THE ELECTRONIC THEORY OF
OXIDATION & REDUCTION
1. Oxidation & Reduction
Simple definitions of oxidation and reduction are based on the loss/gain of oxygen or
the loss/gain of hydrogen. Oxidation is the gain of oxygen or the loss of
hydrogen; reduction is the loss of oxygen or the gain of hydrogen. These
definitions can only be used when a chemical reaction involves hydrogen and
oxygen, and therefore their usefulness is limited.
A more basic and more useful definition of oxidation and
reduction is based on the loss/gain of electrons.
OXIDATION IS LOSS OF ELECTRONS
REDUCTION IS GAIN OF ELECTRONS
In reactions involving simple ions, it is usually easy to tell
whether electrons are lost or gained, but it is less easy to
tell when complex ions or covalent molecules are
involved. Oxidation number is a useful concept for helping
to decide in these more awkward cases.
2. Oxidation Number
The oxidation number is used to express the oxidation state of an element, whether
as the uncombined element or when combined in a compound; it consists of a + or –
sign followed by a number, or it is zero.
Atoms of elements have no overall charge and are therefore given an oxidation
number of zero. When two elements combine, the atoms or ions of the more
electropositive element have a positive oxidation state, and those of the more
electronegative element a negative oxidation state. Elements become more
electronegative the higher their Group number and the lower their Period number;
therefore, the most electronegative element is fluorine.
The oxidation number of an element in a compound is equal to the charge which a
particle of the element would carry in the compound, assuming the compound is
ionic. This is a purely theoretical idea, and it is does not matter whether the
compound in question is really ionic or covalent.
TOPIC 12.8:REDOX REACTIONS 1
e.g.
compound
oxidation numbers
NaCl
CCl4
HBr
H2S
Na
C
H
H
+1
+4
+1
+1
Cl
Cl
Br
S
-1
-1
-1
-2
The following general rules are useful:
All free elements (i.e. those not combined with another element) have an
oxidation number of 0.
e.g. Na, Mg, Br2
In simple ions, the charge on the ion is equal to the oxidation number.
e.g.
ion
oxidation number
Na+
+1
Fe3+
+3
Br-1
O2-2
Since fluorine is the most electronegative element, it always has an oxidation
number in its compounds of –1.
Combined oxygen always has an oxidation number of –2, except when in
combination with fluorine or in peroxides.
e.g.
compound
oxidation numbers
Fe2O3
Mn2O7
CrO3
Fe +3
Mn +7
Cr +3
O -2
O -2
O -2
H2O2
F2O
H +1
O +2
O -1
F -1
BUT ….
Group I elements in their compounds always have an oxidation number of +1.
Group II elements in their compounds always have an oxidation number of +2.
Hydrogen in its compounds always has an oxidation number of +1, except
when it has combined with a reactive metal.
e.g.
compound
oxidation numbers
H2O
HCl
CH4
H +1
H +1
H +1
O -2
Cl -1
C -4
NaH
Na +1
H -1
BUT ….
The sum of the oxidation numbers of all the atoms in an uncharged molecule
is zero: in an ion, the sum is equal to the charge on the ion.
TOPIC 12.8:REDOX REACTIONS 2
e.g.
compound/ion oxidation numbers
NH3
NH4+
H2SO4
H +1 N -3
H +1 N -3
H +1 S +6 O -2
(-3+1+1+1=0)
(-3+1+1+1+1=+1)
(+1+1+6-2-2-2-2=0)
3. Number of Oxidation States
Many elements have several oxidation states:
e.g.
sulphur
chlorine
H2S
S
SCl2
SO2
SO3
+1 -2
0
+2 -1
+4 -2
+6 -2
HCl
Cl2
HOCl ClF3 KClO3 KClO4
+1-1
0
+1-2+1 +3-1
+1+5-2 +1+7-2
4. Redox Reactions
If, during a chemical reaction, the oxidation number of an element increases (i.e.
becomes more positive or less negative), then the element has lost electrons and
has been OXIDISED.
Conversely, if the oxidation number of an element decreases (i.e. becomes less
positive or more negative), then the element has gained electrons and has been
REDUCED.
Oxidation
-5 -4 -3 -2 -1 0 +1 +2 +3 +4 +5
Reduction
REDuction and OXidation occur together in what is called a REDOX reaction.
TOPIC 12.8:REDOX REACTIONS 3
Example:

work out the oxidation numbers for all the elements in the reaction

identify the oxidation and reduction steps

work out the total number of electrons transferred in each step: they should be
equal
oxidation: loss of 2e- x 1
Mg + 2HCl
0
MgCl2 + H2
+1 -1
+2 -1
0
reduction: gain of 1e- x 2
check: 2e- x 1 = 1e- x 2
Complete the equations for the remaining reactions in a similar way.
Mg + 2HCl
0
+1 -1
2Fe2+ + Cl2
+2
0
-2
Cu2+ + Zn
+2
TOPIC 12.8:REDOX REACTIONS 4
+2 -1
0
0
2Fe3+ + 2Cl+3
2Fe3+ + S2+3
MgCl2 + H2
-1
2Fe2+ + S
+2
0
Zn2+ + Cu
+2
0
OXIDATION & REDUCTION
QUESTION SHEET 1
Work out the oxidation number of each element in the following chemical formulae.
1.
Cu
…………………………………………………………………..
2.
NaCl
………………………………………………………………….
3.
CuS
………………………………………………………………….
4.
Cl2
………………………………………………………………….
5.
Fe3+
………………………………………………………………….
6.
Cr2O3
………………………………………………………………….
7.
H2O
………………………………………………………………….
8.
HNO3
………………………………………………………………….
9.
MnO4-
………………………………………………………………….
10.
K2CO3
………………………………………………………………….
11.
NaClO3
………………………………………………………………….
12.
SO42-
………………………………………………………………….
13.
NaNO2
………………………………………………………………….
14.
SOCl2
………………………………………………………………….
15.
Cu(NO3)2
………………………………………………………………….
16.
K2Cr2O7
………………………………………………………………….
17.
Al(OH)3
………………………………………………………………….
18.
K2MnO4
………………………………………………………………….
19.
Na3PO4
………………………………………………………………….
20.
Fe3O4
………………………………………………………………….
TOPIC 12.8:REDOX REACTIONS 5
OXIDATION & REDUCTION
QUESTION SHEET 2
Work out the oxidation number of each element in the following chemical formulae
1.
Mg
…………………………………………………………………..
2.
KBr
………………………………………………………………….
3.
CaS
………………………………………………………………….
4.
Br2
………………………………………………………………….
5.
Ba2+
………………………………………………………………….
6.
Fe2O3
………………………………………………………………….
7.
Na2O
………………………………………………………………….
8.
LiNO3
………………………………………………………………….
9.
MnO42-
………………………………………………………………….
10.
Rb2CO3
………………………………………………………………….
11.
KBrO3
………………………………………………………………….
12.
IO4-
………………………………………………………………….
13.
KNO2
………………………………………………………………….
14.
POCl3
………………………………………………………………….
15.
Sr(NO3)2
………………………………………………………………….
16.
K2CrO4
………………………………………………………………….
17.
Cr(OH)3
………………………………………………………………….
18.
KMnO4
………………………………………………………………….
19.
H3PO4
………………………………………………………………….
20.
CuCl2-
………………………………………………………………….
21.
Pb3O4
………………………………………………………………….
TOPIC 12.8:REDOX REACTIONS 6
OXIDATION & REDUCTION
QUESTION SHEET 3
Work out the oxidation number of each element in the following chemical formulae
1.
Ca
…………………………………………………………………..
2.
LiF
………………………………………………………………….
3.
MgO
………………………………………………………………….
4.
I2
………………………………………………………………….
5.
Cr3+
………………………………………………………………….
6.
Al2O3
………………………………………………………………….
7.
HF
………………………………………………………………….
8.
Ni(NO3)2
………………………………………………………………….
9.
CrO42-
………………………………………………………………….
10.
SrCO3
………………………………………………………………….
11.
NaClO4
………………………………………………………………….
12.
SO32-
………………………………………………………………….
13.
NaIO3
………………………………………………………………….
14.
XeF4
………………………………………………………………….
15.
Pb(OH)2
………………………………………………………………….
16.
K2MnO4
………………………………………………………………….
17.
Al2(SO4)3
………………………………………………………………….
18.
NaVO3
………………………………………………………………….
19.
H3PO3
………………………………………………………………….
20.
NH4+
………………………………………………………………….
TOPIC 12.8:REDOX REACTIONS 7
OXIDATION & REDUCTION
QUESTION SHEET 4
Examine each of the following redox equations. Work out the oxidation number of
each element in all the atoms, ions and molecules. Using these numbers, explain
with reasons which substance is oxidised and which substance is reduced
SO2 + 2Mg
C + H2O
SnCl2 + HgCl2
H2 + Cl2
2Fe3+ + 2I-
TOPIC 12.8:REDOX REACTIONS 8
2MgO + S
CO + H2
Hg + SnCl4
2HCl
2Fe2+ + I2
OXIDATION & REDUCTION
QUESTION SHEET 5
Examine each of the following reactions.

