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1. Oxidation and reduction
definitions and selected simple redox equations described and
explained
The basic,
but inadequate descriptions of oxidation and reduction in terms of oxygen are first described,
followed by some simple examples of the much more important electron
transfer definitions.
It is advisable
to study section 1. before proceeding to section 2. on oxidation
state/number.
Please note that in the first parts of section 1. the
equations are not meant to be balanced, but just focus on the
particular oxidation or reduction change and later the full balanced redox equations
are given and explained.
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These example equations are
not always meant to be balanced or complete, but they do show the
essential oxidation or reduction change. |
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OXIDATION |
REDUCTION |
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Gain/addition of oxygen by an atom, molecule or ion
e.g.
- S ==> SO2
- Burning
sulfur – oxidised.
- CH4 ==> CO2
+ H2O
- Burning methane to water and carbon dioxide, C and H
both gain O.
- NO ==> NO2
- Nitrogen monoxide oxidised to nitrogen dioxide, rapid in air with
NO from
car exhaust fumes.
- SO32– ==> SO42–
- Oxidising the sulfite ion to the sulfate ion e.g. by
chlorine/bromine oxidising agents.
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Loss/removal of oxygen from a
compound or ion
e.g.
- CuO ==> Cu
- Loss of oxygen from
copper(II) oxide to form copper atoms in metal
extraction with or displacement by more reactive metal.
- Fe2O3
==>
Fe
- Iron(III) oxide reduced to iron in blast furnace using carbon or
carbon monoxide.
- NO ==> N2
- Nitrogen(II) oxide reduced to nitrogen.
- SO3 ==> SO2
- Sulfur trioxide reduced to sulfur dioxide.
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The oxygen definitions are
inadequate for the full understanding of redox reactions.
You must
look at the oxidation/reduction (redox) changes by considering electron
transfers (actual or theoretical) and the oxidation state (oxidation
number) of the elements involved in the redox change. |
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OXIDATION
Loss/removal of electrons from atom, ion or molecule
e.g.
- Fe ==> Fe2+ +
2e–
- An iron
atom loses 2 electrons to form the iron(II) ion e.g. in the initial chemistry of
iron rusting or in an iron–acid reaction.
- Fe2+ ==> Fe3+
+ e–
- The iron(II) ion loses
1 electron to form the iron(III) ion, e.g. via chlorine or
manganate(VII) oxidising agents.
- 2Cl– ==> Cl2
+ 2e–
- The loss of electrons by chloride ions to form chlorine molecules
e.g. the +ve anode reaction in electrolysis of chlorides or the action of a very
strong oxidising agent.
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REDUCTION
Gain/addition of electrons by
an atom, ion, or molecule
e.g.
- Cu2+ + 2e– ==>
Cu
- The copper(II) ion gains 2 electrons to form neutral copper
atoms e.g. in electrolysis at the –ve cathode or when copper is
displaced from its salt by a more reactive metal.
- Fe3+ + e–
==> Fe2+
- The iron(III) ion gains an electron
and is reduced to the iron(II) ion e.g. by adding zinc to acidified
iron(III) salt solution.
- 2H+ + 2e– ==>
H2
- Hydrogen ions gain electrons to form neutral hydrogen molecules
e.g. in the –ve cathode reaction in the electrolysis of acids or in a metal–acid reaction.
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An oxidising agent is the species
that gives the oxygen or removes the electrons |
A reducing agent is the
species that removes the oxygen or acts as the electron donor |
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The electron gain/loss
definitions are the most important definitions of reduction/oxidation. |
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REDOX REACTIONS
– in an
overall
reaction, oxidation and reduction must go together
e.g. in terms of
oxygen and these are complete and balanced equations.
BUT by the end of this page you to be able
to analyse all these reactions
in terms of changes in oxidation state for all the
elements whose oxidation state changes and also recognise elements whose
oxidation state does NOT change! |
- copper(II) oxide +
hydrogen
==> copper + water
- CuO(s) + H2(g)
==> Cu(s) + H2O(g)
- Heated copper oxide is reduced to copper
(O loss) when hydrogen is passed over it,
- hydrogen is
oxidised to water (O gain),
- hydrogen is the reducing agent (removes O from CuO),
- and copper oxide is the oxidising agent (donates O to hydrogen).
- iron(III) oxide +
carbon monoxide
==> iron + carbon dioxide
-
Fe2O3(s) +
3CO(g) ==> 2Fe(l) + 3CO2(g)
- In the blast furnace the iron(III) oxide is reduced to
liquid iron
(O loss),
- the carbon monoxide is oxidised to carbon dioxide
(O gain),
- CO is the reducing
agent (O remover/acceptor from Fe2O3),
- and Fe2O3
is the oxidising agent (O donator to CO)]
- nitrogen(II) oxide + carbon monoxide
==> nitrogen + carbon dioxide
- 2NO(g) + 2CO(g) ==> N2(g)
+ 2CO2(g)
- (e.g. in car exhaust catalytic converter)
- Nitrogen(II) oxide is reduced to
nitrogen (O loss),
- carbon
monoxide is oxidised to carbon dioxide (O gain),
- and CO is the reducing agent
(O remover/acceptor)
- and NO is
the oxidising agent (O donor).
- iron(III) oxide + aluminium
==> aluminium oxide + iron
-
Fe2O3(s)
+ 2Al(s) ==> Al2O3(s) + 2Fe(s)
- (the very exothermic Thermit reaction)
- Iron(III) oxide is reduced and is the oxidising
agent (its the oxygen loser i.e. oxygen donor),
- and the aluminium is
oxidised and is the reducing agent (its the oxygen gainer/acceptor/remover).
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Redox
reaction analysis based on the more important and advanced electron definitions
note that the
electron gain must equal the electron loss in the complete and fully balanced equation
e.g.
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-
magnesium + iron(II)
sulfate
==> magnesium
sulfate + iron
- Mg(s) + FeSO4(aq)
==> MgSO4(aq) + Fe(s)
- This is the 'ordinary molecular'
equation for a typical metal displacement reaction, but does not really show what
really happens in terms of atoms,
ions and electrons, so we use ionic equations like the one shown
below.
- The sulfate ion SO42–(aq)
is called a spectator ion, because it doesn't change in the
reaction and can be omitted from the ionic equation (below).
