Chpt. 15
Volumetric Analysis - Oxidation-Reduction
We have already studied how
volumetric analysis can be used
to determine the concentrations
of solutions of acids and bases
(Chpt. 13) and also how many
chemical reactions in chemistry
involve oxidation-reduction
reactions (Chpt. 14). So now, lets
combine this knowledge and
investigate how volumetric
analysis can be used to
determine the concentrations of
oxidising and reducing agents.
Redox Titrations:
titrations using oxidising and
reducing agents are called redox
titrations or oxidation-reduction
titrations.
Potassium Permanganate KMnO4
(Potassium Manganate(VII) )
• The most important oxidising agent to be studied is
potassium permanganate.
• Potassium permanganate is a very useful oxidising
agent in redox titrations. (For this reason it is
important to be very familiar with this substance)
Potassium Permanganate – Key Points:
1) Potassium permanganate is a purple coloured solid.
It is NOT a primary standard as it cannot be obtained
in a state of very high purity and it decomposes in the
presence of sunlight.
2) Potassium permanganate acts as an oxidising agent
by gaining 5e- in acidic solution:
*Must Learn - Half Reaction:
MnO4- + 8H+ + 5e(+7)
Purple
Mn2+ + 4H2O
(+2)
Colourless
As acidified potassium
permanganate is
added to the reducing
agent, the manganese
atom changes from
an oxidation state of
+7 (purple) to an
oxidation state of +2
(colourless)
3) In order to get potassium permanganate to undergo
the above reaction, ie reduce it from Mn(+7) to
Mn(+2), some dilute acid is added to supply the H+
ions. It is usual to use sulphuric acid (H2SO4) in
acidifying KMnO4 for several reasons:
- SO42- is very stable to oxidation
- the Cl- ions in HCl would be oxidised to Cl2 by
KMnO4
- Nitric acid (HNO3) cannot be used as it is a
powerful oxidising agent itself and therefore
would interfere with the reaction.
4) There are two situations where the above reaction
will NOT occur:
a) in neutral solution i.e. no acid
b) in basic solution i.e. not enough acid
In the above cases instead of going from
Mn(VII)
Purple
Mn(II)
Colourless
the manganese ion changes to an intermediate
oxidation state – Mn(IV) formation of MnO2:
Mn(VII)
Purple
Mn(IV)
Brown
When KMnO4 reacts with a reducing agent in
neutral or basic solution, the manganese ion
changes from an oxidation state of +7 (purple) to an
oxidation state of +4(brown)
5) Potassium permanganate acts as its own indicator,
giving a sharp end-point. It changes colour from
purple to colourless when acting as an oxidising
agent.
6) Because of the intense purple colour of potassium
permanganate, read the top rather than the
bottom of the meniscus.
7) Generally potassium permanganate is used in
fairly weak solutions – 0.02M. This is due to its
strong oxidising power and the fact that it is not
very soluble (i.e. hard to get to dissolve).
