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Transcript
Chapter 4
Oxidation- Reduction Reactions
What is a redox reaction?
A reaction that involves a transfer of
electrons.
 Oxidation – lose of electrons



Reduction – gain of electrons


Charge goes up
Charge goes down
OIL RIG or LEO says GER
Oxidizing and Reducing Agents
Oxidizing Agent – the
species that causes the
oxidation (the species
being reduced)
 Reducing Agent – the
species that causes the
reduction (the species
being oxidized)

Assigning Oxidation Numbers







1. The oxidation state of an atom in an element is ZERO.
2. The oxidation state of a monatomic ion is the same as its charge.
3. Alkali metals are always +1 and alkaline earth metals are +2.
4. In its compounds, fluorine is always assigned an oxidation state of
-1.
5. Oxygen is usually assigned an oxidation state of -2. Exceptions to
this rule include peroxides (compound containing the O22- group),
where each oxygen is assigned an oxidation state of -1, as in
hydrogen peroxide (H2O2), and OF2 in which oxygen is assigned a +2
oxidation state.
6. In its covalent compounds with nonmetals, hydrogen is assigned
an oxidation state of +1. Metal hydrides are an exception; H is at
the end of the chemical formula since it has an oxidation state of 1-.
7. The sum of the oxidation states must be zero for an electrically
neutral compound. For a polyatomic ion, the sum of the oxidation
states must equal the charge of the ion.
Exercise 16
Assign oxidation states to all atoms in
the following.
C = 4 O = -2
 a. CO2
S = 6 F = -1
 b. SF6
N = +5 O = -2
 c. NO3Hint: Columns 5A and
7A have odd charges,
columns 4A and 6A
have even charges.
Exercise 17
For this reaction, identify the atoms
that are oxidized and reduced, and
specify the oxidizing and reducing
agents.
2Al(s) + 3I2(s) → 2AlI3(s)
oxidized
reduced
reducing agent oxidizing agent
Exercise 18

Metallurgy, the process of producing a
metal from its ore, always involves
oxidation-reduction reactions. In the
metallurgy of galena (PbS), the principal
lead-containing ore, the first step is the
conversion of lead sulfide to its oxide (a
process called roasting):
2PbS(s) + 3O2(g) → 2PbO(s) + 2SO2(g)
oxidized
reduced
reducing agent oxidizing agent
Exercise 18 - Continued

The oxide is then treated with carbon
monoxide to produce the free metal:
PbO(s) + CO(g) → Pb(s) + CO2(g)
reduced
oxidized
oxidizing agent
reducing agent
Balancing Redox Reactions
by Half Reaction Method
1.
2.
3.
4.
5.
6.
7.
8.
Divide the equation into oxidation and reduction half
reactions. Balance all elements besides hydrogen and
oxygen.
Balance O’s by adding H2O’s to the appropriate side of each
equation.
Balance H’s by adding H+
Balance the charge by adding electrons.
Multiply the half reactions to make electrons equal for both
half-reactions.
Cancel out any common terms and recombine the two half
reactions.
IF BASIC, neutralize any H+ by adding the SAME NUMBER
of OH- to EACH side of the balanced equation. [This creates
some waters that will cancel!]
CHECK!!
Acidic Redox Balance
MnO4-(aq) + I-(aq)  Mn+2(aq) + I2(aq)
(+7)
(+2)
2( 5e- + 8 H+ + MnO4-  Mn2+ + 4 H2O )
5( 2 I-  I2 + 2e- )
(-2)
(0)
10e- + 16 H+ + 2MnO4-  2Mn2+ + 8 H2O
10 I-  5 I2 + 10 e10 I- + 16 H+ + 2MnO4-  2Mn2+ + 8 H2O + 5 I2
Basic Redox Balance
Ag(s) + CN- + O2  Ag(CN)2-(aq)
(-2)
(-1)
4( Ag + 2 CN-  Ag(CN)2- + e-)
4e- + 2 H2O + O2  4 OH(0)
(-4)
4 Ag + 8 CN-  4Ag(CN)2- + 4 e4 e- + 2 H2O + O2  4 OH4 Ag + 8 CN- + 2 H2O + O2  4Ag(CN)2- + 4 OH-
Basic Redox Balance
Ag(s) + CN- + O2  Ag(CN)2-(aq)
(-2)
(-1)
4( Ag + 2 CN-  Ag(CN)2- + e-)
4e- + 4H+ + O2  2H20
(+4)
(0)
4 Ag + 8 CN-  4Ag(CN)2- + 4 e4 e- + 4H++ O2  2H2O
4 Ag + 8 CN- + 4H+ + O2  4Ag(CN)2- + 2 H2O
Now what…

4 Ag + 8 CN- + 4H+ + O2  4Ag(CN)2- + 2 H2O + 4 OH4 OH=
4 H2O


Cancel 2 extra
waters on both
sides.
Every H+ add OH- = H2O
Cancel extra water.
4
Ag + 8 CN- + 2H2O + O2  4Ag(CN)2- + 4 OH-