Download OXIDATION - REDUCTION

Redox definitions
An oxidation-reduction reaction (or
redox reaction) is one that involves the
transfer of electrons from one species to
another.
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
In this reaction the Zn metal has been
oxidised as it has lost electrons, and the
Cu2+ has been reduced as it has gained
electrons.
Zn
 Zn2+ + 2e and
Cu2+ +
2e

Cu
The species containing the atom that
is oxidised, in this case Zn, is called the
reducing agent or reductant.
The species containing the atom that
is reduced, in this case Cu2+, is the
oxidising agent or oxidant.
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Oxidation numbers.
An increase in oxidation number
corresponds to oxidation
A decrease in oxidation number
corresponds to reduction.
Oxidation numbers
The oxidation number (or state) is a
number that can be assigned to each
individual atom in an element,
compound or ion, using a set of rules.
1. The oxidation number of an atom in
an element is zero eg. in H2 the
oxidation number of H is 0.
2. The oxidation number of an atom in a
monatomic ion is the same as the
charge on the ion.
eg. In Na+ the oxidation number is +1,
in O2 the oxidation number is -2.
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In an ionic compound containing ions
the ions have the same oxidation
numbers as they would alone eg. in
Na2O the oxidation numbers are still
+1 for Na+ and -2 for O2.
3. In compounds each hydrogen atom
usually has an oxidation number of +1
(the exception is in the metal hydrides
e.g.NaH where the oxidation number
of H= -1).
4. In compounds each oxygen atom has
oxidation number of -2 (except in
peroxides eg. H2O2 when it is -1).
5. In a molecule the sum of the oxidation
numbers of all the atoms is zero.
6. In polyatomic ions the sum of the
oxidation numbers of all the atoms is
equal to the overall charge on the ion.
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Example:
a)
Find the oxidation number of S in
H2SO4.
(2 x +1) + (1 x ?) + (4 x –2) = 0
Oxidation number of S = +8 - 2 = +6
b) In the ion Cr2O72 the oxidation
number of Cr is calculated as follows:
(2 x ?) + (7 x –2) = -2
(2 x ?) = -2 + 14 = +12
Oxidation number of Cr =
= +6
c) In an ionic compound such as
NH4NO3 the oxidation numbers of N
are determined by first separating into
ions NH4+ and NO3- and then finding
the oxidation numbers. The values
obtained are –3 in NH4+ and +5 in
NO3
 12
2
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Exercise:
1) Find the oxidation numbers of the S
atoms in each of the following
substances.
SO2, H2SO4, SO3, S8, H2SO3,
Na2S2O3
2) Calculate the oxidation number of N
in each of the following ions:
NO3,
NH4+,
NO2,
N3
3) By assigning oxidation numbers show
which of the following reactions is not
a redox reaction.
(a) CuCO3  CuO + CO2
(b) Cu + 2AgNO3  Cu(NO3)2 + 2Ag
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(c) Cr2O72 + 6Fe2+  6Fe3+ + 2Cr3+ +
7H2O
(d) Cr2O72 + 2OH  2CrO42 + H2O
Balancing Redox Equations
Step 1 - Identify the two half reactions
and the appropriate reactant and
product in each case.
When a solution of potassium
dichromate reacts with iron(II) nitrate the
species oxidised is Fe2+ and the species
reduced is Cr2O72
Fe2+
 Fe3+
and
Cr2O72