Two types of redox
titrations to be studied:
- the rxn of MnO4- and
Fe2+ ions
- the rxn of iodine and
thiosulphate ions
Reaction of MnO4- ions and Fe2+ ions
KMnO4 – is not a primary standard (not possible to
make up solution of KMnO4 of accurate concentration
by weighing out a known amount of the solid and
dissolving in a known volume of solution) However, it is
possible to determine the concentration of a KMnO4
solution by titrating it against a solution containing
Fe2+ (iron) ions:
Standardising Potassium Permanganate using Iron
• KMnO4 – oxidising agent – when added to an
acidified (to ensure Mn+7 is reduced to Mn+2)
solution containing Fe2+ ions it converts the Fe2+ ions
to Fe3+ ions. Reactions involved:
Oxidising Agent (itself reduced):
(A) MnO4- + 8H+ + 5eReducing Agent (itself oxidised):
(B) Fe2+ - e-
Fe3+
Mn2+ + 4H2O
Upon inspection of equations A & B it is clear to see
that every MnO4- ion must receive 5e- in order for the
reaction to go to completion but every Fe2+ ion will
only give 1e-. So, the following ratio can be
determined:
MnO4- ion : Fe2+ ion
1
:
5
Rewriting equations A & B:
(A) MnO4- + 8H+ + 5e(B) 5Fe2+ - 5e-
Mn2+ + 4H2O
5Fe3+
Adding these two half equations:
MnO4- + 8H+ + 5Fe2+
Mn2+ + 5Fe3+ + 4H2O
Ammonium Iron(II) Sulphate (Ferrous Ammonium
Sulphate) Vs Iron(II) Sulphate
• Cannot use iron(II) sulphate(FeSO4) crystals as a
primary standard because:
- the crystals are oxidised slightly by the air
- the crystals tend to lose their water of
crystallisation when exposed to the air
(efflorescence)
• Ammonium Iron(II) Sulphate (NH4)2SO4.FeSO4.6H2O
is used as a source of Fe2+ ions because:
- it is a primary standard (crystals not affected by
air)
- it is stable and available in a high state of
purity
Ammonium Iron(II)
Sulphate Crystals
• When preparing ammonium iron(II) sulphate it must
be dissolved in water containing dilute sulphuric
acid:
- the sulphuric acid prevents the Fe2+ ions
reacting with water to form a new substance –
prevents hydrolysis
- the sulphuric acid helps to prevent the Fe2+ ions
being oxidised to Fe3+ ions by the oxygen in the
air or dissolved in the water
Redox Reactions
Volumetric Problems
A formula similar to that used in acid-base titrations is
used in these calculations:
VO x MO
nO
=
Vr x Mred
nr
VO = vol. of oxidising agent (cm3)
MO = molarity of oxidising agent
nO = no. of moles of oxidising agent (from balanced eqn.)
Vr = vol. of reducing agent (cm3 )
Mred = molarity of reducing agent
nr = no. of moles of reducing agent (from balanced eqn.)
Example 1: (Please leave a page for answer)
24cm3 of a potassium manganate(VII) solution required
25cm3 of a 0.0108M ammonium iron(II) sulphate
solution for complete reaction. The equation for the
reaction is:
MnO4- + 8H+ + 5Fe2+
Mn2+ + 5Fe3+ + 4H2O
Calculate the concentration of the potassium
manganate(VII) solution in (a) moles per litre (b) grams
per litre
Mandatory Expt.: To prepare a standard solution of
ammonium iron(II) sulphate and to use this solution
to standardise a solution of KMnO4 by titration
Points to Note:
• KMnO4 is its own indicator. Before the end point,
all added KMnO4 solution is decolourised by the
reaction with Fe2+ ions. At the end point, all of
the Fe2+ ions have reacted, and added KMnO4
solution is no longer decolourised:
Purple
colourless
pink
• Read from the top of the meniscus.
Example 2: (Please leave a page for answer)
A sample of iron(II) sulphate crystals FeSO4. xH2O was
analysed in order to find out the value of x. 6.42g of the
crystals were dissolved in deionised water to which
some dilute sulphuric acid had been added and the
resulting solution was made up to 250cm3 in a
volumetric flask. 25cm3 samples of this solution were
titrated against 0.018M KMnO4 solution. The balanced
equation for the reaction is:
MnO4- + 8H+ + 5Fe2+
Mn2+ + 5Fe3+ + 4H2O
The average titration figure was 25.5cm3. Calculate the
value of x in the above formula.
Applications of Potassium Permanganate – Iron(II)
Titrations
• Iron is an important element in our bodies as it forms
part of the haemoglobin molecule in our blood.
Haemoglobin is the molecule responsible for
transporting oxygen in our blood. The oxygen actually
becomes attached to the iron atom in the haemoglobin
molecule.
