|
5. Using oxidation states
to describe redox changes in a given reaction equation
Ex
5.1 The
reaction between aluminium and a copper(II) salt solution
Ex
5.2 The reaction
between sulphur dioxide/sulphite and halogens
-
If neutral molecules or more
complex ions are involved, a bit more care must be taken e.g. when the
sulphur dioxide is oxidised to sulphate by bromine (or the reduction of
bromine to bromide).
-
SO2(aq)
+ Br2(aq) + 2H2O(l) ==> SO42–(aq)
+ 2Br–(aq) + 4H+(aq)
-
(i) the oxidation half
reaction is: SO2(aq) + 2H2O(l) ==>
SO42–(aq) + 4H+(aq)
+ 2e–
-
(ii) the reduction
half–reaction is: Br2(aq) + 2e– ==> 2Br–(aq)
-
The hydrogen (+1) and
oxygen (–2) do not change oxidation state.
-
(i) + (ii)
equals the balanced equation, 2 electrons gained and lost or an ox.
state rise and fall of 2 units.
-
Bromine is the
oxidising agent (gain/accept e–s, lowered ox. state),
-
and sulphur dioxide is the reducing agent (loses e–s,
inc. ox. state of S).
-
Sulphur dioxide does
ionise to a small extent in water to give the sulphite ion, and adding a
strong non–oxidising acid like dilute hydrochloric acid to sodium
metabisulphite produces the ion, which means another equation can also adequately describe
the redox change in terms of sulphur and bromine.
-
Section 5. Index of examples of redox
equation analysis
Ex
5.3
The
oxidation of ammonia with molecular oxygen
-
The concept of oxidation
state can now be fully applied to reactions which do not involve ions e.g.
-
The
oxidation of ammonia via a Pt catalyst at high
temperature which is part of the chemistry of nitric acid manufacture.
-
4NH3(g) + 5O2(g)
==> 4NO(g) + 6H2O(g)
-
The oxidation number
analysis is:
-
4N at (–3) each
in NH3 and 10O all at (0) in O2
change to ...
-
4N at (+2) each
in NH3, 4O at
(–2) each and 6 O at (–2) each in H2O.
-
H is
+1 throughout i.e. does not undergo an ox. state change.
-
Oxygen is reduced from ox.
state (0) to (–2).
-
Nitrogen is oxidised from
ox. state (–3) to (+2).
-
The total
increase in ox. state change of 4 x (–3 to +2) for nitrogen
is balanced by the total decrease in ox. state change of 10 x (0 to –2) for oxygen
i.e. 20
e– or ox. state units change in each case.
-
Oxygen is the
oxidising agent (gain/accept e–s, lowered ox. state)
and ammonia is the reducing agent (loses e–s, inc. ox.
state of N).
-
Section 5. Index of examples of redox
equation analysis
Ex
5.4 The reaction
between iron and steam at >400oC
-
4Fe(s) +
4H2O(g) ==> Fe3O4(s) + 4H2(g)
-
Oxidation
state analysis:
-
O stays at ox.
state –2
(no change) but for the ....
-
8H's in
H2O, the ox. state decreases from +1 to 0 in
H2
(reduction, 8e–
gained),
-
this oxidation
number decrease is balanced by the oxidation number increases from
...
-
iron, 3Fe ox. state increases from 0 to one at +2 and two at +3
-
both oxidations, total
8e– loss, because the compound consists Fe2+, 2Fe3+ and 4O2–
ions.
-
Water is the
oxidising agent (gain/accept e–s, lowered ox. state of
H),
-
and iron is the reducing agent (loses e–s, inc. ox.
state).
-
Section 5. Index of examples of redox
equation analysis
-
For more on iron chemistry see Part 10b
3d–block Transition Metals Fe to Zn – detailed revision notes
Ex
5.5 The formation
of titanium(IV) chloride from titanium(IV) oxide in the extraction of
titanium metal
-
TiO2(s) +
C(s) + 2Cl2(g) ==> TiCl4(l) + CO2(g)
-
Oxidation number
changes:
-
The carbon,
C
is oxidised (0 to +4 in CO2) and the chlorine is reduced
(0) in Cl2 to (–1) in TiCl4.
-
The
oxidation of 1 x C from (0) to (+4) is balanced by the
reduction of 4 x Cl from (0) to (–1).
-
Titanium (+4) and oxygen
(–2) do not change.
-
Chlorine is
the oxidising agent (gains e–s, lowered ox. state) and
carbon is the reducing agent (loses e–s, inc. ox. state
of C), but not in sense carbon reduces an iron oxide to iron in a
blast furnace because titanium does not change oxidation state and another step is required to obtain the
metal.
-
The titanium itself is
extracted via another redox reaction by displacement with a more reactive
metal.
-
TiCl4(l) +
4Na(s) ==> Ti(s) + 4NaCl(s)
-
Ox. state
changes :
-
The titanium
is reduced from (+4) in TiO2 to (0)
as Ti, and the displacing metal is oxidised (from 0 to
>+1).
-
The
reduction of 1 x
Ti (+4) to (0) is balanced by the oxidation of 4 x Na (0) to (+1)
-
or TiCl4(l)
+ 2Mg(s) ==> Ti(s) + 2MgCl2(s)
-
In both
cases titanium(IV) chloride is the oxidising agent (gain/accept
e–s, lowered ox. state of Ti) and sodium/magnesium
are the reducing agent (lose/donate e–s, inc. their
ox. state).
