Download Chapter 4: Oxidation and Reduction MH5 4

Unit 3
Oxidation and Reduction
Chemistry 020, R. R. Martin
1 Introduction
Another important type of reaction in aqueous solution
involves the transfer of electrons between two
species. This is called an oxidation-reduction or a
redox reaction.
What happens when zinc pellets are added to an
acid? The zinc “dissolves”, and a gas is formed.
The net ionic equation for this process is
Zn(s) + 2H+(aq) Æ Zn2+(aq) + H2(g)
Zn loses electrons, and H+ gains electrons.
You can think of this reaction as two separate
processes (half-equations or half reactions)
Oxidation (loss of electrons): Zn(s) Æ Zn2+ + 2eand
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Reduction (gain of electrons): 2H+(aq) + 2e- Æ H2(g)
OIL RIG: Oxidation is loss, reduction is gain
In reality, free electrons are not involved in redox
reactions; oxidation and reduction always occur
together.
All the electrons produced in the oxidation half
reaction have to be consumed in the reduction half
reaction. (the total number of electrons does not change !!!)
When you add the two half-equations together, the
electrons have to “cancel out”:
Oxidation (loss of electrons): Zn(s) Æ Zn2+ + 2eReduction (gain of electrons): 2H+(aq) + 2e- Æ H2(g)
Zn(s) + 2H+(aq) Æ Zn2+(aq) + H2(g)
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The species that accepts electrons is called the
oxidizing agent (“oxidant”, it oxidizes the other species, but in
doing this it is reduced)
The species that donates electrons is called reducing
agent (“reductant”, it reduces the other species, but in doing that it
is oxidized). The oxidant oxidizes the reductant and the reductant reduces the
oxidant.
red agreduced + ox agoxidized Æ red agoxidized + ox agreduced
Here, Zn is the reducing agent, H+ is the oxidizing
agent.
Zn(s) + 2H+(aq) Æ Zn2+(aq) + H2(g)
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2. Oxidation Numbers are required for the proper
balancing of redox equations. They simplify the
electron bookkeeping. Each atom in a compound can
be assigned an oxidation number.
Rules for assigning Oxidation Numbers :
1. Any allotrope of any element in the free state has an
oxidation number of zero. (i.e. C(Diamond) , C(Graphite) , C(Gas) for
C or O2(Gas) , O3(Gas) for Oxygen)
2. For a monatomic ionic species the oxidation number is equal
to the net charge of the species. (i. e. for O2-, F- , Cu+ , Fe2+ and
Al3+ the oxidation numbers are -2, -1 , +1, +2 and +3
respectively).
3. The following rules are applied to specific atoms in their
chemical compounds. If a conflict occurs, the atom listed first
in the list takes preference:
a) F = -1
b) Li, Na, K, Rb, Cs = +1
c) Be, Mg, Ca, Sr, Ba = +2
d) H = +1
e) O = -2
f) Cl, Br, I = -1
4. The sum of all the oxidation numbers in a polyatomic
chemical species is equal to the net charge on the species.
(i.e. carbon tetrachloride, CCl4 : the net charge is zero, since Cl
is -1, the oxidation number of C must be +4.)
N.B. The rule that occurs first in the list always takes
precedence so that:
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Using these rules the oxidation number of O in OF2 is +2 and
in H2O2 it is -1. In BrF5 the oxidation number for Br is +5.
Example: Zn(s) + 2H+(aq) Æ Zn2+(aq) + H2(g)
Zn is oxidized (0 Æ +2)
H+ is reduced (+1 Æ 0)
An easy way to recognize a redox equation is to note
changes in oxidation number:
2Al(s) + 3Cu2+(aq) Æ 2Al3+(aq) + 3Cu(s)
CO32-(aq) + 2H+(aq) Æ CO2(g) + H2O
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An old Exam Question: Chemistry 020, Nov 2004
The oxidation # of N in NH4+ is 3The oxidation # of N in the product, N2 is zero
The N has lost electrons and acts as a reducing agent
The oxidation # of O goes from 2- to zero in O2 in
other words O is oxidized.
The oxidation # of H does not change.
Ans is D
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3. Balancing Redox Equations
Acidic solution:
Cu + NO3- → Cu2+ + NO2
a) Separate the equation into two half-reactions, an
oxidation and a reduction and balance in all
species except H and O:
→ Cu2+
Cu
NO3- →
NO2
b) Assign oxidation #s and balance electrons:
Cu
→ Cu2+ + 2e-
1e- + (N= 5+) NO3- →
NO2 (N=4+)
c) Make the solution Acidic by adding H+ to balance
charge:
→ Cu2+ + 2e-
(oxidation) Cu
(reduction) 2H+
+ 1e- + NO3- →
d) Add water to balance oxygen:
Cu
→ Cu2+ + 2e-
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NO2
2H+
+ 1e- + NO3- →
NO2 + H2O
e) Multiply each half-reaction by an appropriate
factor so that the # of electrons in each reaction
is the same, add the two reaction together and
tidy up the result (Remove any reactants that
appear on both sides of the resulting equation):
Cu
(2H+
4H+
→ Cu2+ + 2e-
+ 1e- + NO3- →
+
Cu +
2NO3- →
NO2 + H2O) X 2
2NO2 + 2 H2O
f) Check for balance in atoms and charge. Note:
also make sure that the final equation is in terms
of the smallest whole numbers.