- Also, no electrons should show
up in the 'real' balanced equation because the number of electrons lost by some reactant species
equals the electrons gained
by other reactant species.
-
Mg(s) + Fe2+(aq)
==> Mg2+(aq) + Fe(s)
- The magnesium atom, Mg, loses 2 electrons
(oxidation) to form the magnesium ion, Mg2+,
- the iron(II) ion,
Fe2+, gains 2
electrons (reduced) to form iron atoms, Fe,
- Mg is the reducing agent
i.e. the electron donor to the Fe2+ ion,
- and the
Fe2+
ion is the oxidising agent i.e. the electron remover or acceptor from
the Mg atom.
- zinc + hydrochloric
acid
==> zinc chloride + hydrogen
- Zn(s) + 2HCl(aq)
==> ZnCl2(aq)
+ H2(g)
- The chloride ion Cl– is the spectator
ion so the ionic–redox equation is ...
-
Zn(s) +
2H+(aq) ==> Zn2+(aq)
+ H2(g)
- Zinc atoms are oxidised to zinc ions by
losing two electrons,
- and zinc is the reducing agent i.e. the electron donor to
the hydrogen ion, H+.
- Hydrogen ions are the
oxidising agent i.e. two hydrogen ions gain/accept one electron each from
a Zn
atom,
- and so
are reduced to form hydrogen molecules.
- copper + silver nitrate
==> silver + copper(II) nitrate
- Cu(s) + 2AgNO3(aq)
==> 2Ag(s) + Cu(NO3)2(aq)
- The nitrate ion NO3– is the spectator
ion so the ionic–redox equation is
-
Cu(s) + 2Ag+(aq)
==> 2Ag(s) + Cu2+(aq)
- in which copper atoms are oxidised by the silver
ions by a two electron loss,
- these electrons are transferred from
the copper atoms to the silver ions,
- so they are reduced by one electron
gain each to silver atoms.
- The silver ions are the
oxidising agent (e– acceptor),
- and the copper atoms are the reducing agent
(e– donor).
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One other, and
very limited definition, is
oxidation and reduction in terms of
hydrogen, but such redox definitions have little use these days except
in the reduction (H gain) reactions of unsaturated organic compounds such as alkenes, aldehydes, ketones, carboxylic acids and the reduction nitro–aromatic
compounds to amines. However, that is not to say they are
unimportant reactions, in fact all four reactions below are all of
considerable industrial significance.
QUIZ on oxidation state
Application of the electron based
definitions of oxidation and reduction
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Electron definitions
of oxidation or reduction are the much more
useful and important for advanced level chemistry as is the very important
associated concept of
oxidation state/number which is explained in section 2.
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The IMPORTANT
DEFINITIONS you need to know
(covering section 2. as well)
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Definitions and analysis of redox reactions in terms of oxygen or
hydrogen may not get you the marks at advanced level! So,
LEARN and USE the following as soon as possible in your advanced redox analysis.
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Oxidation is
electron loss by a species or increase in the oxidation state/number of an element
in the species involved which can be the element itself, or a covalent
compound or ion containing the element (ditto for 2., 3. and 4.).
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Reduction is
electron gain by a species or decrease in the oxidation state/number of an element
in the species involved.
-
An electron acceptor is
an oxidising agent and therefore gets reduced in its action, so there is a decrease in
the oxidation number/state of one of the atoms of the oxidising agent.
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An electron donor is a
reducing agent and therefore gets oxidised in the process, so there is an increase in
the oxidation
number/state of one of the atoms of the reducing agent.
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You also need to conceive
of 'redox'
reactions as a combination of two half–cell reactions/equations
(see
Equilibria Part 7 for more on half–cell potentials and redox
equilibria.
-
One
half–reaction will be an
oxidation and the other 'half' will be a reduction.
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How to put the two
together to gives the full ionic–redox equation to completely summarise a redox reaction
in simple cases is given in
redox section 2.
below, and see sections 5. and
6.
for lots of more complex examples).
QUIZ on oxidation state
2. Introducing, explaining and defining oxidation
state and so describing redox equations in terms
of oxidation state changes
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Before studying the rest of this page you
must be familiar with ALL the definitions at the end of section
1. above.
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In simple cases,
particularly simple ions, the oxidation state can be defined
as the number of electrons which must be added (+ oxidation
state) or removed (- oxidation state) to
leave the element in its electrically neutral elemental state
which is then considered to be in a zero oxidation state e.g.
-
This is an abstract concept definition
and is quite easy to appreciate when dealing with simple ions
(but not so for covalent molecules and more
complex ions) e.g.
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chlorine, Cl or Cl2,
oxidation state zero (0), the element itself in an electrically
neutral state.
-
chloride ion, Cl–,
oxidation state minus one ( –1), because one electron must be
removed from the ion to leave the electrically neutral element
atoms of chlorine.
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sulfur, S or S8,
oxidation state zero (0), the element itself in an electrically
neutral state.
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sulfide ion, S2–,
oxidation state minus two ( –2), because two electrons must be
removed from the ion to leave the electrically neutral element
atoms of sulfur.
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sodium, Na,
oxidation state zero (0), the element itself in an electrically
neutral state.
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sodium ion, Na+,
oxidation state plus one (+1), because one electron must be
added to the ion to give the electrically neutral atom of the element sodium.
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iron, Fe, oxidation
state zero (0), the element itself in an electrically neutral
state.
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iron(II) ion, Fe2+,
oxidation state plus three (+2), because two electrons must be
added to the ion to give the electrically neutral atoms of the element iron.
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iron(III) ion, Fe3+,
oxidation state plus three (+3), because three electrons must be
added to the ion to give the electrically neutral atoms of the element iron.
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Note:
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Important sign
conventions
(do NOT confuse!):
-
The sign (+/–)
for an oxidation state/oxidation number is quoted first
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e.g. –3, –2, –1, 0,
+1, +2, +3 etc. (obviously no sign for zero)
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In the
symbols/formulae of ions, the numerical value of the ion charge
is quoted first
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e.g. PO43–,
S2–, I–, NH4+, Mg2+,
Al3+ etc.
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Note there is
no number
required if the ionic charge is +/– one, BUT remember
oxidation state always needs a number i.e. +1 or -1.
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In an
electrically neutral compound or molecule, the sum of the
oxidation states/oxidation numbers must add up to zero.
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If it doesn't, then
you have got something wrong in your analysis.