-
Section 5. Index of examples of redox
equation analysis
-
For more on titanium chemistry see
Periodic Table Advanced Inorganic Chemistry Notes Part
10a "3d block Transition Metals Series Introduction and Elements Sc to
Mn on Period 4
– detailed revision notes
Ex
5.6 The combustion of liquid hydrazine and oxygen
-
This has been
used as a liquid rocket fuel mixture.
-
N2H4(l)
+ O2(l) ==> N2(g) + 2H2O(g)
-
Oxidation state
changes:
-
The nitrogen in
hydrazine oxidised (N from –2 to 0) and the oxygen molecules are reduced
(O from 0 to –2).
-
The oxidation of 2 x
N from (–2) to (0) is balanced by the reduction of 2 x O from (0) to
(–2).
-
H at (+1) does not
change in oxidation state and nitrogen is initially (–2) here, NOT (–3) as in ammonia, NH3. -
Oxygen is
the oxidising agent (gains/accepts e–s, lowered ox.
state) and hydrazine is the reducing agent (loses/donates e–s,
inc. ox. state of N).
-
Section 5. Index of examples of redox
equation analysis
Ex
5.7 Converting manganese(IV) oxide into potassium manganate(VI)/manganate(VII)
-
5.7.1: If a mixture of
manganese(IV) oxide, potassium hydroxide and potassium chlorate(V) is
heated strongly and fuse in a crucible, the following redox reaction takes
place:
-
3MnO2 +
6OH– + ClO3– ==> 3MnO42–
+ 3H2O + Cl–
-
The
manganate(VI) ion is formed and the ox. number changes are ...
-
the oxidation of 3 x
Mn from (+4) to (+6), MnO2 ==> MnO4–,
total 6 e's lost
-
Is balanced by the reduction of 1 x
Cl from
(+5) to (–1), ClO3– ==> Cl–,
total 6 e's gained
-
and hydrogen (+1)
and oxygen (–2) do not change oxidation state.
-
The
chlorate(V) ion is the oxidising agent (gains/accepts e–s,
lowered ox. state of Cl) and manganese(IV) oxide is the reducing
agent (loses/donates e–s, inc. ox. state of Mn).
-
5.7.2: When the fused mixture
is dissolved in water, the initially green solution of the manganate(VI)
ion, slowly changes to the purple colour of the manganate(VII) ion and a black
precipitate of manganese(IV) oxide.
-
3MnO42–(aq)
+ 2H2O(l) ==> 2MnO4–(aq)
+ MnO2(s) + 4OH–(aq)
-
Oxidation number
changes:
-
Initially there are three Mn at (+6).
-
Two Mn at (+6) are
oxidised to two Mn(+7), an ox. state total increase of 2 units or 2e–
lost,
-
and one Mn at (+6) is reduced to
Mn(+4), an ox. state total decrease of 2
units or 2e– gained.
-
This is an
example of disproportionation where an element in
one oxidation state simultaneously changes into a higher and
lower oxidation state species.
-
It also
means that the manganate(VI) ions simultaneously acts as a
reducing agent and oxidising agent!
-
As in 5.7.1, hydrogen (+1)
and oxygen (–2) do not change in oxidation state.
-
Section 5. Index of examples of redox
equation analysis
-
For more on manganese chemistry see
Periodic Table Advanced Inorganic Chemistry Notes Part
10a "3d block Transition Metals Series Introduction and Elements Sc to
Mn on Period 4
– detailed revision notes
Ex
5.8 The disproportionation of copper(I) oxide in acid
-
When brown
copper(I) oxide is dissolved in dilute sulphuric acid, a deposit
of copper is formed and a blue solution of copper(II) sulphate.
-
Cu2O(s)
+ H2SO4(aq) ==> Cu(s) + CuSO4(aq)
+ H2O(l)
-
Cu2O(s)
+ 2H+(aq) ==> Cu(s) + Cu2+(aq)
+ H2O(l)
-
The hydrated
copper(I)
ion, Cu+(aq) is unstable and changes into two
oxidation states simultaneously,
-
elemental Cu(0) and the
blue
copper(II) ion (+2), and is another example disproportionation (see
also 5.7.2).
- i.e. 2 x Cu (+1)
changes to 1 Cu at (0) by reduction and 1 Cu at (+2) by oxidation.
-
Oxygen in its (–2) ox.
state and hydrogen in its (+1) ox. state do not change.
-
It also means
that the copper(I) ions simultaneously acts as a reducing agent
and oxidising agent and are simultaneously oxidised and reduced.
-
You also
observe the same disproportionation if copper(I) sulphate is
dissolved in water
-
Cu2SO4(s)
+ aq ==> Cu(s) + Cu2+(aq) + SO42–(aq)
-
Section 5. Index of examples of redox
equation analysis
-
For more on copper chemistry see Part 10b
3d–block Transition Metals Fe to Zn – detailed revision notes
Ex
5.9 The conversion of cobalt(II) to cobalt(III)
via molecular oxygen (air) or hydrogen peroxide
-
When aqueous
ammonia is added to cobalt(II) salt solutions the hexaamminecobalt(II) complex is formed which is readily oxidised to the
cobalt(II) complex by (i) oxygen dissolving from air or (ii) adding hydrogen peroxide
solution.
-
[Co(H2O)6]2+(aq) +
6NH3(aq) ==> [Co(NH3)6]2+(aq) +
6H2O(l)
-
(i) 4[Co(NH3)6]2+(aq) + O2(aq) + 4H+(aq)
==> 4[Co(NH3)6]3+(aq) + 2H2O(l)
-
Oxidation:
In the complexes 4 Co at (+2) change to 4 Co at (+3),
total increase of 4 ox. state
units.
