Chem 1011
Chem 1011
Dr. L. Dawe
Dr. L. Dawe
Winter 2010
January 11, 2010 – Lecture 2
•
Lecture 2
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18.2 Balancing Oxidation-Reduction Reactions
– Half-reaction method of balancing aqueous redox
equations in acidic and basic solutions
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Chem 1011
Dr. L. Dawe
Winter 2010
Identifying Redox Reactions
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In a chemical reaction, the change in oxidation
number can help you identify which species is
oxidized, and which is reduced.
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An increase in oxidation number of an atom
signifies oxidation (LEO)
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A decrease in oxidation number of an atom
signifies reduction (GER)
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Chem 1011
Dr. L. Dawe
___________________________________
Winter 2010
Identifying Redox Reactions
Oxidizing Agent: causes something to be oxidized. It
does this by taking electrons from it. When it takes
these electrons it undergoes reduction.
Oxidizing agents gain electrons and become reduced.
Reducing Agent: causes something to be reduced. It
does this by giving electrons. When it gives these
electrons it undergoes oxidation.
Reducing agents lose electrons and become oxidized.
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1
Chem 1011
Dr. L. Dawe
Chem 1011
Dr. L. Dawe
Winter 2010
Identifying Redox Reactions
Problem: For each of the following write the net ionic
equation, and the half-reactions. For each of the
half reactions state which is the oxidation halfreaction, which is the reduction half-reaction, what
is the oxidizing agent and what is the reducing
agent.
Lecture 2
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(a) Pb(s) + Cu(NO3)2(aq) Cu(s) + Pb(NO3)2(aq)
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(b) Cu(s) + 2 AgNO3(aq) 2 Ag(s) + Cu(NO3)2(aq)
Chem 1011
Dr. L. Dawe
Winter 2010
Identifying Redox Reactions
Pb(s) +
+2
NO3-
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Solution:
Total Ionic Equation:
Cu2+(aq)
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(aq) Cu(s) +
Pb2+(aq)
+2
NO3- (aq)
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Chem 1011
Dr. L. Dawe
Winter 2010
Identifying Redox Reactions
Solution:
Total Ionic Equation:
Pb(s) +
Cu2+(aq)
+2
NO3-
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Spectator Ions
(aq) Cu(s) +
Pb2+(aq)
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+2
NO3- (aq)
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2
Chem 1011
Dr. L. Dawe
Chem 1011
Dr. L. Dawe
Winter 2010
Identifying Redox Reactions
Solution:
Total Ionic Equation:
Pb(s) +
Cu2+(aq)
+2
NO3-
(aq) Cu(s) +
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+2
NO3- (aq)
Net Ionic Equation:
Pb(s) + Cu2+(aq)
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Spectator Ions
Pb2+(aq)
Lecture 2
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Cu(s) + Pb2+(aq)
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Chem 1011
Dr. L. Dawe
Winter 2010
Identifying Redox Reactions
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Solution:
Oxidation Half-Reaction:
Pb(s)
O.N. = 0
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Pb2+(aq) + 2 eO.N. = +2
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Chem 1011
Dr. L. Dawe
Winter 2010
Identifying Redox Reactions
Solution:
Oxidation Half-Reaction:
Pb(s)
O.N. = 0
Pb(s) loses two electrons
Pb2+(aq) + 2 eO.N. = +2
Note: Increase in oxidation number loss of electrons
Reducing agents lose electrons and become oxidized.
Pb(s) is the reducing agent.
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Chem 1011
Dr. L. Dawe
Chem 1011
Dr. L. Dawe
Winter 2010
Identifying Redox Reactions
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Solution:
Reduction Half-Reaction:
Cu2+(aq) + 2 eO.N. = +2
Lecture 2
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Cu(s)
O.N. = 0
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Chem 1011
Dr. L. Dawe
Winter 2010
Identifying Redox Reactions
Solution:
Reduction Half-Reaction:
Cu2+(aq) + 2 eO.N. = +2
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Cu2+(aq) gains two electrons
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Cu(s)
O.N. = 0
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Note: Decrease in oxidation number gain of electrons
Oxidizing agents gain electrons and become reduced.