For each equation, work out the oxidation number of each element in all the
atoms, ions and molecules. Use these numbers to decide whether the change
taking place is a redox reaction or not.

Where a redox reaction occurs, indicate, with reasons, which species is
oxidised and which is reduced

Where the change is not a redox reaction, describe in one word the type of
change taking place.
1.
Mg(s) + Cl2(g)
2.
NaCl(s)
3.
2Fe3+(aq) + 2I-(aq)
4.
C(s) + H2O(g)
5.
Ag+(aq) + Cl-(aq)
MgCl2(s)
Na+(l) + Cl-(l)
TOPIC 12.8:REDOX REACTIONS 9
2Fe2+(aq) + I2(s)
CO(g) + H2(g)
AgCl(s)
6.
NaOH(aq) + HCl(aq)
7.
2H2(g) + O2(g)
8.
Cu(s) + 4HNO3(aq)
TOPIC 12.8:REDOX REACTIONS 10
NaCl(aq) + H2O(l)
2H2O(l)
Cu(NO3)2(aq) + 2H2O(l) + 2NO2(g)
REDOX EQUATIONS
Constructing Half –Equations
The half-equation shows either the oxidation or the reduction step of a redox change.
In a half-equation:

only one element changes its oxidation state

The loss/gain of electrons responsible for the change in oxidation state is
shown

the atoms of each element must balance

the total charge on both sides of the equation must be the same
When constructing half-equations for reactions which take place in aqueous solution:

Water can be used as a source of oxygen atoms on the reactant side of the
equation, the hydrogen appearing in the products as H+.

Any extra oxygen atoms on the reactant side of the equation can be
converted into water by reaction with hydrogen ions from an acid.