-
Reduction: 2
O in O2 at (0) change to 2 O at (–2) in 2H2O,
total decrease of 4 ox. state units.
-
No redox
change involving the NH3 ligand or the H+ ions, N
stays at –3 and H at +1.
-
(ii) 2[Co(NH3)6]2+(aq) + H2O2(aq) +
2H+(aq) ==> 2[Co(NH3)6]3+(aq) + 2H2O(l)
-
Oxidation: 2
Co at (+2) change to 2 Co at (+3),
total increase of 2 ox. state units.
-
Reduction: 2
O
at (–1) in H2O2 change to 2 O at (–2) in 2H2O,
total decrease of 4 ox. state units.
-
No redox
change involving the NH3 ligand or the H's of the hydrogen
peroxide molecule or hydrogen ions.
-
Oxygen/hydrogen peroxide act as the oxidising agent (gains/accepts
e–s, lowered ox. state of O) and the cobalt(II) complex
is the reducing agent (loses/donates e–s, inc. ox.
state of Co).
-
Section 5. Index of examples of redox
equation analysis
-
For more on cobalt chemistry see Part 10b
3d–block Transition Metals Fe to Zn – detailed revision notes
Ex
5.10 The conversion of chromium(III) to chromium(VI) compounds
Ex
5.11 The
oxidation of hydrogen sulphide with iron(III) ions
-
If an iron(III) salt (old name,
ferric salt) is
added to hydrogen sulphide solution a precipitate of sulphur forms and the
orange–brown solution turns pale green.
-
H2S(aq)
+ 2Fe3+(aq) ==> 2Fe2+(aq)
+ 2H+(aq) + S(s)
-
Oxidation: 1
S
at (–2) change to 1 S at (0), H2S
==> S, a loss of 2 electrons, inc. 2 ox. state units.
-
Reduction: 2
Fe at (+3) change to 2 Fe at (+2), gain in total of 2 electrons,
decrease in 2 ox. state units.
-
No change in
the oxidation state of the 2H's (+1) involved.
-
The iron(III)
ion acts as the oxidising agent (gains/accepts e–s,
lowered ox. state of Fe) and the hydrogen sulphide is the reducing agent (loses/donates e–s, inc. ox. state of
S).
-
Section 5. Index of examples of redox
equation analysis
-
For more on iron chemistry see Part 10b
3d–block Transition Metals Fe to Zn – detailed revision notes
Ex
5.12 The reaction
of copper(II) ions with iodide ion
-
When potassium
iodide solution is added to a copper salt solution, a white
precipitate of copper(I) iodide is formed which is masked by a brown solution/black ppt. of
iodine. If sodium thiosulphate solution is added dropwise
carefully, it reacts to remove the iodine giving the colourless
iodide ion (see Ex 5.12), hence the
CuI precipitate is better scene.
-
2Cu2+(aq)
+ 4I–(aq) ==> 2CuI(s) + I2(aq/s)
-
Oxidation: 2
I–
at (–1) of the 4I– change to 2 at (0) in
I2,
total 2e– loss, inc. 2 ox. state units.
-
Reduction: 2
Cu at (+2) change to 2 Cu (+1) in the CuI, total 2e– gain,
decrease 2 ox. state units.
-
Two of the I–
iodide
ions do not change oxidation state, but they do change their
physical–chemical situation.
-
The copper(II)
ion acts as the oxidising agent (gains/accepts e–s,
lowered ox. state of O) and the iodide ion is the reducing
agent (loses/donates e–s, inc. ox. state of I).
-
Section 5. Index of examples of redox
equation analysis
-
For more on copper chemistry see Part 10b
3d–block Transition Metals Fe to Zn – detailed revision notes
Ex
5.13 More
examples of disproportionation and vice versa (involving O and N
ions/compounds)
-
5.13.1
Hydrogen peroxide decomposition, catalysed by the black solid
manganese(IV) oxide, MnO2.
-
2H2O2(aq)
==> O2(g) + 2H2O(l)
-
Ox. state
changes: 4O at (–1) change to 2O at (0) in O2 and 2O
at (–2) in H2O
-
and H is unchanged at (+1).
-
A case of
disproportionation where an element in a species simultaneously
changes into a higher and lower oxidation state i.e. here two
oxygen atoms increase their oxidation state and two oxygen atoms
decrease their oxidation state.
-
It also
means that hydrogen peroxide simultaneously acts as a reducing
agent and oxidising agent.
-
5.13.2
The reaction between ammonium and nitrate(III) (nitrite)
ions
-
NH4+(aq)
+ NO2–(aq) ==> 2H2O(l)
+ N2(g)
-
Here its the
opposite of disproportionation where two species of an element in
different oxidation states react to produce one species of a
single
intermediate oxidation state.
-
Ox. state
changes: Nitrogen in a (–3) and a (+3) state both end up in the
(0) state.
-
Oxygen at
(–2) and hydrogen (+1) remain unchanged in oxidation state.
-
The nitrite
ion acts as the oxidising agent and gets reduced (N +3 to 0,
3e's gained, decrease of 3 ox. state units)
-
and
the ammonium ion acts as the reducing agent and gets oxidised (N
–3 to 0, 3 e's lost, inc. ox. state 3 units).
-
The nitrite
ion acts as the oxidising agent (gains/accepts e–s,
lowered ox. state of N) and the ammonium ion acts as the reducing agent (loses/donates e–s, inc. ox. state of
N).