• Our bodies can easily become deficient in iron resulting
in anaemia. This condition can be remedied with a
change in diet and a course of iron tablets (FeSO4)
• In the next experiment we will attempt to calculate the
amount of iron(II) sulphate in iron tablets
*Do Not take Diagram
Haemoglobin Molecule
Mandatory Expt.: To determine the amount of iron in
an iron tablet
Example 1: (Please leave a page for your answer)
A student was asked to analyse an iron tablet to
determine if the amount of iron sulphate stated on the
packet was correct. She dissolved five iron tablets of
total mass 1.8g in dilute sulphuric acid and made the
solution up to 250cm3 with deionised water. 25cm3 of
this solution required 16.5cm3 of a 0.005M potassium
manganate(VII) solution for complete reaction.
Calculate (a) the mass of anhydrous FeSO4 in each tablet
(b) the mass of iron in each tablet (c) the percentage of
FeSO4 in each tablet and (d) the percentage of iron in
each tablet
Sodium Thiosulphate
(Na2S2O3)
• Sodium thiosulphate (hypo) is an important reducing
agent used in chemistry
• Used in developing photographs and in analytical
chemistry
• Most commonly encountered as a colourless
crystalline solid of the form:
Na2S2O3.5H2O (pentahydrate)
Reaction of Iodine and Thiosulphate Ions
• Sodium thiosulphate is a reducing agent and reacts
with iodine molecules (I2), to convert them to iodide (Iions):
I2(aq) +
2S2 O32-(aq)
Iodine
Thiosulphate
Ion
0
Oxidising
Agent
+2
Reducing
Agent
S4 O62-(aq) + 2I-(aq)
Tetrathionate
Ion
+2.5
Iodide
Ion
-1
Ratio
I2 : S2O321 :
2
Thiosulphate Ion
S2O32-
*Take Diagram
Tetrathionate Ion
S4O62-
• Sodium Thiosulphate is NOT usually a primary standard
because it is difficult to obtain in the anhydrous form
(no water) and the crystals lose water of crystallisation
to the atmosphere and so are unstable. However, it is
possible to determine the concentration of a sodium
thiosulphate solution by titrating it against a standard
iodine solution:
Standardising Sodium Thiosulphate using Iodine
• Iodine is NOT usually a primary standard. A standard
solution of iodine cannot be made up by direct
weighing because:
- it is too volatile (vaporises slightly at room temp.)
- it does NOT dissolve in water
• A STANDARD SOLUTION of IODINE (0.06M) is made up
by reacting a standard solution of acidified potassium
permanganate (0.02M) with excess potassium iodide.
Reaction equations for formation of standard
solution of iodine
(A) MnO4- + 8H+ + 5eMn2+ + 4H2O
(Each permanganate ion needs 5e-)
(B)
2I-
I2 + 2e(Each iodide loses 1e- (2I- = 2e- and
forms iodine as a result)
Upon inspection of equations A & B it is clear to see
that every MnO4- ion must receive 5e- in order for the
reaction to go to completion. But the 2I- only results in
the loss of 2e-, so in order to balance the e- gain and
loss we must:
(A) x2: 2MnO4- + 16H+ + 10e(B) x5: 10I-
2Mn2+ + 8H2O
5I2 + 10e-
Adding these two half equations:
2MnO4- + 10I- + 16H+
2Mn2+ + 5I2 + 8H2O
In the previous equation the I- ions (from potassium
iodide) are kept in excess. As a result the amount of
iodine formed in this reaction depends on the amount
of permanganate present. So, by using a standard
solution of KMnO4 we can determine the amount of
iodine formed.
This known amount of iodine formed can then be used to
standardise a sodium thiosulphate solution
**Note:
Iodine is non-polar and water is a polar solvent.
• For this reason iodine will NOT dissolve in water but
will dissolve in a solution of potassium iodide (KI).
• Adding I2 to KI:
- I2 molecule reacts with I- to form I3- (tri-iodide ion)
I2 + I-
I3- state of equilibrium
- I3- is identical in its chemical behaviour to I2 and
because of its charge it dissolves easily in water
- As a result a solution of iodine in potassium iodide
may be treated as if it were a solution of iodine in
water
• Determining the end point of an iodine/thiosulphate
titration
I2(aq) +
Reddish
Brown
2S2 O32-(aq)
(Colourless)
S4O62-(aq) + 2I-(aq)
Colourless
As thiosulphate is added to the iodine the
reddish/brown colour gradually turns yellow which
gradually turns paler and paler until it turns
colourless – very difficult to detect end point.