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BUT, the 'zero
sum rule' provides a way of working out another unknown
oxidation state of an element, if you know the oxidation states of
the other element atoms in the compound.
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In the case of
ions, the sum of the oxidation states equals the overall charge
on the ion.
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When dealing with atoms and
simple ions, redox analysis is usually quite straight forward e.g. the
three
electron analysed' examples in section 1.
above and
the examples
below where the concept of oxidation state/number is now further introduced
via detailed examples for
simple ions or molecules. It is best to study them in the
order given.
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After this, the more
awkward situations are dealt with.
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Below is the section
2 reaction sub–index:
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2.1 metal + acid
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2.2 reactive metal +
non–metal
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2.3 halogen–halogen displacement
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2.4 metal–metal displacement
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2.5
what about more complex molecule/ion situations?
QUIZ on oxidation state
Ex
2.1
Reacting a metal with an
acid
-
magnesium +
sulfuric acid
==> magnesium sulfate + hydrogen
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Since the
chloride ion SO42– is
a spectator
ion i.e. it doesn't change chemically or physically in
state, so the ionic–redox equation is
-
Magnesium atoms are oxidised to
magnesium ions by the loss of two electrons.
-
Hydrogen ions
are reduced by electron gain to form hydrogen molecules.
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The oxidation states
of the elements are shown in () for the species involved in the actual chemical change
and would be
quoted as follows:
-
Mg (0)
+ 2 x H(+1) ==>
Mg(+2) + 2 x
H (0) and the half–cell reactions are ...
-
the
oxidation: Mg(s) ==> Mg2+(aq)
+ 2e– (oxidation state increases by 2 units per Mg
atom by loss of two electrons)
-
the
reduction: 2H+(aq) + 2e– ==>
H2(aq) (oxidation state decreases by 1 unit per H+
ion by gain of one electron)
-
The oxidation state for
magnesium
is increased by two units per atom, is balanced by the oxidation state of
two hydrogens (as H+ ion) decreasing by one unit each.
-
Having now quoted some
oxidation states, now is the time to explain the concept further and
connect it with electron transfers in simple ionic–redox reactions
like the one above.
-
Important
points and ideas to take forward into section 2.2 onwards ...
-
The oxidation
state defines the theoretical electron change needed to convert the oxidised or
reduced 'state' of
the element in an ion (or in a molecule later), back to its 'elemental
state' where the oxidation state is considered as zero (0).
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Conventions reminders to be strictly
adhered to ...
-
In quoting
oxidation states, the sign (apart from 0) is placed before the number (e.g. ...
–2, –1, 0, +1, +2 ... etc.), and there is always a number after the
sign.
-
For ions not 0), the charge sign comes
after the number (e.g. ... 3–, 2–, –, +, 2+, 3+ ... etc.), but the
case of a single +/– charge no number is required, the '1' is
assumed.
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The 'free element' is considered to have an
oxidation state of zero (0) i.e. not combined with another
element in an ionic or covalent compound. It can of course be
combined with itself e.g. H2, but its still in a zero
oxidation state.
-
So, in terms
of the magnesium–acid reaction ....
-
The magnesium (Mg
atom) is the reducing agent
or electron donor, so its oxidation state is increased.
-
The hydrogen ion (H+)
is the oxidising agent or electron acceptor, so its oxidation state is decreased.
-
The total/net
electron transfers or oxidation state changes must be zero
i.e. for example above
-
For every
magnesium atom losing two electrons, two hydrogen ions gain an
electron each, i.e.
-
magnesium's oxidation
state increases by two units balanced by two hydrogens (ions) each decreasing
their oxidation state by one unit.
-
Oxidation (e– loss) = increase in oxidation state and
reduction (e– gain) = decrease in oxidation state.
-
So, numerically, electrons lost = electrons gained
-
and
increase in oxidation states = decrease in oxidation states.
Section 2. sub-index
QUIZ on oxidation state
Ex
2.2
Heating reactive metals with non–metals
-
When reactive
metals are heated with reactive non–metals, ionic compounds are
readily formed exothermically in a combustion reaction. MgS, Li2O , Na2O2,
the major product of burning sodium in excess
air/oxygen, and Ca3P2
respectively.
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Note that
strict ionic formulae are also given to make the redox analysis clearer.
-
In each case
the metal is the reducing agent (electron donor/loser,
oxidised in process),
-
and the non–metal is the oxidising agent (electron
acceptor/gainer, reduced in process).
-
Ex 2.2.1:
2Al(s)
+ 3Cl2(g) ==>
2Al3+(Cl–)3(s) (or
2AlCl3, aluminium chloride,
has an ionic lattice)
-
Oxidation
state changes:
-
Aluminium atom
==> aluminium ion: oxidation, e– loss, oxidation state
change ..
-
2 x Al
increasing, each from (0) to (+3), losing 3 electrons per atom,
-
chlorine
molecule ==> chloride ion, reduction, e– gain,
oxidation
state change is ..
-
6 x Cl
decreasing, each from (0) to (–1), gaining 1 electron per atom
(or 2 e– per molecule),
-
and the
half–reactions* are written as:
-
(i) oxidation:
Al ==> Al3+ + 3e–
and (ii) reduction: Cl2 + 2e–
==> 2Cl–
-
adding
2 x (i) + 3 x (ii) gives the full equation (iii)
2Al + 3Cl2 ==>
2AlCl3
-
In terms of
the full balanced equation, 6 electrons lost i.e. a rise of 6
oxidation
state 'units', which is balanced by 6 electrons gained i.e. a
decrease of 6 oxidation state 'units'.
-
Note this is an simple example of to use
oxidation states to balance a redox equation.
-
*Note these also
called half–cell potential equations, and is a way of
representing the 'separate' oxidation or reduction in equation
form.
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Ex 2.2.2:
Mg(s)
+ S(s) ==> Mg2+S2–(s)
(or MgS,
ionic magnesium
sulfide)
-
Oxidation state
changes:
-
Magnesium
atom ==> magnesium ion: oxidation, e– loss, oxidation state change ...
-
1 x Mg increasing from
(0) to (+2),
loses two electrons per Mg atom,
-
which is
'electronically' balanced by the ...
-
sulfur atom
==> sulfide ion: reduction, e– gain, oxidation state change
...