-
See other
disproportionation reactions 5.7.2 MnO42–,
5.8 Cu2O,
5.14 chlorine and an
organic example.
-
and the
opposite of disproportionation! reactions 6.6
iodate(V) + iodide.
-
Section 5. Index of examples of redox
equation analysis
5.14 Some
chlorine, chlorates and chloride redox changes
-
In all the
reactions quoted in section 5.14, (i) the oxidation states of
hydrogen (+1) and oxygen (–2) remain unchanged and (ii) the
process descriptions are over simplified but the main reactions
described provide good examples of the redox chemistry of
chlorine.
-
5.14.1
With cold dilute sodium hydroxide solution alkali sodium chlorate(I) (NaClO, the bleach sodium
hypochlorite) is formed as well as sodium chloride.
-
2NaOH(aq)
+ Cl2(aq) ==> NaCl(aq) + NaClO(aq)
+ H2O(l)
-
2OH–(aq)
+ Cl2(aq) ==> Cl–(aq) +
ClO–(aq) + H2O(l)
-
The chlorine
disproportionates from 2Cl(0) to 1Cl
(–1, chloride ion) plus 1Cl(+1, chlorate(I)
ion).
-
Overall 1
electron gained, (1 ox. state unit decrease) balanced by 1
electron lost (1 ox. state unit increase).
-
5.14.2
However, with hot concentrated sodium hydroxide solution, above 75oC, the formation of sodium
chlorate(V) predominates as well as sodium chloride.
-
6NaOH(aq)
+ 3Cl2(aq) ==> 5NaCl(aq) + NaClO3(aq)
+ 3H2O(l)
-
6OH–(aq)
+ 3Cl2(aq) ==> 5Cl–(aq)
+ ClO3–(aq) + 3H2O(l)
-
The chlorine
disproportionates from 6Cl(0) to 5Cl(–1,
chloride ion) plus 1Cl(+5, chlorate(V)ion).
-
Overall 5
electrons gained, (5 ox. state unit decrease) balanced by 5
electrons lost (5 ox. state unit increase).
-
5.14.3
The change in reaction mode from 5.14.1 to 5.14.2 is due to the
instability of the chlorate(I) ion, which at higher temperatures
disproportionates into the chloride ion and the chlorate(V) ion.
-
3NaClO(aq)
==> 2NaCl(aq) + NaClO3(aq)
-
3ClO–(aq)
==> 2Cl–(aq) + ClO3–(aq)
-
The 'chlorine'
in the chlorate ion disproportionates from 3Cl(+1)
to 2Cl(–1, chloride ion) plus 1Cl(+5,
chlorate(V) ion).
-
Overall 4
electrons gained, (4 ox. state unit decrease) balanced by 4
electrons lost (4 ox. state unit increase).
-
5.14.4 A concentrated solution of
sodium chlorate(I) is a useful source
of chlorine in the laboratory because it readily reacts with conc.
hydrochloric acid to give off the gas.
-
NaClO(aq)
+ 2HCl(aq) ==> NaCl(aq) + H2O(l)
+ Cl2(aq/g)
-
ClO–(aq)
+ Cl–(aq) + 2H+(aq)
==> H2O(l) + Cl2(aq/g)
-
The
'chlorine' here does the opposite of disproportionation and
changes from 1Cl(+1, chlorate(I) ion) plus 1Cl(–1,
chloride ion) to give 2Cl(0, chlorine molecule).
-
Overall 1
electron lost, (1 ox. state unit increase) balanced by
electron gained (1 ox. state unit decrease).
-
Section 5. Index of examples of redox
equation analysis
6.
Constructing full ionic–redox equations
from half–cell equations
REDOX EQUATION
CHECKS
-
Use of the
correct 'species'
of the half–cell equations to be put together, the right way round
AND in the correct ratio based on the number of electrons
transferred or oxidation state changes.
-
Getting the right
ratio of the oxidation/reduction half–cell reactions should ensure the electrons
are 'hidden' and you can add them up in a simple algebraic way
– see the tabular expression of the way of thinking.
-
If not told,
you must
decide on the direction
of change – which is oxidised or reduced? from EØ data
supplied (half–cell potentials) and the most +ve
(least –ve) half–cell potential indicates the
reduction half–equation. Much more on this in Equilibria Part 7.
- The
total increase in oxidation states of elements = the total decrease in
oxidation states of elements,
-
or, total electrons gained
by species = total electrons lost by the
species involved.
-
Add up the ion charges,
the totals should be the same on both sides of the equation. I find this a
handy extra check especially with stray H+'s and H2O's.
-
The 'traditional' atom count
–
do last because its not completely reliable with redox equations!
Ex
6.1 The reaction between
zinc metal and a silver salt solution
All the
half–cell equations are presented as a reduction – electron gain, so
one must be reversed!
See also Equilibria Part 7 Redox Reactions
for Half cell equilibria, electrode potential, standard hydrogen electrode, Simple cells and notation,
Electrochemical Series, EØcell for reaction feasibility, 'batteries' and fuel cell systems
etc.
-
Half–cell reaction
data:
-
(i) Zn2+(aq) + 2e– ==>
Zn(s) (EØ = –0.76V*,
Zn will act as reducing agent, EØ less positive)
-
(ii) Ag+(aq) + e– ==>
Ag(s) (EØ = +0.80V*, reduction of the oxidising agent with the
more positive EØ)
-
*
It doesn't matter here if you haven't
yet studied EØ, half–cell potentials in detail, but the more
+ve half–cell species acts as the oxidising agent and so is the reduction
half of the reaction.
