Download 4.6 Oxidation-Reduction (Redox) Reactions Oxidation Reduction

4.6 Oxidation-Reduction (Redox) Reactions
Oxidation Reduction Reactions
Oxidation-Reduction Reactions are commonly called
Redox reactions and are those involving the oxidation
of one chemical species with simultaneous reduction of
another species.
Oxidation and reduction must occur together. They
cannot exist alone
Is a reaction in aqueous solution involving an exchange
of electrons between two species
oxidation - the loss of one or more electrons by a
substance
reduction - the gain of one or more electrons by a
substance.
e.g., Zn (s) + 2H+ (aq)
Zn (s) + 2H+ (aq)
Zn2+ (aq) + H2 (g)
Zn2+ (aq) + H2 (g)
can be broken down into two half reactions:
oxidation (loss of e-):
Zn (s)
Zn 2+ (aq) + 2e-
reduction (gain of e-):
2H+ (aq) + 2e-
H2 (g)
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Oxidation Number
can help identify a redox reaction by looking for
change in oxidation number of an element in a
compound
oxidation number
provides a measure of whether the atom is
neutral, electron-rich or electron-poor
comparing the oxidation number of an atom before
and after reaction allows us to determine whether
the atom has gained or lost electrons
convenient way to keep track of electrons in a
redox reaction
Assigning Oxidation Numbers: All atoms have an
oxidation number regardless of whether it carries an
ionic charge.
1.
An atom in its elemental state has an oxidation number of
zero.
2.
An atom in a monatomic ion has an oxidation number
identical to its charge.
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--Assigning Oxidation Number
3. Hydrogen, oxygen and halogens in a polyatomic ion
or in a molecular compound usually has the same
oxidation number it would have if it were a monatomic
ion.
Hydrogen has an oxidation number of +1
BUT in hydrides , hydrogen is
1 (i.e, NaH)
Oxygen usually has an oxidation number of 2.
BUT in peroxides, oxygen is 1 (i.e., H2O2)
Halogens usually have an oxidation number of 1.
BUT when bonded to oxygen, they have positive oxidation
numbers (i.e, ClO2,,, BrO3- )
--assigning oxidation number
4.
The algebraic sum of the oxidation numbers in a
neutral compound must be zero.
for H2SO4
2(+1) + (?) + 4( 2) = 0 (net charge)
oxid. # of S = ? = 0
5.
2(+1)
4( 2) = +6
in a polyatomic ion, the sum must be equal to the
ion charge
for ClO4
(?) + 4( 2) = 1 (net charge)
Oxid. # of Cl = ? = 1
4( 2) = +7
PROBLEM: assign an oxidation # for each atom in the following
(a) Cr2O72-
(b) Fe SO3
(c) AlH3
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Other examples
?
?
?
?
Identifying Redox Reactions
First determine oxidation numbers of each species in the reaction and
then identify the oxidation and reduction processes
A. Oxidation and reduction occur together.
Whenever an atom loses electrons (is oxidized) another atom
must gain those electrons (be reduced).
B. Reducing Agent- the substance that causes reduction to occur.
loses one or more electrons and undergoes oxidation
oxidation # of atom increases
metals act as good reducing agents [ Na (s) , Fe (s) , Ca (s)]
- Group 1A metals give up one electron to become monopositive ion (Na+)
- Group 2A metals give up two electrons to become a dipositive ion (Ca2+)
- Group 3A give up three electrons (Al3+) ;
- Transition metals give up a variable number of electrons (hard to predict.
C. Oxidizing Agent- the substance that causes oxidation to occur
gains one or more electrons and undergoes reduction
oxidation # of atom decreases
reactive non-metals act as oxidizing agents [ Cl2 (g), O2 (g) ]
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Take, for example, the following reaction:
eg.4 Fe (s)
+
4 Fe (s)
3 O2 (g)
+
2 Fe2O3 (s)
3 O2 (g)
2 Fe (s)
Identify what is oxidized, what is reduced.
EXAMPLE: Identify the species oxidized, the species reduced, and the oxidizing
and reducing agents in this net ionic equation.
16 H+ (aq) + 5 Sn2+ (aq) + 2 MnO4- (aq)
Oxid. #
+1
+2
2 Mn2+ (aq) + 5 Sn4+ (aq) + 8 H2O (l)
+7 -2
+2
+4
+1 -2
reduction
oxidation
Here, Sn is being oxidized . Sn2+ is the reducing agent
Mn is being reduced. MnO4- is the oxidizing agent
PROBLEM: Identify the species oxidized and reduced and the oxidizing and
reducing agent in the following net ionic equation:
3 SO42- (aq) + 12 H+ (aq) + 2 Al (s)
Answer:
2 Al3+ (aq) + 3 H2SO3 (aq) + 3 H2O (l)
S is being reduced; SO42- is the oxidizing agent
Al is being oxidized; Al (s) is the reducing agent
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