Cr3+
Step 2 - Balance all atoms undergoing
a change in oxidation number.
Fe2+  Fe3+
and
Cr2O72  2Cr3+
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Step 3 - Balance the number of O
atoms by adding the appropriate
number of water molecules.
Fe2+  Fe3+
and
Cr2O72  2Cr3+ + 7H2O
Step 4 - Balance the H atoms by adding
H+ ions.
Fe2+
 Fe3+
and
Cr2O72 + 14H+  2Cr3+ + 7H2O
Step 5 - Balance the charge by adding
electrons, e-. This gives 2 balanced
half-equations.
Fe2+  Fe3+ + e and
Cr2O72 + 14H+ + 6e  2Cr3+ + 7H2O
Note: In the oxidation half-equation the
Fe2+ loses electrons and in the reduction
half-equation the Cr2O72 gains electrons.
OIL
RIG
oxidation is loss reduction is gain
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Step 6 - To obtain an overall balanced
equation add the two half equations
together. Before doing this the
equations may have to be multiplied so
that the number of electrons in each
half-equation is the same.
Fe2+  Fe3+ + e (x6) =>
6Fe2+  6 Fe3+
+ 6e. Added to
Cr2O72 + 14H+ + 6e  2Cr3+ + 7H2O
gives the final equation
6Fe2+ + Cr2O72 + 14H+ 
2Cr3+ + 7H2O + 6 Fe3+
Finally check that the equation is
balanced, particularly for charge!!
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Exercise - Balance each of the
following equations.
1. MnO4 + SO2  Mn2+ + SO42
2. S2O32 + I2
I
+
S4O62
Common Reductants
1. Metals, especially those high on the
activity series, are oxidised to metal
ions. Common examples are Zn
(forms Zn2+), Mg (forms Mg2+) and Fe
(forms Fe2+)
Zn(s)  Zn2+(aq) + 2e
Both Zn2+ and Mg2+ are colourless
while Fe2+ is pale green in solution.
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2. Iron(II) ion, Fe2+, can be easily
oxidised to Fe3+. The colour change
observed is from pale green Fe2+ to
orange Fe3+.
3. Sulfur dioxide, SO2, and sulfite ion,
SO32 are both oxidised to sulfate ion,
SO42. All of these species are
colourless. Solutions of SO2 in water
form sulfurous acid, H2SO3 which is a
weak acid and dissociates to form
HSO3 and SO32 ions.
4.
Iodide ions, I and bromide ions, Br
are both colourless and are oxidised
to the halogens I2 and Br2. When I2 is
formed the solution goes a
brown/yellow colour in aqueous
solution. Solid iodine may form as a
black precipitate of I2. The formation
of Br2 results in the solution turning an
orange colour.
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5. Thiosulfate ions, S2O32 - This
colourless ion is commonly used in
the redox titration which determines
the concentration of iodine in a
solution. The thiosulfate ion is
oxidised to S4O62 (tetrathionate ion).
I2 + S2O32  S4O62 + 2I
5. Carbon, C, and carbon monoxide,
CO - Carbon is commonly used as a
reductant as graphite or coal.
In a plentiful supply of air it is oxidised
to CO2 (also a colourless gas).
CO (made from combustion of coal)
is used as a
Fe2O3 + 3CO  2Fe + 3CO2
7. Hydrogen gas, H2 - This colourless
gas can be used as a reductant,
commonly combining with oxygen to
form water, H2O.
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8. Oxalic acid, H2C2O4 – This
colourless, posonous substance is a
weak acid that is oxidised to carbon
dioxide CO2. Its anion is the oxalate
ion C2O42, found in salts such as
sodium oxalate.
9. Hydrogen sulfide, H2S – This gas
can be used as a reductant as it can
be oxidised to sulfur species in a
higher oxidation state eg SO2.
10. Hydrogen peroxide, H2O2 –
When acting as a reductant it is
oxidised to O2. This may be seen as
bubbles of gas. Note that H2O2 can
also act as an oxidant (see below).
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Common Oxidants
1. Oxygen gas, O2 is involved in all
burning reactions producing the oxide
ion, O2. Both species are colourless
so no colour change is observed.
2. Hydrogen ions, H+ present in dilute
acids is reduced to hydrogen gas, H2.
3. Halogens - Chlorine, Cl2 (a yellowgreen gas), bromine, Br2 (an orange
liquid), and iodine, I2 (a shiny black
solid) are all reduced to their
respective colourless halide ions, Cl,
Br, I. Because of its oxidising
properties Cl2 is used as a disinfectant
and to sterilise swimming pools.
4. Permanganate ion, MnO4- and
manganese dioxide MnO2 - The
purple ion MnO4- and the brown solid
MnO2 are both reduced to colourless
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Mn2+ ion if the reaction is carried out
in acidic solution.
MnO4 + 8H+ + 5e Mn2+ + 4H2O
MnO2 + 4H+ + 2e  Mn2+ + 2H2O
If the reaction of MnO4- is carried out
in neutral or slightly basic conditions
then the reduction product is a brown
solid, MnO2.
MnO4 + 2H2O + 3e  MnO2 + 4OH
In strongly basic conditions the
permanganate ion is reduced to the
green manganate ion
MnO4 + e  MnO42
Dichromate ion, Cr2O72 - This orange
ion is reduced to green Cr3+.
Cr2O72 + 14H+ + 6e  2Cr3+ + 7H2O
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5. Hydrogen peroxide, H2O2 - When
colourless H2O2 acts as an oxidant it
is reduced to water.
H2O2 + 2H+ + 2e  2H2O
6. Metal ions (Fe3+ and Cu2+) –
The orange Fe3+ ion undergoes
reduction to the pale green Fe2+ ion.
Blue Cu2+ may be reduced to either Cu+
or pink Cu metal depending on the
reducing agent used. The reduction of
Cu2+ in the redox reaction of aldehydes
with Benedict’s solution produces the
red precipitate of Cu2O (copper I oxide).
The reduction of Cu2+ using I- produces
a white precipitate of CuI (copper I
iodide).
Cu2+ + CH3CHO  Cu2O + CH3CO2H
Cu2+
+
I

CuI
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+
I2
8. Concentrated nitric acid, HNO3 When colourless conc nitric acid is
used as an oxidant. When reacted
with copper metal, the observed
product is brown nitrogen dioxide gas,
NO2.
Cu + 2HNO3 + 2H+  2NO2 + Cu2+ + 2H2O
Iodate ion, IO3 - This ion is colourless
and can be reduced to form the
halogen, I2 (a black solid or when
dissolved in water a brown aqueous
solution).
2IO3 + 12H+ + 10e  I2 + 6H2O
9. Hypochlorite ion, OCl - A colourless
ion that can be reduced to form
chlorine, Cl2, a yellow-green gas.
2OCl + 4H+ + 2e  Cl2 + 2H2O
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