-
The more reactive
metal Zn, displaces the less reactive metal (Ag) from one of its compounds, which is
the reaction feasibility rule at lower academic levels for such a redox reaction (see also halogen
displacement reaction 6.2 below).
-
With redox analysis
of the reaction we can now say:
-
The zinc is
oxidised from 0 to +2 in ox. state, 2e–
loss,
-
and the
two silver ions
are reduced from –1 to 0 ox. state, 2 x 1e–
gain.
-
The zinc metal is
a stronger reducing agent (more powerful e– donor, less +ve
EØ) than silver,
-
or to put it
another way,
-
the Ag+
ion is a stronger oxidising agent (more powerful e– acceptor,
more +ve EØ) than the Zn2+
ion.
-
So one of
the Zn half–cell equations will be balanced by two of the
silver half–cell equations giving the complete
ionic–redox equation, showing NO electrons.
-
|
1 x
oxidation half–cell, (i) reversed |
Zn(s) ==> Zn2+(aq) + 2e–
|
|
2 x
reduction half cell, (ii) |
2Ag+(aq) +
2e– ==>
2Ag(s) |
|
added gives
full redox equation |
Zn(s) +
2Ag+(aq) ==> Zn2+(aq)
+ 2Ag(s) |
-
This sort of
displacement reaction can be used to plate more reactive metals with a
less reactive metal without the need for electrolysis–electroplating e.g.
dipping iron/steel into copper(II) sulphate to give a pink–brown
coating of copper.
-
Section 6. Index of examples of constructing
balanced ionic redox reaction equation from half–cell/EØ
data
-
See also Periodic Table Advanced Inorganic Chemistry Notes Part
10a "3d block Transition Metals Series Introduction and Elements Sc to
Mn on Period 4
– detailed revision notes
Ex
6.2 The reaction between
aqueous chlorine and potassium iodide solution
-
Half–cell reaction
data:
-
(i)
1/2Cl2(aq) +
e– ==> Cl–(aq) (EØ
= +1.36, can be Cl2(aq) +
2e– ==> 2Cl–(aq) but EØ
still +1.36)
-
(ii)
1/2I2(aq) + e– ==>
I–(aq) (EØ = +0.54V,
can be I2(aq) + 2e– ==> 2I–(aq)
but EØ still +0.54V)
-
Chlorine
molecules are reduced from ox. state (0) to (–1) of the chloride
ion, 1 electron gain.
-
Iodide ions
are oxidised from ox. state (–1) to (0) of the iodine molecule,
1 electron loss.
-
So chlorine
molecules are the oxidising agent (more powerful e– acceptor, more
+ve EØ) and iodide ions are the reducing agent (e–
donor, less +ve EØ).
-
|
2 x
oxidation half–cell, (ii) reversed |
2I–(aq) ==>
I2(aq) + 2e– |
|
2 x
reduction half–cell, (i) |
Cl2(aq) +
2e– ==> 2Cl–(aq) |
|
added gives
full redox equation |
Cl2(aq)
+ 2I–(aq) ==>
Cl2(aq) + I2(aq) |
-
One method of
estimating chlorine in water e.g. from bleaches, is to add excess
potassium iodide and titrating the liberated iodine with
standardised sodium thiosulphate, which itself is another redox
reaction (see Ex 6.10)
-
Section 6. Index of examples of constructing
balanced ionic redox reaction equation from half–cell/EØ
data
Ex
6.3 The reaction between
hydrogen peroxide and iron(II) ions
-
Half–cell reaction
data:
-
(i) H2O2(aq)
+ 2H+(aq) + 2e– ==> 2H2O(l) (EØ
= +1.36, reduction of oxidising agent with the more positive EØ)
-
(ii) Fe3+(aq)
+ e– ==> Fe2+(aq) (EØ = +0.77,
less positive, so Fe3+ can't oxidise hydrogen
peroxide)
-
Both the iron(III) ion
and hydrogen peroxide molecule can act as oxidising agents, but hydrogen
peroxide is stronger and so oxidises the iron(II) ion to the iron(III) ion.
-
Oxidation: Two
iron(II) ions at (+2) lose an electron each to give
iron(III) ions
at (+3) ox. state.
-
Reduction: 2
O
at (–1) in each H2O2 are reduced to the (–2)
state in the 2H2O.
-
The hydrogens
(+1) do not change oxidation state.
-
|
2 x
oxidation half–cell, (ii) reversed |
2Fe2+(aq)
– 2e– ==> 2Fe3+(aq) |
|
1 x
reduction half–cell, (i) |
H2O2(aq)
+ 2H+(aq) + 2e– ==> 2H2O(l) |
|
added gives
full redox equation |
2Fe2+(aq)
+ H2O2(aq) + 2H+(aq) ==>
2Fe3+(aq) + 2H2O(l)
|
-
This reaction
is used to convert e.g. iron(II) sulphate, FeSO4, into
iron(III) sulphate, Fe2(SO4)3,
because dissolving iron in dil. sulphuric acid gives the Fe(II)
salt.
-
Section 6. Index of examples of constructing
balanced ionic redox reaction equation from half–cell/EØ
data
-
For more on iron chemistry see Part 10b
3d–block Transition Metals Fe to Zn – detailed revision notes
Ex
6.4 The reaction between acidified manganate(VII) ions and iron(II) ions
-
Half–cell reaction
data:
-
(i) MnO4–(aq) +
8H+(aq) + 5e– ==> Mn2+(aq)
+ 4H2O(l) (EØ = +1.52, reduction of
oxidising agent)
-
(ii) Fe3+(aq)
+ e– ==> Fe2+(aq) (EØ = +0.77,
Fe2+ acts as reducing agent, Fe2+ gets
oxidised, EØ less positive)
-
Oxidation:
Iron(II) ions, Fe2+, (+2) lose an electron, so
oxidised to the iron(III)
ion, Fe3+, (+3), Fe +2 to +3 ox. state.