The oxidation numbers of hydrogen and oxygen do not change.
Example 1: Deduce the half-equation for the reduction of VO3- to V2+ in
acid solution.
Vanadium is changing its oxidation state from +5 in VO 3- to +2 in V2+ and
therefore needs to gain 3e-.
VO3- + 3eV2+
The acid solution provides the six hydrogen ions needed to react with the three
extra oxygen atoms in VO3- to form water.
VO3- + 6H+ + 3e-
V2+ + 3H2O
Finally, a check shows that the total charge on each side of the equation is
the same.
l.h.s. -1 + 6 - 3 = +2
TOPIC 12.8:REDOX REACTIONS 11
r.h.s. +2
Example 2:
Deduce the half-equation for the reduction of MnO4- to Mn2+ in
acid solution.
Manganese is changing its oxidation state from +7 in MnO 4- to +2 in Mn2+
and therefore needs to gain 5e-.
MnO4- + 5e-
Mn2+
The acid solution provides the eight hydrogen ions needed to react with
the four extra oxygen atoms in MnO4- to form water.
MnO4- + 8H+ + 5e-
Mn2+ + 4H2O
Finally, a check shows that the total charge on each side of the equation
is the same.
l.h.s. -1 + 8 - 5 = +2
r.h.s. +2
Example 3: Deduce the half-equation for the reduction of NO3- to NO in acid
solution.
Nitrogen is changing its oxidation state from +5 in NO3- to +2 in NO and
therefore needs to gain 3e-.
NO3- + 3e-
NO
The acid solution provides the four hydrogen ions needed to react with
the two extra oxygen atoms in NO3- to form water.
NO3- + 4H+ + 3e-
NO + 2H2O
Finally, a check shows that the total charge on each side of the equation
is the same.
l.h.s. -1 + 4 - 3 = 0
TOPIC 12.8:REDOX REACTIONS 12
r.h.s. 0
Combining Half –Equations
The overall equation for a redox change can be obtained by combining the halfequations for the oxidation and reduction steps in such a way that the number of
electrons donated by the reducing agent is equal to the number accepted by the
oxidising agent.
Any molecules or ions which appear on both sides of the overall equation, such as
H+ and H2O, then need to be cancelled down.
Examples:
4. Manganate(VII) ions (MnO4-) oxidise iron(II) ions (Fe2+) in acid solution,
forming iron(III) ions (Fe3+), and are reduced to manganese(II) ions (Mn2+).
Deduce the two half-equations for this reaction, and hence derive an overall
equation.
Manganese is changing its oxidation state from +7 in MnO 4- to +2 in Mn2+ and
therefore needs to gain 5e-.
MnO4- + 5eMn2+
The acid solution provides the eight hydrogen ions needed to react with the
four extra oxygen atoms in MnO4- to form water.
MnO4- + 8H+ + 5e-
Mn2+ + 4H2O
Iron is changing its oxidation state from +2 in Fe 2+ to +3 in Fe3+ and therefore
needs to lose 1e-.
Fe2+
Fe3+ + eTo balance the electrons, the first half-equation needs to be multiplied x1:
MnO4- + 8H+ + 5e-
Mn2+ + 4H2O
and the second x5:
5Fe2+
5Fe3+ + 5e-
The two half-equations are now added together:
5Fe2+ + MnO4- + 8H+ + 5e-
Mn2+ + 4H2O + 5Fe3+ + 5e-
The electrons cancel out to give the overall equation:
5Fe2+ + MnO4- + 8H+
Mn2+ + 4H2O + 5Fe3+
Finally, a check shows that the total charge on each side of the equation is the
same.
l.h.s. +10 -1 + 8 = +17
TOPIC 12.8:REDOX REACTIONS 13
r.h.s. +2 +15 = +17
5. Dichromate(VI) ions (Cr2O72-) oxidise sulphate(IV) ions (SO32-) in acid
solution, forming sulphate(VI) ions (SO42-), and are reduced to chromium(III)
ions (Cr3+). Deduce the two half-equations for this reaction, and hence derive
an overall equation.
Chromium is changing its oxidation state from +6 in Cr2O72- to +3 in Cr3+ and
therefore needs to gain 6e.
Cr2O72- + 6e2Cr3+
The acid solution provides the fourteen hydrogen ions needed to react with
the seven extra oxygen atoms in Cr2O72- to form water.
Cr2O72- + 14H+ + 6e-
2Cr3+ + 7H2O
Sulphur is changing its oxidation state from +4 in SO32- to +6 in SO42- and
therefore needs to lose 2e-.
SO32-
SO42- + 2e-
One water molecule is needed to provide the extra oxygen atom:
SO32- + H2O
SO42- + 2H+ + 2e-
To balance the electrons, the first half-equation needs to be multiplied x1:
Cr2O72- + 14H+ + 6e-
2Cr3+ + 7H2O
and the second x3:
3SO32- + 3H2O
3SO42- + 6H+ + 6e-
The two half-equations are now added together:
3SO32-+ 3H2O + Cr2O72- + 14H+ + 6e-
2Cr3+ + 7H2O + 3SO42- + 6H+ + 6e-
The electrons cancel out, and the numbers of H+ ions and H2O molecules are
simplified to give the overall equation:
3SO32- + Cr2O72- + 8H+
2Cr3+ + 4H2O + 3SO42-
Finally, a check shows that the total charge on each side of the equation is
the same.
l.h.s. -6 -2 + 8 = 0
TOPIC 12.8:REDOX REACTIONS 14
r.h.s. +6 -6 = 0
Sometimes the same species oxidises and reduces itself simultaneously. This
process is called disproportionation.
6. Under alkaline conditions, chlorine disproportionates to form chloride ions
(Cl-) and chlorate(V) ions (ClO3-). Deduce the two half-equations for this
reaction, and hence derive an overall equation.
Chlorine is changing its oxidation state from 0 in Cl2 to -1 in Cl- and therefore
needs to gain 1e-.
½ Cl2 + eClChlorine is changing its oxidation state from 0 in Cl2 to +5 in ClO3- and
therefore needs to lose 5e-.
1/ Cl
ClO3- + 5e2
2
Six hydroxide ions in the alkaline solution provide the three extra oxygen
atoms needed to form ClO3-, and also form water.
1/
2
Cl2 + 6OH-
ClO3- + 3H2O + 5e-
To balance the electrons, the first half-equation needs to be multiplied x5:
21/2 Cl2 + 5e-
5Cl-
and the second x1:
1/
2
Cl2 + 6OH-
ClO3- + 3H2O + 5e-
The two half-equations are now added together:
3Cl2 + 6OH- + 5e-
5Cl- + ClO3- + 3H2O + 5e-
The electrons cancel out to give the overall equation:
3Cl2 + 6OH-
5Cl- + ClO3- + 3H2O
Finally, a check shows that the total charge on each side of the equation is
the same.
l.h.s. -6
TOPIC 12.8:REDOX REACTIONS 15
r.h.s. -5 -1 = -6