Cu2+(aq) is the oxidizing agent.
Chem 1011
___________________________________
Dr. L. Dawe
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Winter 2010
Identifying Redox Reactions
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http://cwx.prenhall.com/petrucci/medialib/
Solution:
(b) Cu(s) + 2 AgNO3(aq) 2 Ag(s) + Cu(NO3)2(aq)
media_portfolio/text_images/024_SILVER
CRYSTA.MOV
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http://cwx.prenhall.com/petrucci/medialib/media_portfolio/text_images/024_SILVERCRYSTA.MOV
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Chem 1011
Dr. L. Dawe
Chem 1011
Dr. L. Dawe
Winter 2010
Identifying Redox Reactions
Lecture 2
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Combustion reactions are also a type of redox reaction!
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CH4 + 2 O2 → CO2 + 2 H2O
−4 +1
0
+4 –2
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+1 −2
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Chem 1011
Dr. L. Dawe
Winter 2010
Identifying Redox Reactions
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Combustion reactions are also a type of redox reaction!
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CH4 + 2 O2 → CO2 + 2 H2O
−4 +1
0
+4 –2
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+1 −2
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oxidation
reduction
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• Oxidation and reduction must occur simultaneously.
• If an atom loses electrons, another atom must take them.
Chem 1011
Dr. L. Dawe
Winter 2010
Balancing Oxidation-Reduction Reactions
•
•
•
Balancing redox reactions by inspection can be
challenging because we need to balance for both
mass and charge.
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Redox reactions taking place in aqueous solution
can be balanced by applying a special procedure
knows as the half-reaction method of balancing.
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The steps involved differ slightly for reactions
taking place in acidic and basic solutions. We will
look at both cases.
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Chem 1011
Dr. L. Dawe
Chem 1011
Dr. L. Dawe
Winter 2010
Balancing Redox Reactions – Acid Solution
Example: Write the balanced equation for this reaction
in acidic solution.
SO32-(aq) + MnO4-(aq) → SO42-(aq) + Mn2+(aq)
1. Assign
oxidation
numbers to all
species and
identify your
redox couples.
SO32-(aq) + MnO4-(aq) → SO42-(aq) + Mn2+(aq)
2. Separate the
overall reaction
into two halfreactions.
Oxidation: SO32-(aq) → SO42-(aq)
Chem 1011
+4 -2
+7 -2
+6 -2
+2
oxidation
Lecture 2
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reduction
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Reduction: MnO4-(aq) → Mn2+(aq)
Dr. L. Dawe
Winter 2010
Balancing Redox Reactions – Acid Solution
3.(a) Balance all SO32-(aq) → SO42-(aq)
elements except MnO -(aq) → Mn2+(aq)
4
O and H.
(Both already balanced, except O and H)
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3.(b) Balance O SO32-(aq) + H2O(l) → SO42-(aq)
atoms by adding
H2O with the
MnO4-(aq) → Mn2+(aq) + 4 H2O(l)
appropriate
coefficients
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3.(c) Balance H SO32-(aq) + H2O(l) → SO42-(aq) + 2H+(aq)
atoms by adding
H+ with the
MnO4-(aq) + 8 H+(aq) → Mn2+(aq) + 4 H2O(l)
appropriate
coefficients
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Do these steps in this order!!!
Chem 1011
Dr. L. Dawe
Winter 2010
Balancing Redox Reactions – Acid Solution
4. Add the
SO32-(aq) + H2O(l) → SO42-(aq) + 2H+(aq) + 2 enumber of
electrons
MnO4-(aq) + 8 H+(aq) + 5 e-→ Mn2+(aq) + 4 H2O(l)
necessary to get
the same
electric charge
on either side of
each halfreaction
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6
Chem 1011
Chem 1011
Dr. L. Dawe
Dr. L. Dawe
Winter 2010
Balancing Redox Reactions – Acid Solution
5. Multiply
the halfreactions to
obtain the
lowest
common
multiple for
the
electrons.