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Basic solution:
Br2(aq) → Br-(aq) + BrO3-(aq)
(Note this is a disproportionation reaction: A
substance in an intermediate oxidation state goes to
both higher and lower states.)
a) Separate the equation into two half-reactions, an
oxidation and a reduction (By assigning oxidation
#s) and balance in all species except H and O:
(reduction) Br2(aq) → 2Br-(aq)
(oxidation) Br2(aq) →
2BrO3-(aq)
b) balance electrons:
2e- + Br2(aq) → 2Br-(aq)
(ox # + 0) Br2(aq) → (ox # = 5+) 2BrO3-(aq) + 10e-
c) Make the solution Basic by adding OH- to
balance charge:
2e- + Br2(aq) → 2Br-(aq)
12OH- +
Br2(aq) →
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2BrO3-(aq) + 10e-
d) Add water to balance oxygen:
2e- + Br2(aq) → 2Br-(aq)
12OH- +
Br2(aq) →
2BrO3-(aq) + 10e- + 6H2O
e) Multiply each half-reaction by an appropriate
factor so that the # of electrons in each reaction
is the same, add the two reaction together and
tidy up the result (Remove any reactants that
appear on both sides of the resulting equation):
(2e- + Br2(aq) → 2Br-(aq) ) x 5
12OH- +
Br2(aq) →
2BrO3-(aq) + 10e- + 6H2O
10e- + 5Br2(aq) 12OH- + Br2(aq) → 10Br-(aq) +
2BrO3-(aq) + 10e- + 6H2O
yields:
5Br2(aq) 12OH- +
Br2(aq) → 10Br-(aq) + 2BrO3-(aq) +
6H2O
6Br2(aq) + 12OH- → 10Br-(aq) + 2BrO3-(aq) + 6H2O
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f) Check for balance in atoms and charge. Note:
also make sure that the final equation is in terms
of the smallest whole numbers.
NB: Divide by 2
To get the final answer:
3Br2(aq) + 6OH- → 5Br-(aq) + BrO3-(aq) + 3H2O
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An old exam question:
Separate into an oxidation and a reduction:
(oxidation)
→ Zn2+
Zn
(reduction) VO3-
→ V2+
Balance electrons:
Zn
→ Zn2+ +
VO3- +
3e-
2e-
→ V2+
Balance charge with H+:
Zn
→ Zn2+ +
VO3- +
2e- + H+
3e- + 6H+
→ V2+
Balance oxygen with H2O
Zn
→ Zn2+ +
VO3- +
2e- + H+
3e- + 6H+
→ V2+ + 3 H2O
Eliminate electrons:
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(Zn
→ Zn2+ +
(VO3- +
3 Zn
2e- + H+) x 3
3e- + 6H+
→ 3 Zn2+ +
2 VO3- +
→ V2+ + 3 H2O)
x 2
6e- + 3 H+
6e- + 12 H+ → 2 V2+ + 6 H2O
Adding yields:
3 Zn + 2 VO3- + 12 H+ → 3 Zn2+ + 2 V2+ + 6 H2O
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Separate into an oxidation and a reduction:
(oxidation) SO3- → SO4(reduction) MnO4- →
MnO2
Balance electrons:
SO3- → SO4- + 2eMnO4- + 3e- →
MnO2
Balance charge with OH-:
SO3- + 2 OH- → SO4- + 2eMnO4- + 3e- →
MnO2 + 4 OH-
Balance oxygen with H2O
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SO3- + 2 OH- → SO4- + 2e- +
MnO4- + 3e- + 2 H2O →
H2O
MnO2 + 4 OH-
Eliminate electrons:
3 x (SO3- + 2 OH- → SO4- + 2e- +
2 x (MnO4- + 3e- + 2 H2O →
H2O)
MnO2 + 4 OH-)
3 SO3- + 6 OH- → 3 SO4- + 6 e- + 3 H2O
2 MnO4- + 6e- + 4 H2O → 2 MnO2 + 8 OHadding yields:
3 SO3- + 2 MnO4- +
H2O → 3 SO4- + 2 MnO2 + 2 OH-
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4. Redox Titrations
Recall what we said about acid/base titrations:
Acid-base titrations involves four quantities:
1. Concentration of acid
2. concentration of base
3. volume of acid
4. volume of base (“volume” means “volume to reach the equivalence point).
Three of these are required to calculate the fourth!
With redox titrations, it is very similar:
1. Concentration of oxidant,
2. concentration of reductant,
3. volume of oxidant,
4. volume of reductant (“volume” means “volume to reach the
equivalence point). Three of these are required to calculate
the fourth!
The equivalence point is reached when oxidant and reactant are
present in stoichiometric amounts.
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Consider the titration of an acidic Fe2+ solution with
permanganate:
MnO4- + Fe2+ Æ Fe3+ + Mn2+ (not balanced)
The balanced equation for this redox reaction is
MnO4-(aq) + 8H+(aq) + 5Fe2+(aq)
Æ 5Fe3+(aq) + Mn2+(aq) + 4H2O
What volume of 0.684 M KMnO4 solution is required
to react completely with 27.50 mL of 0.250 M
Fe(NO3)2?
Moles Fe(NO3)2 = M x V = 0.25 moles/liter x 0.0275
liters
= 6.785 x 10-3 moles
Moles KMnO4 required = Moles Fe(NO3)2 /5 = 1.375 x
10-3 moles = M x V = 0.684 moles/liter x a liters
a = volume KMnO4 required = 2.01 x 10-3 liters = 2.02
ml
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