Conservation of
Matter, by placing coefficients in front of the formulas containing the atoms
oxidized and reduced to both sides of the equation.
e. The remaining atoms are balanced by inspection
f. Balance oxygen, or hydrogen by adding H2O
g. Balance remaining hydrogen atoms by adding H+
h. Simplify
i. For basic reactions add the same number of OH- ions to both sides of the
equation as there are H+ ions.
j. Combine H+ and OH- ions to make water
k. Simplify again if necessary.
1+ 5+ 3(2-)=0 2(1+)+2-=0
2HNO3
+
3 Cu2O
2+ 2(1-)=0
→ 3 (2) Cu(NO3)2
Oxidized3( -2e)
Reduced 2(+3)e
2 + 2- =0
+
2NO + H2O
Balancing REDOX Reactions
I. Oxidation Number Method
a.
b.
c.
d.
Assign oxidation numbers to each element
Determine the elements oxidized and reduced
Balance the atoms that are oxidized and reduced
Balance the electrons lost or gained, to conform to the Law of Conservation of
Matter, by placing coefficients in front of the formulas containing the atoms
oxidized and reduced to both sides of the equation.
e. The remaining atoms are balanced by inspection
f. Balance oxygen, or hydrogen by adding H2O
g. Balance remaining hydrogen atoms by adding H+
h. Simplify
i. For basic reactions add the same number of OH- ions to both sides of the
equation as there are H+ ions.
j. Combine H+ and OH- ions to make water
k. Simplify again if necessary.
1+ 5+ 3(2-)=0 2(1+)+2-=0
14HNO3
+
3 Cu2O
2+ 2(1-)=0
→ 3 (2) Cu(NO3)2
Oxidized3( -2e)
Reduced 2(+3)e
2 + 2- =0
+
2NO + 7 H2O
Balancing REDOX Reactions
I. Oxidation Number Method
a.
b.
c.
d.
Assign oxidation numbers to each element
Determine the elements oxidized and reduced
Balance the atoms that are oxidized and reduced
Balance the electrons lost or gained, to conform to the Law of Conservation of
Matter, by placing coefficients in front of the formulas containing the atoms
oxidized and reduced to both sides of the equation.
e. The remaining atoms are balanced by inspection
f. Balance oxygen, or hydrogen by adding H2O
g. Balance remaining hydrogen atoms by adding H+
h. Simplify
i. For basic reactions add the same number of OH- ions to both sides of the
equation as there are H+ ions.
j. Combine H+ and OH- ions to make water
k. Simplify again if necessary.
1+ 5+ 3(2-)=0 2(1+)+2-=0
14 HNO3 +
3 Cu2O
→
2 + 2- =0
2+ 2(1-)=0
6 Cu(NO3)2
Oxidized3( -2e)
Reduced 2(+3)e
+
2 NO + 7 H2O
OX # BALANCING EXAMPLE
MnO4 -
+ Cl- → Mn2+
+
Cl2
OX # BALANCING EXAMPLE
7+
MnO4 -
+ Cl- → Mn2+
+
Cl2
OX # BALANCING EXAMPLE
7+
MnO4 -
1-
+ Cl- → Mn2+
+
Cl2
OX # BALANCING EXAMPLE
7+
MnO4 -
1-
+ Cl- → Mn2+
0
+
Cl2
OX # BALANCING EXAMPLE
7+
MnO4 -
1-
+ Cl- → Mn2+
0
+
Cl2
OX # BALANCING EXAMPLE
1-
7+
MnO4 -
+ Cl- → Mn2+ +
reduced
0
Cl2
oxidized
OX # BALANCING EXAMPLE
1-
7+
MnO4 -
+ Cl- → Mn2+
0
+
Cl2
oxidized
reduced
Step C, balance atoms oxidized or reduced
OX # BALANCING EXAMPLE
1-
7+
MnO4 -
+
2
0
Cl- → Mn2+
+ Cl2
oxidized
reduced
Step C, balance atoms oxidized or reduced
OX # BALANCING EXAMPLE
1-
7+
MnO4 -
+
2 Cl-
0
→ Mn2+ +
- 2 e-
oxidized
Cl2
+ 5 e- reduced
Step d, balance electrons lost or gained.
common denominator between 5 and
2 is 10. Therefore multiply Mn on both
sides of the equation by two and Cl on
both sides by 5.