-
1 x S decreasing from
(0) to (–2), gaining two
electrons per S atom,
-
and the
half–reactions are written as:
-
(i) oxidation:
Mg ==> Mg2+ + 2e–
and (ii) reduction: S + 2e–
==> S2–
-
adding
(i) + (ii) gives the full equation (iii)
Mg + S ==> MgS
-
In terms of
the full balanced equation, 2 electrons lost i.e. a rise of 2
oxidation
state 'units', which is balanced by 2 electrons gained i.e. a
decrease of 2 oxidation state 'units'.
-
Ex 2.2.3:
4Li(s)
+ O2(g) ==> 2(Li+)2O2–(s)
(or 2Li2O, lithium oxide)
-
Oxidation state
changes:
-
Lithium atom
==> lithium ion: oxidation, e– loss, oxidation state change
for lithium is from 0 to +1
-
4 x Li
increasing, each from
(0) to (+1), losing one
electron per Li atom, 4 electrons change in terms of the
balanced equation
-
which is
'electronically' balanced by the ...
-
two oxygen
molecules ==> two oxide ions: reduction, e– gain,
oxidation state change,
oxidation state change of 0 to –2
-
2 x O decreases,
each from
(0) to (–2),
gaining two electrons per O atom (or four electrons per O2
molecule),
-
and the
half–reactions are:
-
adding
and balancing gives 4 x (i) + (ii) giving the full equation
-
In terms of
the full balanced equation, 4 electrons lost i.e. a rise of 4
oxidation
state 'units' by the lithium atoms, which is balanced by 4 electrons gained i.e. a
decrease of 4 oxidation state 'units' by the oxygen atoms.
-
Ex 2.2.4:
2Na(s)
+ O2(s) ==> (Na+)2O22–(s)
-
(or
Na2O2, sodium peroxide is the
major product in excess air/oxygen)
-
Oxidation state
changes: This is very similar to example 2.2.3
-
Sodium atom
==> sodium ion: oxidation, e– loss, oxidation state change is
from 0 to +1
-
2 x Na increases,
each from
(0) to (+1), losing one
electron per Na atom,
-
which is
'electronically' balanced by the ...
-
oxygen
molecule ==> peroxide ion: reduction, e– gain,
oxidation
state change ...
-
2 x O decreases,
each from
(0) to (–1),
gaining two electrons per O atom (or two electrons per O2
molecule),
-
and the
half–reactions are:
-
adding
and balancing 2 x (i) + (ii) gives the full equation
-
In terms of
the full balanced equation, 2 electrons lost i.e. a rise of 2
oxidation
state 'units' for the sodium atoms, which is balanced by 2 electrons gained i.e. a
decrease of 2 oxidation state 'units' of the oxygen atoms of the oxygen
molecule.
-
Ex 2.2.5:
6Ca(s)
+ P4(s) ==> 2(Ca2+)3(P3–)2(s)
(or 2Ca3P2, calcium phosphide)
-
Oxidation state
changes:
-
Calcium atom
==> calcium ion: oxidation, e– loss, oxidation state change ...
-
6 x Ca increasing,
each from (0) to (+2), losing two
electrons per Ca atom (total 12 e- change),
-
which is
'electronically' balanced by the ...
-
phosphorus
molecule ==> phosphide ion: reduction, e– gain,
oxidation state change
...
-
4 x P decreasing,
each from (0) to (–3),
gaining three electrons per P atom (or twelve electrons per P4
molecule),
-
and the
half–reactions are:
-
(i) oxidation: Ca ==>
Ca2+ + 2e–
and (ii) reduction: P4 + 12e– ==>
4P3–
-
adding
and balancing 6 x (i) + (ii) gives the full equation (iii)
6Ca
+ P4 ==> 2Ca3P2
-
In terms of
the full balanced equation, 12 electrons lost i.e. a rise of 12
oxidation state 'units', which is balanced by 12 electrons gained i.e. a
decrease of 12 oxidation state 'units'.
-
You can also
work out this example based on (iv) P + 3e– ==> P3–,
-
in which case the balanced equation is based on 3 x (i) + 2 x (iv)
==> Ca3P2.
Section 2. sub-index
QUIZ on oxidation state
Ex
2.3
A halogen displacement
reaction
-
A 'more reactive'** halogen displacing a less reactive
halogen (X > Y).
-
halogen X
+ Y halide salt ==>
X halide salt
+ halogen Y
-
X2(aq) + 2KY(aq)
==> 2KX(aq) + Y2(aq)
-
since the potassium ion (K+)
is a spectator, the ionic–redox equation is
-
X2(aq) + 2Y–(aq)
==> 2X–(aq) + Y2(aq)
-
where halogen
X is (in simple terms) more
reactive than halogen Y (F > Cl > Br > I).
-
In more advanced
terms, the EθX2/X-
electrode potential of X is more positive than that of Y
and is therefore the stronger oxidising agent.
-
X2
is the oxidising agent or electron acceptor, X has
an oxidation state of
(0) in the halogen molecule and accepts an electron from Y–, and
so X gets reduced to oxidation state(–1) in
the X– ion.
-
Y–
is the reducing agent or electron donor, Y has an
oxidation state
of
(–1) in the halide ion and donates an electron to X2,
and so Y is oxidised to oxidation state (0) in the Y2
molecules.
-
so in terms
of oxidation states: 2 x X(0)
+ 2 x Y(–1)
==> 2 x X (–1) + 2 x Y (0)
-
and the
half–cell reactions are ...
-
(i) reduction*: X2(aq) + 2e– ==>
2X–(aq)
-
and (ii) oxidation*: 2Y–(aq) ==>
Y2(aq)
+ 2e–
-
adding* (i) + (ii) gives (iii)
X2(aq)
+ 2Y–(aq) ==> 2X–(aq) + Y2(aq)
-
* Note
that the equations can be written on the basis of half–mole of
the halogen molecule,
-
*i.e. (iv)
½X2(aq)
+ e– ==>
X–(aq), (v) Y–(aq) ==>
½Y(aq)
+ e–
-
and
(iv) + (v) gives the equally valid balanced equation
-
(vi)
½X2(aq)
+ Y–(aq) ==> X–(aq) +
½Y2(aq)
-
** More reactive here means the
half–cell potential for X2/X– is more
positive than for Y2/Y–,
-
i.e. X2
is a stronger oxidising agent than Y2. (see
Equilibria Part 7 on half-cell
potentials).