5 x [SO32-(aq) + H2O(l) → SO42-(aq) + 2H+(aq) + 2 e- ]
----------------------------------------------------------------------5SO32-(aq) + 5H2O(l) → 5SO42-(aq) + 10H+(aq) + 10e-
Lecture 2
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2 x [MnO4-(aq) + 8 H+(aq) + 5 e-→ Mn2+(aq) + 4 H2O(l) ]
----------------------------------------------------------------------2MnO4-(aq) + 16 H+(aq) + 10 e-→ 2 Mn2+(aq) + 8H2O(l)
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Chem 1011
Dr. L. Dawe
Winter 2010
Balancing Redox Reactions – Acid Solution
6. Add the
two halfreactions
together,
cancelling
electrons
and other
species as
necesssary.
5SO32-(aq) + 5H2O(l) → 5SO42-(aq) + 10H+(aq) + 10e-
7. Verify
mass and
charge
balance
Reactants (Overall)
5S
2 Mn
23 O
6H
Charge = -6
Chem 1011
2MnO4-(aq) + 16 H+(aq) + 10 e-→ 2 Mn2+(aq) + 8 H2O(l)
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5 SO32-(aq) + 2 MnO4-(aq) + 6 H+(aq) →
5 SO42-(aq) + 2 Mn2+(aq) + 3 H2O(l)
Products (Overall)
5S
2 Mn
23 O
6H
Charge = -6
Dr. L. Dawe
Winter 2010
Balancing Redox Reactions – Acid Solution
Problem: Balance the following equation in acidic
solution.
Fe2+(aq) + MnO4-(aq) → Fe3+(aq) + Mn2+(aq)
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Chem 1011
Dr. L. Dawe
Lecture 2
Problem: Balance the following equation in acidic solution.
Fe2+(aq) + MnO4-(aq) → Fe3+(aq) + Mn2+(aq)
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Chem 1011
Dr. L. Dawe
Chem 1011
Dr. L. Dawe
Winter 2010
Balancing Redox Reactions – Basic Solution
Example: Write the balanced equation for this reaction
in basic solution.
CN-(aq) + MnO4-(aq) → OCN-(aq) + MnO2(s)
1. Assign
oxidation
numbers to all
species and
identify your
redox couples.
2. Separate the
overall reaction
into two halfreactions.
Chem 1011
CN-(aq) + MnO4-(aq) → OCN-(aq) + MnO2(s)
+2 -3
+7 -2
-2 +4 -3
+4 -2
oxidation
Lecture 2
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reduction
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Oxidation: CN-(aq) → OCN-(aq)
Reduction: MnO4-(aq) → MnO2(s)
Dr. L. Dawe
Winter 2010
Balancing Redox Reactions – Basic Solution
3.(a) Balance all CN-(aq) → OCN-(aq)
elements except MnO4-(aq) → MnO2(s)
O and H.
(Both already balanced, except O and H)
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3.(b) Balance O CN-(aq) + H2O(l) → OCN-(aq)
atoms by adding
H2O with the
MnO4-(aq) → MnO2(s) + 2 H2O(l)
appropriate
coefficients
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Chem 1011
Dr. L. Dawe
Winter 2010
Balancing Redox Reactions – Basic Solution
3.(c) Balance CN-(aq) + H2O(l) → OCN-(aq) + 2 H+(aq)
H atoms by
adding H+ with MnO -(aq) + 4 H+(aq) → MnO (s) + 2 H O(l)
4
2
2
the
appropriate
coefficients
3.(d) Convert
from acidic to
basic medium
by adding
sufficient OH-,
to neutralize
all the H+, to
each side of
the halfreactions.