Therefore need to use an indicator – starch indicator
• Starch forms a blue-black colour with iodine
therefore when ALL of the iodine has reacted with
the thiosulphate the blue-black colour suddenly
goes colourless.
Note: starch is only added when most of iodine has
disappeared i.e. when reddish/brown colour of
iodine has gone pale(straw) yellow.
Why???
• If too much added or added
too early starch participates in
rxn and removes to much
iodine
• When pale colour is observed it
indicates end point is near
Colour Change
End Point
Reddish/brown
Straw Yellow
Add
Starch
Blue/Black
Colourless
Colour Change
Points to Note:
1. A standard solution of sodium thiosulphate can be
used to standardise an iodine solution
2. In an iodine-thiosulphate titration:
- iodine in conical flask
- sodium thiosulphate in burette
3. Starch indicator must be prepared fresh as it
deteriorates quickly on standing.
4. A 0.02M standard solution of potassium iodate may
be used in place of potassium permanganate in
preparing the iodine solution.
Mandatory Expt.: To prepare a solution of sodium
thiosulphate and standardise it by titration against a
solution of iodine
Two reactions:
1. Formation of iodine in conical flask:
2MnO4- + 10I- + 16H+
2Mn2+ + 5I2 + 8H2O
2. Main rxn – Reaction of Iodine with Sodium
Thiosulphate
I2(aq) + 2S2O32-(aq)
S4O62-(aq) + 2I-(aq)
So,
2MnO4- : 5I2 : 10S2O32Ratio
1 : 2
Example 1: (Please leave a page for answer)
25cm3 of 0.018M potassium permanganate is pipetted
into a conical flask. Some dilute sulphuric acid and
excess potassium iodide solution are then added and
the liberated iodine is titrated against a solution of
sodium thiosulphate of unknown concentration. The
average titration figure is 15.6cm3. The reactions may
be represented as:
2MnO4- + 10I- + 16H+
I2(aq) +
2S2O32-(aq)
2Mn2+ + 5I2 + 8H2O
S4O62-(aq) + 2I-(aq)
Calculate the concentration of the sodium thiosulphate
solution in a) moles per litre and b) grams per litre of
crystalline Na2 S2O3.5H2O
Applications of Iodine –Thiosulphate Titrations
• Most household bleaches contain chlorine in the
form of sodium hypochlorite, NaClO.
• NaClO contains the hypochlorite ion (chlorate(I) ion)
– ClO• Hypochlorite ion is an oxidising agent and converts
iodide ions to iodine:
ClO- + 2I- + 2H+
I2(aq) + 2S2O32-(aq)
Cl- + I2 + H2O
S4O62-(aq) + 2I-(aq)
Ratio:
ClO- : I2 : 2S2 O32-
1
: 1 :
2
Where, ClO- is the oxidising agent and 2S2O32- is the
reducing agent
Experiment: To determine the percentage (w/v) of
sodium hypochlorite in household bleach
• To determine the strength of household bleach a
sample of the bleach is reacted with a solution of
iodide ions. The liberated iodine is then titrated
against a solution of sodium thiosulphate of known
concentration.
• Results obtained will allow calculation of the
concentration of sodium hypochlorite in bleach.
*Note: As household bleach is fairly concentrated it
must be diluted first otherwise an excessive amount of
reagents would be used in titration
Example 1: (Please leave a page for answer)
25cm3 of household bleach was diluted to 250cm3. A
25cm3 portion of the diluted solution was added to an
excess of acidified potassium iodide solution and
titrated against 0.12M sodium thiosulphate solution
(standardised). The average titration figure was
32.1cm3. Calculate the concentration of sodium
hypochlorite in the bleach in (a) moles per litre (b)
grams per litre and (c) % w/v. The equations for the
reaction are:
ClO- + 2I- + 2H+
I2(aq) + 2S2O32-(aq)
Cl- + I2 + H2O
S4O62-(aq) + 2I-(aq)