Section 2. sub-index
QUIZ on oxidation state
Ex
2.4 A
metal displacing a less reactive metal
-
A 'more
reactive'*
metal displacing a less reactive metal.
-
copper + silver nitrate ==> silver + copper(II) nitrate
-
Cu(s) + 2Ag+(aq)
==> 2Ag(s) + Cu2+(aq)
-
One copper is
oxidised by losing two electrons, so its oxidation number changes and
increases from (0) to (+2).
-
Two silver ions
are reduced by gaining one electron each, so their oxidation state
decreases from (+1) to (0).
-
and the
half–cell reactions are ...
-
(i) reduction:
Ag(aq) + e– ==> 2X–(aq)
-
and (ii) oxidation: Cu(s) ==> Cu2+(aq)
+ 2e–
-
adding 2 x (i)
+ (ii) gives the full balanced equation.
-
*More reactive here means the
half–cell potential for the more reactive metal is the less
positive or more negative potential, or in other words, the Ag/Ag+ is more positive than
for Cu/Cu2+, i.e. Ag+ is a stronger oxidising
agent than Cu2+. (see
Equilibria Part 7).
Section 2. sub-index
QUIZ on oxidation state
Ex 2.5 What about
oxidation state and the
'not so simple'
situations?
-
How in redox
terms do we deal with the structure and reactions of electrically neutral covalent molecules like H2O
or more complex ions like PO43– which are held together with
covalent bonds but carry an overall electrical charge?
-
These situations
will be redox
analysed by extending the concept of oxidation number or
oxidation state introduced in
section 2.1, but first a recap of the simple
ion situation.
-
As pointed out
already, for simple ions of one atom, you can think of it as the
electron change required to return the atom to an uncombined electrically neutral
atom or molecule, because this relates numerically to the relative
electron loss or gain involved in forming the ion from the free element e.g.
-
Magnesium is in the
+2 oxidation state in ionic compounds containing the magnesium
ion Mg2+
-
Phosphorus is in the
–3 oxidation state
in ionic compounds containing the phosphide ion P3–
-
Oxygen
is in the –2 oxidation state e.g. in ionic compounds containing the
oxide ion O2–
-
Important
conventions,
reminders and trends before proceeding to the 'not so
simple' situations:
-
In writing oxidation
states, the + or – sign is written before the number and there is always
a sign except for zero (0) for the uncombined element itself.
-
There is always a
sign for the electrical charge on an ion (charged particle) symbol/formula,
since it can't be zero by definition, but no number is written for a single
+/– charge, but must have 2, 3 etc. for double or triple
charges etc., and the number comes before the electric charge sign on an
ion.
-
These two conventions must
be strictly adhered to e.g. ion formula Al3+,
oxidation state is
+3.
-
Sometimes
()
are needed in ionic formulae, e.g.
the correct ionic formula (Na+)3P3– is NOT the same as
the incorrect Na+3P3–, because there is no
such ion as Na+3 which implies three sodium atoms
combined carrying an overall charge of single +, rather than the correct three separate
correct Na+ ions, each with a single plus charge.
-
Now, very
importantly conceptually for the more awkward cases ...
-
the more
electropositive elements tend to have a positive oxidation state
even in
a covalently bonded compound or ion e.g. because most metals more readily lose electrons than
non–metals,
-
and the
more electronegative elements tend to have a negative oxidation
state in
a covalently compound or ion because most non–metals more readily gain electrons
compared to metals.
-
BUT, this is where
oxidation state becomes much more abstract.
-
This idea,
based on the relative electronegativity of an element
(table below), is exploited in a way in
the next section 3. to derive the oxidation state of any element in any
molecule or ion consisting of at least two atoms.
-
BUT beware,
initially you may have to
conceive of a molecule/molecular ion as if it is composed of
a combination of simple one atom ions, even if that's not the case in reality
it does help to work through the abstract nature of this topic!
-
Its a useful
conceptualisation of the 'redox' situation i.e. a method of
problem solving at the start of dealing with the 'not so simple'
chemical species. So, read on in section 3 where, by the
end, hopefully you will be able to analyse species without
thinking 'falsely' in terms of combining simple ions that don't
really exist in the situation. You may need to refer to the
tabulated values of electronegativity below.
-
Electronegativity is
a measure of the power of an atom to attract electron
charge from another atom it is covalently bonded to.
i.e. in the context of a covalent bond.
-
Some Pauling
values of electronegativity are quoted below.
-
|
element |
Na |
Mg |
Al |
Mn |
Fe |
H |
Si |
P |
C |
S |
I |
Br |
Cl |
N |
O |
F |
|
electronegativity |
0.9 |
1.2 |
1.5 |
1.5 |
1.8 |
2.1 |
1.8 |
2.1 |
2.5 |
2.5 |
2.5 |
2.8 |
3.0 |
3.0 |
3.5 |
4.0 |
-
Generally
speaking electronegativity increases from left to right across a
period of the periodic table and decreases down a group of the
periodic table.
Section 2. sub-index
QUIZ on oxidation state
3. Oxidation state rules and
guidelines and application to inorganic molecules/ions
-
Its really
important that you have studied at least
section
2.5 before reading this section, and have some idea
on the relative electronegativity
of elements commonly encountered in your inorganic chemistry. It
will greatly help in understanding the assignment of oxidation states
of elements in the quoted compounds on these three web pages.
-
Further more, as much as
possible of the following guidelines and examples, 3.1 to 3.4 should be
learned as quickly as possible and they will help you
throughout your chemistry studies.
-
Oxidation
states in
naming inorganic compounds with
respect to oxidation state is explained in section 4. but they are
unavoidably quoted in here in section
3.
-
3.1: Atoms in elements,
i.e. not covalently or ionically combined with atoms of another elements, are
considered to have an oxidation state of zero or 0
irrespective of the physical, molecular or allotropic* state of the element,
-
e.g.
in Xe(g), Fe(s), H2(g),
Cl2(g), Br2(l), O2(g),
O3(g), P4(s), S8(s), C60(a
fullerene), Cn(graphite) etc.
-
*Allotropes are different
physical forms of the same element in the same physical state
e.g. solid carbon (graphite, diamond, fullerenes), gaseous
oxygen (dioxygen, trioxygen–ozone). The different forms arise
from differences in the bonding BUT they are all in an
oxidation state
of zero)
-
3.2: Irrespective of being
in an ionic or covalent compound or the physical state of the
compound, for
neutral molecules or simple/complex molecular ions of selected
elements the following 'rules' 3.2.1 to 3.2.4 apply ...
