Download Common Oxidation–Reduction Reactions

Common Oxidation–Reduction Reactions
1.
2.
3.
4.
Combination reaction
Decomposition reaction
Displacement reaction
Combustion reaction
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Combination Reaction
A reaction in which two substances combine to form a
third substance.
For example:
2 Na(s) + Cl2(g)  2 NaCl(s)
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Decomposition Reaction
A reaction in which a single
compound reacts to give two or
more substances.
For example:
2HgO(s)  2Hg(l) + O2(g)
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Displacement Reaction
A reaction in which an element
reacts with a compound, displacing
another element from it.
For example:
Zn(s) + 2HCl(aq) 
H2(g) + ZnCl2(aq)
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Combustion Reaction
A reaction in which a substance reacts
with oxygen, usually with the rapid
release of heat to produce a flame.
For example:
4 Fe(s) + 3 O2(g)  2 Fe2O3(s)
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Combustion of Organic Compounds
When organic compound reacts with oxygen, O2,
carbon dioxide, CO2, and water, H2O, are usually
produced.
If an organic substance contains nitrogen,
nitrogen gas, N2, is generated in addition to
carbon dioxide, CO2, and water, H2O.
If an organic substance contains sulfur, sulfur
dioxide gas, SO2, is generated in addition to
carbon dioxide, CO2, and water, H2O.
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Combustion of Organic Compounds
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g)
C2H5OH(l) + 3 O2(g)  2 CO2(g) + 3 H2O(g)
4 CH3NH2(g) + 9 O2(g)  4 CO2(g) + 10 H2O(g) + 2 N2(g)
2 C2H5SH(l) + 9 O2(g)  4 CO2(g) + 6 H2O(g) + 2 SO2(g)
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Oxidation Number
The actual ionic charge or a hypothetical charge
assigned to an element in the substance using a
set of rules.
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Rules for Assigning Oxidation Numbers
1. Pure Elements: The oxidation number of an
atom in an element is zero.
2. Monatomic ions: The oxidation number of an
atom in a monatomic ion equals the charge on
the ion.
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?
What are the oxidation numbers of
phosphorus in the following:
a) P4
b) P3c) P
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?
What are the oxidation numbers of
phosphorus in the following:
0
a) P4 (white phosphorus): element
-3
b) P3- (phosphide ion): monoatomic ion
0
c) P (red phosphorus): element
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Rules for Assigning Oxidation Numbers
3. Fluorine: The oxidation number of fluorine is −1 in
all of its compounds.
4. Oxygen: The oxidation number of oxygen is −2 in
most of its compounds.
Exceptions:
1) O in H2O2 and other peroxides:
the oxidation number of oxygen is −1;
2) binary compounds of O and F: OF2 and O2F2
the oxidation number of oxygen has to be calculated.
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Rules for Assigning Oxidation Numbers
5. Hydrogen: The oxidation number of hydrogen is +1
in most of its compounds.
Exceptions:
H in metal hydrides (binary compounds of H and any
metal):
the oxidation number of hydrogen is −1.
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Rules for Assigning Oxidation Numbers
6. Chlorine, bromine and iodine: The oxidation
number of these elements is usually −1 (similar to
fluorine)
Exceptions:
1) binary compounds of O and a halogen:
the halogen’s oxidation number has to be calculated;
2) binary compounds of two different halogens:
more electronegative halogen gets the oxidation
number −1, oxidation number of the less
electronegative halogen has to be calculated.
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Rules for Assigning Oxidation Numbers
7. Compounds: The sum of the oxidation numbers of
the atoms in a compound is zero.
8. Ions: The sum of the oxidation numbers of the
atoms in a polyatomic ion equals the charge on the
ion.
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?
Assign the oxidation numbers for every
element in the following substances:
OF2
KMnO4
O2F2
K2MnO4
CaH2
NH4+
CO2
Cr2O72−
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x
-1
O
F2
1∙x + 2(−1) = 0
x-2=0
x=2
Oxidation number of O in OF2 = +2
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x
-1
O2
F2
2∙x + 2(−1) = 0
2x - 2 = 0
x=1
Oxidation number of O in OF2 = +1
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CaH2 is a binary ionic compound composed
of Ca2+ and H- ions.