CN-(aq) + H2O(l) + 2 OH-(aq) →
OCN-(aq) + 2 H+(aq) + 2 OH-(aq)
MnO4-(aq) + 4 H+(aq) + 4 OH-(aq) →
MnO2(s) + 2 H2O(l) + 4 OH-(aq)
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Chem 1011
Dr. L. Dawe
Chem 1011
Dr. L. Dawe
Winter 2010
Balancing Redox Reactions – Basic Solution
3.(e)
Combine
H+ and
OH- where
appropriate
to form
H2O.
Simplify
the halfreactions.
CN-(aq) + H2O(l) + 2 OH-(aq) → OCN-(aq) + 2 H2O(l)
------------------------------------------------------------------------CN-(aq) + 2 OH-(aq) → OCN-(aq) + H2O(l)
Lecture 2
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OH-(aq)
MnO4-(aq)
+ 4 H2O(l) → MnO2(s) + 2 H2O(l) + 4
------------------------------------------------------------------------MnO4 (aq) + 2 H2O(l) → MnO2(s) + 4 OH-(aq)
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Chem 1011
Dr. L. Dawe
Winter 2010
Balancing Redox Reactions – Basic Solution
4. Add the
CN-(aq) + 2 OH-(aq) → OCN-(aq) + H2O(l) + 2 enumber of
electrons
MnO4-(aq) + 2 H2O(l) + 3 e- → MnO2(s) + 4 OH-(aq)
necessary to
get the same
electric charge
on either side
of each halfreaction
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Chem 1011
Dr. L. Dawe
Winter 2010
Balancing Redox Reactions – Basic Solution
5. Multiply
the halfreactions to
obtain the
lowest
common
multiple for
the
electrons.
3 x [CN-(aq) + 2 OH-(aq) → OCN-(aq) + H2O(l) + 2 e-]
----------------------------------------------------------------------3 CN-(aq) + 6 OH-(aq) → 3 OCN-(aq) + 3 H2O(l) + 6 e-
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2 x [MnO4-(aq) + 2 H2O(l) + 3 e- → MnO2(s) + 4 OH-(aq)]
----------------------------------------------------------------------2 MnO4-(aq) + 4 H2O(l) + 6 e- → 2 MnO2(s) + 8 OH-(aq)
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Chem 1011
Chem 1011
Dr. L. Dawe
Dr. L. Dawe
Winter 2010
Balancing Redox Reactions – Basic Solution
6. Add the
two halfreactions
together,
cancelling
electrons
and other
species as
necesssary.
3 CN-(aq) + 6 OH-(aq) → 3 OCN-(aq) + 3 H2O(l) + 6 e-
7. Verify
mass and
charge
balance
Reactants (Overall)
2 Mn
3C
3N
9O
2H
Charge = -5
Chem 1011
2 MnO4-(aq) + 4 H2O(l) + 6 e- → 2 MnO2(s) + 8 OH-(aq)
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2 MnO4-(aq) + 3 CN-(aq) + H2O(l) →
2 MnO2(s) + 3 OCN-(aq) + 2 OH-(aq)
Products (Overall)
2 Mn
3C
3N
9O
2H
Charge = -5
Dr. L. Dawe
Winter 2010
Balancing Redox Reactions – Basic Solution
Problem: Balance the following equation in basic
solution.
Lecture 2
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S(s) + OCl-(aq) → SO32-(aq) + Cl-(aq)
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Chem 1011
Dr. L. Dawe
Winter 2010
Looking Ahead: January 13, 2010 – Lecture 3
•
•
9.5 Covalent Bonding: Lewis Structures
– Single Covalent Bonds
– Double and Triple Covalent Bonds
– Covalent Bonding: Models and Reality
9.6 Electronegativity and Bond Polarity
– Electronegativity
– Bond Polarity, Dipole Moment and Percent Ionic
Character
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11
Chem 1011
Dr. L. Dawe
Lecture 2
Problem: Balance the following equation in basic solution.
S(s) + OCl-(aq) → SO32-(aq) + Cl-(aq)
12