-
BUT please note
that ...
-
the total oxidation numbers of all atoms in a compound is zero
(more details
guideline 3.3),
-
the total oxidation
numbers of all atoms in an ion equals the overall charge on the ion
(see
guideline 3.4) and watch the sign
conventions for ion charge and oxidation state/number,
-
the most electronegative element in the compound or ion carries
the negative oxidation state (see above for
Pauling values of electronegativity),
-
there are
important exceptions to the usual oxidation state of an element
mentioned, in terms of the advanced chemistry you will encounter
– so watch out and away we go!
-
Ex
3.2.1:
fluorine: F is always (–1) in all compounds,
-
because it
has the highest electronegativity of any element and only
one electron short of a stable noble gas structure
electronically e.g. formation of fluoride ion F [2.7]
+ e– ==> F– [2.8]
-
The (–1)
state relates to either gaining an electron to form the
fluoride ion, F–, e.g. in ionic sodium fluoride, NaF
(Na+ Cl– ionic bond, Na is +1),
-
or
share with one other electron to form a covalent bond e.g. in
the molecules hydrogen fluoride, HF (δ+H–Fδ–
polar bond, H is +1), or tetrafluoromethane (carbon
tetrafluoride), CF4 (δ+C–Fδ–
polar bond, C is +4).
-
Ex
3.2.2:
hydrogen H is (+1) in most compounds.
-
This
relates to the one outer electron either being lost in
the formation of a hydrogen ion, H+, as in acids,
or being shared with another electron to form a covalent bond pair,
when bonding with a more electronegative atom.
-
e.g. HCl
(δ+H–Clδ– polar bond, Cl is –1), H2O
(δ+H–Oδ– polar bond, O is –2), NH3
(δ+H–Nδ– polar bond, N is –3).
-
One exception
is in the formation of hydrides with metals
where hydrogen's oxidation state is (–1)
-
e.g. in ionic sodium
hydride, Na+H–, Na is +1) or the covalent beryllium
hydride, BeH2 (δ+Be–Hδ–
polar bond, Be is +2) because most metals are more
electronegative than hydrogen.
-
This
situation happens if the other atom of the bond has a very
low electronegativity, i.e. with very electropositive metals
of low ionisation energy such as Group 1 (Li, K, Rb, Cs), Group
2 (Ca, Sr) and also in many covalent metal hydrides of less
electropositive metals than Gps1/2, but still less electronegative than
hydrogen e.g. beryllium BeH2 and aluminium
AlH3.(δ+Al–Hδ–
polar bond, Al is +3)
-
Ex
3.2.3:
oxygen: O
is (–2) in most compounds,
-
Since oxygen is the 2nd most
electronegative element.
-
This
readily relates to the formation of the oxide ion, O2–
with electropositive metals (very low electronegativity) e.g.
MgO (Mg2+O2– ionic bond, Mg is
+2), Al2O3 (Al3+ O2–
ionic bond, Al is +3) etc.,
-
or by sharing
two electrons from other less electronegative atoms to form
two single bonds or one double bond e.g. water, H2O,
δ+H–Oδ2––Hδ+ (H
is +1), or carbon dioxide, CO2, δ2–O=Cδ4+=Oδ2–
(C is +4) molecules represented in terms of partial charges
of the polar bonds due to the electronegativity difference.
-
BUT there
are two important oxidation state exceptions for O
-
(i) In peroxides H2O2
(covalent, H is +1)) or
Na2O2 (ionic, Na is +1)/O22–,where it
is (–1)
-
Here two
oxygen atoms are linked together e.g. H–O–O–H,
-
which changes the
situation compared to H–O–H.
-
and (ii)
O
is (+2) in F2O
-
(There is
also the curious situation regarding the 'super–oxide ion' O2– as in KO2
(K is +1), where each
oxygen has an average oxidation state of 0.5, since the ion is
the equivalent of an dioxygen molecule plus one electron.)
-
Ex
3.2.4: chlorine Cl is –1 in many simple
compounds
-
This is the case when chlorine is
combined with a less electronegative
elements.
-
This is
related electronically to the formation of the chloride ion by
chlorine gaining one electron.
-
or
sharing with one electron from a less electronegative
element in covalently bonding
-
BUT when combined with
the more electronegative O or F, Cl has a (+) oxidation state
-
e.g. Cl is
(+1) in chlorine(I) oxide, Cl2O (δ+Cl–Oδ–
polar bond, O is –2)
-
or (+3) in
chlorine(III) fluoride, ClF3
(δ+Cl–Fδ–
polar bond, F is –1)
-
There dozens
of similar so–called 'interhalogen compounds where the most
electronegative halogen carries the negative oxidation state and
the positive oxidation state of the less electronegative halogen
is shown with (Roman numerals) and it retains its
'elemental' name.
-
There
are also a whole series of 'chlorate ions' and 'oxides', where the more
electronegative oxygen results in chlorine having oxidation states
of (+1), (+3), (+4), (+5) and the maximum possible (+7). The latter
is the maximum oxidation state for any halogen, and is dictated by using all 7 outer electrons in bonding
e.g.
-
chlorate(I), ClO–, chlorate(III), ClO2–,
chlorate(V), ClO3–, chlorate(VII), ClO4–,
-
These sort
of ion examples are explained in more detail in
section 3.4
-
Note that
the name of a non–metallic oxy–anion changes from ...ide to
...ate when the non–metal is combined with oxygen to form
the polyatomic anion and the oxidation state is denoted after
the name by a (Roman numerals number) in brackets.
-
chlorine(IV)
oxide,
ClO2 (chlorine dioxide), bromine(IV) oxide, BrO2,
-
(for more
details and examples see
3.4 charge on ions and
3.5
maximum oxidation state patterns)
-
These
guidelines for chlorine also apply to bromine and iodine.
-
3.3: Since compounds
have no net overall electrical charge, the sum of the oxidation states
of all the atoms in a compound is also zero,
again, this is easy to follow in simple ionic compounds e.g.
-
Ex 3.3.1: Magnesium oxide,
MgO, Mg2+O2–,
-
One Mg2+
balances one O2–, or ...
-
one Mg in the +2
oxidation state balances one O in the –2 oxidation state.