-
Reduction: Manganate(VII)
ions, MnO4–,
(+7) are reduced to manganese(II) ions, Mn2+, (+2), 5e– gain, so five Fe2+
ions can be oxidised, Mn +7 to +2 ox. state.
-
Hydrogen (+1)
and oxygen (–2) do not change oxidation state.
-
|
5 x
ox. half–cell, (ii) reversed |
5Fe2+(aq) ==>
5Fe3+(aq) + 5e– |
|
1 x
reduction half–cell, (i) |
MnO4–(aq) +
8H+(aq) + 5e– ==> Mn2+(aq)
+ 4H2O(l) |
|
added
full redox equation |
MnO4–(aq) +
8H+(aq) + 5Fe2+(aq) ==>
Mn2+(aq)
+ 5Fe3+(aq) + 4H2O(l) |
-
This reaction is
used to quantitatively estimate iron(II) ions and is self–indicating. On the addition of standardised potassium manganate(VII) to the iron
solution, decolourisation occurs as the almost colourless Mn(II)
ion (a VERY pale pink) is formed from the reduction of the intensely
purple manganate(VII) ion, and the end–point is the
first permanent pale pink with =< 1 drop excess of the
oxidising agent.
-
The presence
of dilute sulfuric ('supplier' of the proto, H+ ion), ensures the desired
sole reduction of the
manganate(VII) ion to the Mn(II) ion, thereby preventing the
formation of a manganese(IV) oxide precipitate. Formation of MnO2
which would not give a good end point and cause a duality in the
redox reactions occurring, so introducing errors and
quantitative complications.
-
There is a
2nd good reason for using dilute sulphuric acid, as opposed to
using other common mineral acids. Dil. sulphuric acid does not
undergo any redox reactions under the conditions of this
titration.
-
Dilute
hydrochloric acid cannot be used because the manganate(VII) ion
will oxidise the chloride ion (see 6.11)
and dil. nitric acid, via the nitrate(V) ion, will oxidise the
iron(II) ion, i.e. both acids will lead to false titration
results.
-
Section 6. Index of examples of constructing
balanced ionic redox reaction equation from half–cell/EØ
data
-
For more on manganese chemistry see
Periodic Table Advanced Inorganic Chemistry Notes Part
10a "3d block Transition Metals Series Introduction and Elements Sc to
Mn on Period 4
– detailed revision notes
Ex
6.5
The reaction between
acidified potassium dichromate(VI) and iron(II) ions
-
Half–cell reaction
data:
-
(i) Cr2O72–(aq)
+ 14H+(aq) + 6e– ==> 2Cr3+(aq)
+ 7H2O(l) (EØ = +1.33,
reduction of the oxidising agent)
-
(ii) Fe3+(aq)
+ e– ==> Fe2+(aq) (EØ = +0.77,
Fe2+ will act as reducing agent, EØ
less positive)
-
Oxidation:
Iron(II) ions at (+2) lose an electron each to give an iron(III)
ion at (+3), Fe +2 to +3 ox. state.
-
Reduction:
Each Cr at (+6) is reduced by gaining 3e– to give Cr at
(+3) ox state, Cr +6 to +3 ox. state.
-
the Cr2O72–
ion is the
oxidising agent and each can oxidise 6 Fe2+ ions.
-
Hydrogen (+1)
and oxygen (–2) do not change oxidation state.
-
|
6 x
ox. half–cell, (ii) rev. |
6Fe2+(aq) ==>
6Fe3+(aq) + 6e– |
|
1 x
red'n half–cell, (i) |
Cr2O72–(aq) + 14H+(aq)
+ 6e– ==> 2Cr3+(aq) + 7H2O(l)
|
|
added
full equation |
Cr2O72–(aq)
+
14H+(aq) + 6Fe2+(aq)
==>
2Cr3+(aq)
+ 6Fe3+(aq) + 7H2O(l) |
-
Like with
potassium manganate(VII), standardised potassium dichromate(VI)
solution can be used to estimate quantitatively iron(II) ions in
solution, though a special redox organic dye* indicator
which must be used to
detect the end point.
-
The organic
dye changes colour when oxidised to another form, but only after
the iron is oxidised i.e. it is not as easily oxidised as Fe2+,
i.e. the dye's EØ is more +ve than Fe2+
but lees than for the manganate(VII) ion, hence it is capable of being oxidized by the dichromate(VI) ion
to show the end–point.
-
Section 6. Index of examples of constructing
balanced ionic redox reaction equation from half–cell/EØ
data
-
For more on chromium chemistry see
Periodic Table Advanced Inorganic Chemistry Notes Part
10a "3d block Transition Metals Series Introduction and Elements Sc to
Mn on Period 4
– detailed revision notes
Ex
6.6
The reaction
between iodate(V) and iodide ions in acidified aqueous solution
-
Half–cell reaction
data:
-
(i)
1/2I2(aq) + e– ==>
I–(aq) (EØ = +0.54V,
iodide gets oxidised, acts as reducing agent, less positive
EØ)
-
(ii) IO3–(aq) +
6H+(aq) + 5e– ==>
1/2I2(aq) + 3H2O(l)
(EØ = +1.19V, reduction of
oxidising agent, more positive EØ)
-
The iodide
ions (I at –1) are oxidised to iodine molecules (I at 0) by electron
loss to the iodate(V) ion, I –1 to 0 ox. state.