OX # BALANCING EXAMPLE
1-
7+
2 MnO4 -
+
5(2) Cl-
0
2 Mn2+ +
- 2 eoxidized
→
5 Cl2
+ 5 e- reduced
Step d, balance electrons lost or gained. The
common denominator between 5 and
2 is 10. Therefore multiply Mn on both
sides of the equation by 2 and Cl on
both sides by 5.
OX # BALANCING EXAMPLE
1-
7+
2 MnO4 -
+
5(2) Cl-
0
2 Mn2+
+
- 2 eoxidized
→
5 Cl2
+ 5 e- reduced
Step e, balance remaining elements by inspection.
There are 8 oxygen atoms on the left. Oxygen is
balanced by adding water to the appropriate side.
In this case since there are 8 oxygen atoms on the
reactant side which requires adding 8 water
molecules to the product side of the equation.
OX # BALANCING EXAMPLE
7+
1-
2 MnO4- + 10
Cl- →
2 Mn2+
- 2 e- oxidized
+ 5 e- reduced
+
0
5 Cl2 + 8H2O
Step e, balance remaining elements by inspection.
There are 8 oxygen atoms on the left. Oxygen is
balanced by adding water to the appropriate side.
In this case since there are 8 oxygen atoms on the
reactant side which requires adding 8 water
molecules to the product side of the equation. Now
the hydrogen atoms need to be balanced by
adding 16 H+ to the reactant side.
OX # BALANCING EXAMPLE
7+
1-
16 H++ 2 MnO4-+ 10 Cl- → 2 Mn2++
0
5 Cl2 + 8H2O
- 2 e- oxidized
+ 5 e- reduced
Step e, balance remaining elements by inspection.
There are 8 oxygen atoms on the left. Oxygen is
balanced by adding water to the appropriate side.
In this case since there are 8 oxygen atoms on the
reactant side which requires adding 8 water
molecules to the product side of the equation. Now
the hydrogen atoms need to be balanced by
adding 16 H+ to the reactant side.
Balancing REDOX Equations
by
The Half Reaction Method
Half Reaction Steps
1. Write separate equations (Half-reactions) for oxidized and reduced
substances.
2. For each half-reaction balance all elements, except hydrogen and
oxygen
a. Balance oxygen using H2O
b. Balance hydrogen using H+
c. Balance charge in each half-reaction by adding electrons
(reduction), or removing electrons (oxidation) to the appropriate half
reaction.
3. Multiply each half-reaction by an integer so that the number of
electrons lost equal the number of electrons gained
a. Add half-reactions, and simplify
b. For basic reactions add the same number of OH- ions to both
sides of the equation as there are H+ ions.
c. Combine H+ and OH- ions to make water
d. Simplify again if necessary.
Half Reaction Example
MnO4-
+
Fe2+
→
Mn2+
+
Fe3+
Half Reaction Example
MnO4-
+
Fe2+
→
Mn2+
Step 1, Write half reactions
+
Fe3+
Half Reaction Example
MnO4-
+
Fe2+
→
Mn2+
Step 1, Write half reactions
MnO4-
→
Mn2+
Fe2+
→
Fe3+
+
Fe3+
Half Reaction Example
MnO4-
+
Fe2+
→
Mn2+
+
Fe3+
Step 2a, Balance Oxygen by adding water.
MnO4-
→
Mn2+ + 4 H2O
Fe2+
→
Fe3+
Half Reaction Example
MnO4-
+
Fe2+
→
Mn2+
+
Fe3+
Step 2b, Balance hydrogen by adding H+.