-
Ex 3.3.2: Aluminium
sulfide, Al2S3, ionically (Al3+)2(S2–)3,
-
two 3+ ions
electrically balance three 2– ions, or
-
two Al's
in the +3 oxidation state balance three S's in the –2
oxidation state.
-
But it is
not so easy at first in covalent molecules because you
need to know how to assign the relevant oxidation states. This
is again is readily done by using the difference in electronegativity
guideline, and, particularly at the start, thinking in a sort of 'ionic' conceptual way,
even though the atoms in such an ion are covalently bonded
together and the 'ionic' nature is really due to the overall net
electrical charge
carried by the particle.
-
In covalent molecules the most
electronegative elements carry the negative oxidation states e.g.
using the guidelines
3.2.1 to 3.2.4
-
In the following examples,
I use the word 'carry'
in terms of the oxidation state, but it does NOT mean an ionic
state, but an abstract oxidation state concept as if it was
ionic, but isn't !!!
-
Ex 3.3.3:
Water, H2O
-
Two hydrogens in the
+1 state balance one oxygen in the –2
state because oxygen's electronegativity is higher than
hydrogen's.
-
In other
words, the assignment of oxidation states will coincide with
bond polarity
δ+H–Oδ2––Hδ+
based on electronegativity differences
and you can crudely think of water made up of two H+ ions
and one O2– ion, even though in reality water is neutral covalent
molecule!
- Ex 3.3.4:
Phosphorus
oxychloride, POCl3.
- The two most
electronegative elements are oxygen and chlorine and they will
carry the negative oxidation states of –2 and –1 respectively.
- Phosphorus is the least electronegative and will carry
the positive oxidation sate.
- You can think of this
covalent compound as if it is made up of one P at (+5), one O
at (–2) and three Cl at (–1), so (+5) + (–2) + (3 x –1) = 0.
- Note that by using the
guidelines you can deduce that P is in the +5 oxidation state even
though there is no rule specified to phosphorus quoted here.
- Ex 3.3.5:
Pure liquid
sulfuric acid, H2SO4
- From the guidelines hydrogen is +1,
oxygen is –2, so sulfur will be sulfur in the +6
oxidation
state.
- so, (2 x +1) + (+6) +
(4 x –2) = 0
- Note again, that
by following the guidelines you can deduce that S is in the +6
oxidation state even though there is no rule specified for sulfur.
-
3.4: In
simple ions of one atom the
oxidation state is numerically the same as the charge on the ion.
-
Ex 3.4.1:
Sodium as Na+
has an oxidation state of (+1), oxygen as the oxide ion O2– has an
oxidation state of (–2), for aluminium in the ion Al3+
it is (+3), chlorine as the chloride ion Cl– (–1),
sulfur as the sulfide ion S2– (–2) etc.
-
For ions
of at least two atoms, the charge on an ion is numerically equal to the sum of the oxidation
numbers of all the atoms in the ion.
-
Therefore
you can now deal
with non–electrically neutral molecular ions like PO43–
in the same way outlined for neutral covalent molecules in 3.3,
BUT you must take into account the overall charge on the ion.
-
Ex 3.4.2:
For ions like
phosphate(V),
PO43–
-
Oxygen is more
electronegative than phosphorus, so ..
-
P is +5, O is –2,
and therefore +5 + (4 x –2) = –3 in total, = 3– for
overall charge on the ion.
-
You can
conceive of it made up of one fictitious P5+
combined with four fictitious O2– ions,
even if they don't exist here, but (5+) + (4 x 2–) =
3–.
-
Note again
that the Roman numeral in brackets denotes the positive oxidation
state of phosphorus,
-
the –2 of oxygen is correctly assumed (more on
naming ions/compounds later).
-
Note how oxidation states can be 'partially'
deduced from the charge on the ion or deduce the charge on
the ion from given oxidation states.
-
3.5:
The maximum oxidation state possible is
often the number of outer electrons in the valency shell of the
original neutral atom
-
For Groups 1
to 7 and even the Noble Gases (group 0/18) there is, not
surprisingly, a nice simple pattern related to the Group number
of the Periodic Table e.g.
-
Groups 1–4:
Na is +1 e.g. in NaCl, Ca is +2 e.g. in CaO, Al is +3 e.g. in AlCl3,
C is +4 e.g. in CO2,
-
Groups 5–8:
P is +5 e.g. in H3PO4, S is +6 e.g. in SO42–,
Cl is +7 e.g. in ClO4–, and Xe is +8 e.g. in XeO4.
-
After K and
Ca on Period 4 of the Periodic Table come
the 3d block of metals,
where things are not so clear cut.
-
Up to manganese, the maximum
oxidation state is still dictated by the
number of valency electrons e.g.
-
Sc +3 in
ScCl3, Ti +4 in TiO2, V +5 in VO43–,
Cr +6 in K2Cr2O7, Mn +7 in
MnO4–,
-
but for
Fe, Co, Ni, Cu and Zn the pattern is complex, and generally
one of decreasing max. oxidation state,
-
e.g.
effectively a max. of +3 for Fe since only a few unstable
compounds with over +3 oxidation state exist (e.g. FeO42–
[ferrate(VI) oxyanion], Co max. +3 and usually
in complex ions, for Cu there are a few unstable complex ions showing the
+3 state, but effectively the max. is +2 in e.g. CuSO4
etc. and Zn only exhibits the +2 oxidation state in its compounds
e.g. ZnO, ZnCl2 etc.
-
Apart from
Sc (+3) and Zn (+2), all the rest of the 3d block are true
transition metals exhibiting variable oxidation states, a fact
that is important in the catalytic activity of there compounds
e.g. the
use of V2O5
in the Contact Process for making sulfur trioxide in
the manufacture of sulfuric acid or the
iron(III) ion catalysis
of iodide oxidation by peroxodisulfate..
-
3.6: Other Periodic Table 'patterns'
-
e.g. for those elements
encountered at UK advanced AS–A2 and IB academic level.
-
For Group 1
(+1) and group 2 (+2) elements there is only one stable oxidation
state,
-
Group 3/13 mainly
+1 and +3,
-
Group 4/14 mainly
+2 and +4,
-
Group 5/15 mainly
–3, +3 and +5, (though see the variation for nitrogen in
section 3.7 below)
-
Group 6/16 mainly
–2, +4 and +6,
-
Group 7/17 can be
very variable from –1 all the way to +7 (see
3.2.4 for Cl and Br examples).