Oxidation number of Ca in CaH2 = +2
Oxidation number of H in CaH2 = -1
1(+2) + 2(-1) = 0
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x
-2
C
O2
1∙x + 2(−2) = 0
x-4=0
x=4
Oxidation number of C in CO2 = +4
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+1
K
x
Mn
-2
O4
1(+1) + 1∙x + 4(−2) = 0
1+x-8=0
x=7
Oxidation number of Mn in KMnO4 = +7
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+1
K2
x
Mn
-2
O4
2(+1) + 1∙x + 4(−2) = 0
2+x-8=0
x=6
Oxidation number of Mn in K2MnO4 = +6
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x
+1
N
H4
1∙x + 4(+1) = +1 (ion charge)
x+4=0
x = −3
Oxidation number of N = −3
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x
Cr2
-2
O7
2∙x + 7(−2) = −2 (ion charge)
2x − 12 = 0
x=6
Oxidation number of Cr = +6
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Balancing Simple Oxidation−Reduction
Reactions: Half−Reaction Method
1. Determine the oxidation numbers for every
element involved in the reaction.
2. Identify the elements that change their oxidation
numbers.
Example:
0
+1
+2
0
Zn(s) + Ag+(aq)  Zn2+(aq) + Ag(s)
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Balancing Simple Oxidation−Reduction
Reactions: Half−Reaction Method
3. Split the reaction into two half-reactions:
• oxidation half-reaction (involves a loss of
electrons and an increase in oxidation number):
0
+2
Zn(s)  Zn2+(aq) + 2 e• reduction half-reaction (involves a gain of
electrons and a decrease in oxidation number):
+1
0
Ag+(aq) + 1 e-  Ag(s)
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Balancing Simple Oxidation−Reduction
Reactions: Half−Reaction Method
3. OIL RIG mnemonic:
• Oxidation Is Loss (of electrons):
0
+2
Zn(s)  Zn2+(aq) + 2 e•
Reduction Is Gain (of electrons):
+1
0
Ag+(aq) + 1 e-  Ag(s)
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Balancing Simple Oxidation−Reduction
Reactions: Half−Reaction Method
4. Determine the reducing agent (species that is
itself oxidized and the oxidizing agent (species
that oxidizes another species; it is itself reduced):
Zn is oxidized from 0 to +2. Zn is the reducing
agent.
Ag+ is reduced from +1 to 0. Ag is the oxidizing
agent.
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Balancing Simple Oxidation−Reduction
Reactions: Half−Reaction Method
5. Since the electrons lost in oxidation are the same
as those gained in reduction, we need each
half−reaction to have the same number of
electrons. To do this, multiply the reduction
half−reaction by 2:
0
+2
Zn(s)  Zn2+(aq) + 2 e+1
0
2 Ag+(aq) + 2 e-  2 Ag(s)
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Balancing Simple Oxidation−Reduction
Reactions: Half−Reaction Method
6. Add the two half−reactions together. When the
half−reactions are added, the electrons cancel:
Zn(s) + 2 Ag+(aq) + 2 e-  Zn2+(aq) + 2 e- + 2 Ag(s)
Zn(s) + 2 Ag+(aq)  Zn2+(aq) + 2 Ag(s)
The equation is now balanced!
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?
Balance the following oxidation−reduction
reaction. Identify the oxidizing agent and
the reducing agent.
FeI3(aq) + Mg(s)  Fe(s) + MgI2(aq)
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?
Balance the following oxidation−reduction
reaction. Identify the oxidizing agent and
the reducing agent.
+3 −1
0
0
+2 −1
FeI3(aq) + Mg(s)  Fe(s) + MgI2(aq)
The oxidation numbers are given above each of the
elements.
Mg is oxidized from 0 to +2. Mg is the reducing agent.
Fe3+ is reduced from +3 to 0. Fe3+ is the oxidizing agent.
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Now, write the half−reactions. Since iodide, I−, does not change
its oxidation number, treat it as a spectator ion and omit at this
point:
Mg(s)  Mg2+(aq) + 2 e−
Fe3+(aq) + 3 e−  Fe(s)
(oxidation half−reaction)
(reduction half−reaction)
To have the same number of electrons gained and lost, multiply
the oxidation half−reaction by 3 and the reduction half−reaction
by 2:
3 Mg(s)  3 Mg2+(aq) + 6 e−
(oxidation half−reaction)
2 Fe3+(aq) ) + 6 e−  2 Fe(s)
(reduction half−reaction)
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Add the half−reactions together:
2 Fe3+(aq) + 6 e− + 3 Mg(s)  2 Fe(s) + 3 Mg2+(aq) + 6 e−
2 Fe3+(aq) + 3 Mg(s)  2 Fe(s) + 3 Mg2+(aq)
Now, return the spectator ion, I−:
2 FeI3(aq) + 3 Mg(s)  2 Fe(s) + 3 MgI2(aq)
oxidizing
agent
reducing
agent
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