8 H+ + MnO4-
→
Mn2+ + 4 H2O
Fe2+
→
Fe3+
Half Reaction Example
MnO4-
+
Fe2+
→
Mn2+
+
Fe3+
Step 2c, Balance charge by adding/removing e’s
8 H+ + MnO4-
→
Mn2+ + 4 H2O
Fe2+
→
Fe3+
In the top half equation the reactants have 7+ and the products
2+, adding 5 e’s to the reactant side gives 2+ on both sides.
In the bottom half equation the reactants have 2+ and the
products have 2+, removing 1 e from the reactant side gives 2+
on both sides.
Half Reaction Example
MnO4-
+
Fe2+
→
Mn2+
+
Fe3+
Step 2c, Balance charge by adding/removing e’s
8 H+ + MnO4-+ 5e-→
Fe2+
→
Mn2+ + 4 H2O
Fe3+
In the top half equation the reactants have 7+ and the products
2+, adding 5 e’s to the reactant side gives 2+ on both sides.
In the bottom half equation the reactants have 2+ and the
products have 2+, removing 1 e from the reactant side gives 2+
on both sides.
Half Reaction Example
MnO4-
+
Fe2+
→
Mn2+
+
Fe3+
Step 2c, Balance charge by adding/removing e’s
8 H+ + MnO4-+ 5e-→
Fe2+ - e →
Mn2+ + 4 H2O
Fe3+
In the top half equation the reactants have 7+ and the products
2+, adding 5 e’s to the reactant side gives 2+ on both sides.
In the bottom half equation the reactants have 2+ and the
products have 2+, removing 1 e from the reactant side gives 2+
on both sides.
Half Reaction Example
MnO4-
+
Fe2+
→
Mn2+
+
Fe3+
Step 2c, Balance charge by adding/removing e’s
8 H+ + MnO4-+ 5e-→
Fe2+ - e →
Mn2+ + 4 H2O
Fe3+
In the top half equation the reactants have 7+ and the products
2+, adding 5 e’s to the reactant side gives 2+ on both sides.
In the bottom half equation the reactants have 2+ and the
products have 2+, removing 1 e from the reactant side gives 2+
on both sides.
Half Reaction Example
MnO4- + Fe2+ → Mn2+ + Fe3+
Step 3, The common denominator between 5 and
1 is 5. Multiply the bottom half equation by 5
8 H+ + MnO4-+ 5e-→
Fe2+ - e →
Mn2+ + 4 H2O
Fe3+
Half Reaction Example
MnO4- + Fe2+ → Mn2+ + Fe3+
Step 4, Add the two half equations together
8 H+ + MnO4-+ 5e-→
2+ - e- →
Fe
5(
Mn2+ + 4 H2O
Fe3+)
Half Reaction Example
MnO4- + Fe2+ → Mn2+ + Fe3+
Step 4, Add the two half equations together
8 H+ + MnO4-+ 5e-→
2+ - e- →
Fe
5(
Mn2+ + 4 H2O
Fe3+)
8 H+ + MnO4- + 5 Fe2+ → 5 Fe3+ + Mn2+ + 4 H2O
Other REDOX Examples
HNO2
CN-
+
+
Cr2O72MnO4-
→
→
Cr2+
CNO-
+
+
NO3- (acidic)
MnO2 (basic)
Al(s) + OH- (aq) → Al(OH)4- (aq) + H2 (g) (acidic or basic)
Cl2 (g) → Cl- (aq)
+ ClO- (aq)
(basic)
Ag (s) + CN- + O2 → Ag(CN)2 - (aq) (basic)
Real Life Examples of REDOX
•REDOX reactions can be used to generate
electricity.
•REDOX reactions can be used to protect metals
from oxidation.
•REDOX reactions can be used to plate metals on
to other metals or surfaces.
The End