-
Group 0/18 for xenon can be +2, +4, + 6 and even
the maximum of +8.
-
3d Block –
Transition metals show values from +1 to +7 depending on the
element.
-
Scandium
(+3) and zinc (+2) have only one stable oxidation state in
compounds and are not true transition metal elements.
-
From titanium
to copper, all of the elements exhibit at least two stable
oxidation states in simple compounds or complex ions and are
considered true transition metals.
-
Lastly a
'curiosity', nickel(0) carbonyl is Ni(CO)4, a sort of neutral
complex formed on heating nickel in carbon monoxide (C +2, O –2)
so Ni is in a zero oxidation state!
-
3.7: Oxidation
state can be quite variable even for a single element
-
e.g. nitrogen's oxidation state in various compounds or
ions:
-
for ammonia NH3,
ammonium ion
NH4+, ammonium chloride NH4Cl,
-
for hydrazine
N2H4 it is (–2),
-
for azo
compounds R–N=N–R it is (–1), R is usually aromatic,
-
for nitrogen N2 it is (0),
-
for
nitrogen(I) oxide N2O (+1),
-
for
nitrogen(II) oxide NO (+2),
-
for sodium
nitrite NaNO2,
nitrite/nitrate(III) ion NO2– and nitrogen
(III) oxide N2O3 it is
(+3)
-
for
nitrogen(IV) oxide NO2 (+4)
-
for nitrogen(V)
oxide N2O5,
nitronium cation NO2+, nitrate ion NO3–, NaNO3 and HNO3
it is
(+5)
-
and just to
complicate matters further in ammonium nitrate NH4NO3,
-
This
sequence illustrates why the concept of oxidation number/state
is so important in redox chemistry e.g. the
nitrogen cycle, which is all about electron transfer via
enzymes!
4.
Oxidation states and naming inorganic compounds
-
4.1: Different compounds
or ions
can be formed from the same two (or more) different elements, so at least one of the elements is
in a different oxidation state in each compound or ion.
-
Therefore there is a need to
indicate this when writing the name of the compound or ion.
-
(Roman numerals) are used
to indicate the value of the oxidation state but with no preceding +
sign.
-
Some examples
have already been dealt with in
section 3.4,
so section 4. acts as a collective recap and extension of using
Roman numeral notation depicting different oxidation states of the
same element in two or more compounds.
-
A
positive oxidation state
is assumed, and if you have worked through the
guidelines in section 3. you should have
no trouble in analysing the examples below and connecting the
(II) etc. in the name with an oxidation state analysis of the
compound or ion.
-
4.2: For
metallic or non–metallic elements the name of the element is used if
NOT in an anion
-
Some of the
old names are still in common use, but try to use the
correct systematic name
e.g.
-
copper(I) oxide Cu2O and
copper(II) oxide CuO
-
iron(II)
chloride FeCl2 and iron(III) chloride FeCl3
-
iron(II)
oxide FeO, iron(III) oxide Fe2O3 and
diiron(II) iron(III) oxide, Fe3O4
-
(once called
ferrous oxide and ferric oxide and triiron
tetroxide)
-
Historic
note: ...ous was the lower oxidation state,
...ic the higher.
-
vanadium(II) sulfate for
VSO4 or V2+SO42–
and vanadium(III) sulfate V2(SO4)3
-
sulfur(IV) oxide SO2
(sulfur dioxide) and sulfur(VI) oxide, SO3
(sulfur trioxide)
-
nitrogen(I)
oxide N2O (dinitrogen oxide) nitrogen(II)
oxide NO (nitrogen monoxide), nitrogen(IV) oxide NO2
(nitrogen dioxide) and nitrogen(V) oxide N2O5
(nitrogen pentoxide).
-
transition
metal complex cations e.g.
-
4.3:
For elements (metal or non–metal) combined with oxygen or other
more electronegative element, giving an anion, the ion name
ends in ...ate with the prefix derived from the elements
name.
In such cases the oxygen carries the negative oxidation state of
(–2) or chlorine (–1) e.g.
-
vanadate(V)
ion, VO43–,
-
manganate(VI)
ion,
MnO42–, manganate(VII) ion,
MnO4–, (was called the permanganate
ion)
-
sulfate(IV)
ion, SO32–
(sulfite) and sulfate(VI)
ion,SO42–
(sulfate)
-
nitrate(III),
NO2– (nitrite) and nitrate(V), NO3– (nitrate)
-
chlorate(I),
ClO–, chlorate(VII), ClO4–
etc. oxygen is more electronegative than chlorine.
-
transition
metal complex anions e.g.
QUIZ on oxidation state
Learning objections for
oxidation state and oxidation numbers
Know what is meant by oxidation state and know
how to assign oxidation state values.
Know that oxidation is a process of electron
loss from an atom, neutral molecule or ion.
Know that an oxidising agent is an electron
acceptor and is reduced in the process.
Know that that oxidation produces an increase in
the oxidation state of an element in any chemical context.
Know that reduction is a process of electron
gain by an atom, neutral molecule or ion.
Know that a reducing agent is an electron donor
and is oxidised in the process.
Know that that reduction produces a decrease in
the oxidation state of an element in any chemical context.
Know how to interpret redox equations by
deducing the oxidation and reduction in terms of electron loss
and electron gain.
Know how to interpret redox equations by
deducing which element is oxidised to increase its oxidation
state and which element has been reduced, reducing its oxidation
state.
Be able to write, as well as analyse,
displacement reactions of metals added to other metal salt
solutions and be able to predict whether a reaction is
feasible.
Know that if a compound molecule is neutral, the
sum of the oxidation states of the atoms is zero.
Know that for an ion, simple or complex, the sum
of the oxidation states equals of all the atoms equals the
overall charge on the ion (the theoretical 'net charge').
Be able to use Pauling electronegativity values
to assign an element's oxidation state in more complex molecules
and ions as well as simple binary ionic compounds.
Know how to deduce the likely maximum oxidation
state of an element in its compound or ions, for the elements in
groups 1-2 and 13 to 18 (old groups 3 to 0) it is limited to the
maximum number of outer shell electrons, but more complex for
d-block elements, especially the transition metals series.
[SEARCH
BOX]
QUIZ on oxidation state
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