-
The iodate(V)
ions (I at +5) are reduced to iodine molecules (I at 0) by
electron gain from the iodide ions (the reducing agent), I +5
to 0 ox. state.
-
Hydrogen (+1)
and oxygen (–2) do not change oxidation state.
-
|
5 x
ox'n half–cell, (i) reversed |
5I–(aq) ==>
5/2I2(aq) + 5e– |
|
1 x
reduction half–cell, (ii) |
IO3–(aq) +
6H+(aq) + 5e– ==>
1/2I2(aq) + 3H2O(l)
|
|
added gives
full equation |
IO3–(aq) + 6H+(aq)
+ 5I–(aq) ==>
3I2(aq) + 3H2O(l) |
-
The reaction
can be used to estimate iodate(V) by adding excess potassium
iodide and titrating the liberated iodine with standardised sodium
thiosulphate or using the liberated iodine from a known quantity
of potassium iodate(V) salt with excess KI(aq) salt
solution to standardise the sodium thiosulphate.
-
Section 6. Index of examples of constructing
balanced ionic redox reaction equation from half–cell/EØ
data
Ex
6.7
The reaction between
acidified potassium manganate(VII) and hydrogen peroxide solutions aqueous
Ex
6.8
The titration of
ethanedioate with acidified potassium manganate(VII) solution
Ex
6.9
Reduction of
the vanadium(IV) oxo–cation by tin(II) salts
-
Half–cell reaction
data:
-
(i) VO2+(aq)
+ 2H+(aq) + e– ==> V3+(aq)
+ H2O(l) (EØ = +0.34,
reduction of the oxidising agent)
-
(ii) Sn4+(aq)
+ 2e– ==> Sn2+(aq) (EØ = +0.15,
Sn2+ will act as a reducing agent, EØ
less positive)
-
V at (+4) in
VO2+ is reduced to V at (+3), 1e–
gain per V,
-
Sn at (+2,
tin(II) ion) is
oxidised to Sn at (+4, tin(IV) ion), 2e–
loss per Sn.
-
Hydrogen (+1)
and oxygen (–2) do not change oxidation state.
-
|
1 x
oxi'n half–cell, (ii) rev. |
Sn2+(aq) ==> Sn4+(aq)
+ 2e– |
|
2 x
reduction half–cell, (i) |
2VO2+(aq) + 4H+(aq) +
2e– ==> 2V3+(aq) + 2H2O(l)
|
|
added –
full equation |
2VO2+(aq)
+ 4H+(aq) + Sn2+(aq)
==>
2V3+(aq) + 2H2O(l) + Sn4+(aq) |
-
Tin(II) salts
are used as reducing agents.
-
Section 6. Index of examples of constructing
balanced ionic redox reaction equation from half–cell/EØ
data
-
For more on vanadium chemistry see
Periodic Table Advanced Inorganic Chemistry Notes Part
10a "3d block Transition Metals Series Introduction and Elements Sc to
Mn on Period 4
– detailed revision notes
Ex
6.10
Titrating iodine with
standardised sodium thiosulphate solution
-
Half–cell reaction
data:
-
The iodine is reduced
by the thiosulphate ion to form iodide, ox. state of I (0)
to (–1).
-
The thiosulphate ion
is oxidised to the tetrathionate ion. In doing so the sulphur atom changes
ox. state from an average of four at (+2) in the two S2O32–
ions to an average of four at (+2.5) in the single S4O62–
ion.
-
|
2 x
reduction half–cell, (i) |
I2(aq)
+ 2e– ==>
2I–(aq) |
|
2 x
oxidation half–cell, (ii) rev. |
2S2O32–(aq) ==>
S4O62–(aq) + 2e–
|
|
added gives
full equation |
2S2O32–(aq)
+ I2(aq) ==>
S4O62–(aq) + 2I–(aq)
|
-
This is used to
quantitatively estimate iodine in aqueous solution. The indicator is a few
drops of starch solution which forms a blue–black complex with iodine. The
end–point is when the solution first becomes colourless with no remaining
iodine to form the coloured complex. The iodine to be titrated may arise from
a variety of reactions for analysis purposes, see 6.2, 6.6 and 6.12.
-
Section 6. Index of examples of constructing
balanced ionic redox reaction equation from half–cell/EØ
data
Ex
6.11 The
oxidation of chloride by acidified potassium manganate(VII)
Ex
6.12 The
reduction of acidified dichromate(VI) with iodide ions
-
Half–cell reaction
data:
-
(i) Cr2O72–(aq)
+ 14H+(aq) + 6e– ==> 2Cr3+(aq)
+ 7H2O(l) (EØ = +1.33, the
reduction of the oxidising agent)
-
(ii)
1/2I2(aq) + e– ==>
I–(aq) (EØ = +0.54V,
lower EØ, so cannot act as oxidising agent)
-
Oxidation:
Iodide ions at (–1) lose electron to give iodine molecules at
I(0).
-
Reduction:
Each Cr at (+6) is reduced by gaining 3e– to give
Cr at
(+3), so the Cr2O72– ion is the
oxidising agent.
-
Hydrogen (+1)
and oxygen (–2) do not change oxidation state.
-
|
6 x
oxidation half–cell, (ii) rev. |
3I2(aq) + 6e– ==>
6I–(aq) |
|
1 x
reduction half–cell, (i) |
Cr2O72–(aq) + 14H+(aq)
+ 6e– ==> 2Cr3+(aq) + 7H2O(l)
|
|
added
full redox equation |
Cr2O72–(aq)
+
14H+(aq) + 6I–(aq)
==>
2Cr3+(aq)
+ 3I2(aq) + 7H2O(l) |
-
This reaction
can be used to quantitatively measure chromium(VI) in dichromates, Cr2O72–,
or chromates, CrO42– (which change to
dichromate(VI) on acidification, yellow ==> orange). Excess
potassium iodide is added and the liberated iodine is titrated
with standardised sodium thiosulphate solution (starch indicator,
blue ==> colourless), see Ex 6.10.
-
Section 6. Index of examples of constructing
balanced ionic redox reaction equation from half–cell/EØ
data
-
For more on chromium chemistry see
Periodic Table Advanced Inorganic Chemistry Notes Part
10a "3d block Transition Metals Series Introduction and Elements Sc to
Mn on Period 4
– detailed revision notes
Ex
6.13 The
oxidation of hydrogen sulphide by acidified potassium manganate(VII)
-
Half–cell
reaction data:
-
(i) MnO4–(aq) +
8H+(aq) + 5e– ==> Mn2+(aq)
+ 4H2O(l) (EØ
= +1.52, reduction of the oxidising agent)
-
(ii)
SO42–(aq) + 10H+(aq)
+ 8e– ==> H2S(aq) +
4H2O(l) (EØ
= ?, oxidation of the reducing agent )
-
(ii)
assumes that the sulphur is initially in a covalent
molecular state and NOT sulphide ions, which would be
readily protonated by the dilute sulphuric acid. It further
assumes the sulphur is completely oxidised from –2 in H2S,
to its maximum oxidation state of +6 as the sulphate(VI)
ion.
-
So you
have to balance up an 8 e–/8 ox. no. units
reduction with a 5 e–/5 ox. no. units half–cell
-
|
5 x
oxi'n half–cell, (ii) rev. |
5H2S(aq)
+ 20H2O(l)
==> 5SO42–(aq) + 50H+(aq)
+ 40e– |
|
8 x
red'n half–cell, (i) |
8MnO4–(aq) +
64H+(aq) + 40e– ==>
8Mn2+(aq)
+ 32H2O(l) |
|
added – full redox equation |
8MnO4–(aq)
+ 14H+(aq) + 5H2S(aq)
==>
8Mn2+(aq)
+ 5SO42–(aq) + 12H2O(l) |
-
This is
quite a tricky one to do with awkward numbers!
-
I'm not sure
exactly what happens in practice, so above is theoretical,
therefore in addition to the above 'construction' if the
hydrogen sulphide is just oxidised to a sulphur precipitate, the
equation would be ...
-
2MnO4–(aq)
+ 6H+(aq) + 5H2S(aq)
==>
2Mn2+(aq)
+ 5S(s) + 8H2O(l)
-
2 x
Mn(VII) ==> 2 x Mn(II) of equation (i) (10 e–
change), balanced by 5 x S(–2) to 5 x S(0) of equation (iii)
reversed (below).
-
(iii)
S(s) + 2H+(aq) + 2e–
==>
H2S(aq) (EØ
= +0.14V, less positive, so hydrogen sulphide gets oxidised)
Ex 6.14
The conversion of
chromium(III) to chromium(VI)
| 3 x (a) for 6e change |
3HO2– + 3H2O
+ 6e– |
===> |
9OH– |
| 2 x (b) reversed for 6e
change |
2Cr(OH)3 + 10OH– |
===> |
2CrO42– + 8H2O
+ 6e– |
| = initial total, 6e
cancel out |
3HO2– + 3H2O
+ 2Cr(OH)3 + 10OH– |
===> |
2CrO42– + 8H2O
+ 9OH– |
|
3H2O and 9OH– cancel out
to give |
3HO2– +
2Cr(OH)3 + OH– |
===> |
2CrO42– + 5H2O |
|
leaving the final equation (d)! |
2Cr(OH)3(s) + 3HO2–(aq) + OH–(aq) ===>
2CrO42–(aq)
+ 5H2O(l) |
7.
Redox titration
questions
-
Using the
correct redox equation is obviously important when problem solving
and performing
calculations from redox titrations.
-
A set of
problems involving some of these redox reactions, complete with worked out answers
is available.
-
Redox titration
questions
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REDOX section–index: 1.
Basic redox definitions * 2.
Introducing oxidation state (with sub–index) * 3.
Oxidation state rules–guidelines &
inorganic examples * 4. Naming
inorganic compounds * 5. Using oxidation states to describe redox changes
in a given inorganic reaction equation (with
sub–index) * 6. Constructing full
inorganic redox
equations from half–equations (with
sub–index) * 7. Redox
titrations * 8. Organic synthesis reductions (with summary table) * 9.
Organic synthesis oxidations (with
summary table) * 10.
Other Organic Redox Reactions (with sub–index)
* 11.
Carbon's ox. state in selected
organic compounds and functional group level * You are advised to study
sections 1. to 6. in strict order and covers the requirements of AS–A2
* See also Equilibria Part 7
Redox Reactions for Half cell equilibria, electrode potential, standard hydrogen electrode, Simple cells and notation,
Electrochemical Series, EØcell for reaction feasibility, 'batteries' and fuel cell systems etc. |
2Cr3+(aq) + 7H2O(l)
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structure, concept, equation, 'phrase', homework question! anything
of chemical interest!