Skip this Video
Download Presentation
Chpt. 15 Volumetric Analysis - Oxidation-Reduction

Loading in 2 Seconds...

play fullscreen
1 / 47

Chpt. 15 Volumetric Analysis - Oxidation-Reduction - PowerPoint PPT Presentation

  • Uploaded on

Chpt. 15 Volumetric Analysis - Oxidation-Reduction.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
Download Presentation

PowerPoint Slideshow about ' Chpt. 15 Volumetric Analysis - Oxidation-Reduction' - emmanuel-manning

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

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- Mn2+ + 4H2O

(+7) (+2)

Purple 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) Mn(II)

Purple Colourless

the manganese ion changes to an intermediate

oxidation state – Mn(IV) formation of MnO2:

Mn(VII) Mn(IV)

Purple 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


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):
  • Reducing Agent (itself oxidised):
  • (B) Fe2+ - e- Fe3+
  • MnO4- + 8H+ + 5e- 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:

Adding these two half equations:

  • MnO4- + 8H+ + 5e- Mn2+ + 4H2O

(B) 5Fe2+ - 5e- 5Fe3+

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 = Vr x Mred

nO 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:

Calculate the concentration of the potassium manganate(VII) solution in (a) moles per litre (b) grams per litre

MnO4- + 8H+ + 5Fe2+ Mn2+ + 5Fe3+ + 4H2O


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:

The average titration figure was 25.5cm3. Calculate the value of x in the above formula.

MnO4- + 8H+ + 5Fe2+ Mn2+ + 5Fe3+ + 4H2O


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

Haemoglobin Molecule

*Do Not take Diagram


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 (I-
  • ions):
  • I2(aq) + 2S2 O32-(aq) S4 O62-(aq) + 2I-(aq)

Thiosulphate Ion

Tetrathionate Ion

Iodide Ion






Oxidising Agent

Reducing Agent


Thiosulphate Ion


Tetrathionate Ion


*Take Diagram


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

  • MnO4- + 8H+ + 5e- Mn2+ + 4H2O
  • (Each permanganate ion needs 5e-)
  • 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- 2Mn2+ + 8H2O

(B) x5: 10I- 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



  • 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
  • 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

I2(aq) + 2S2 O32-(aq) S4O62-(aq) + 2I-(aq)


Reddish Brown



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

Reddish/brown Straw Yellow Blue/Black Colourless

End Point

Colour Change

Add Starch


Points to Note:

  • A standard solution of sodium thiosulphate can be used to standardise an iodine solution
  • In an iodine-thiosulphate titration:
  • - iodine in conical flask
  • - sodium thiosulphate in burette
  • Starch indicator must be prepared fresh as it deteriorates quickly on standing.
  • 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:
  • Formation of iodine in conical flask:
  • 2. Main rxn – Reaction of Iodine with Sodium Thiosulphate
  • So,
  • 2MnO4- : 5I2 : 10S2O32-
  • Ratio 1 : 2

2MnO4- + 10I- + 16H+ 2Mn2+ + 5I2 + 8H2O

I2(aq) + 2S2O32-(aq) S4O62-(aq) + 2I-(aq)


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:

Calculate the concentration of the sodium thiosulphate solution in a) moles per litre and b) grams per litre of crystalline Na2 S2O3.5H2O

2MnO4- + 10I- + 16H+ 2Mn2+ + 5I2 + 8H2O

I2(aq) + 2S2O32-(aq) S4O62-(aq) + 2I-(aq)


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+Cl- + I2 + H2O

I2(aq)+ 2S2O32-(aq) S4O62-(aq)+ 2I-(aq)



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+Cl- + I2 + H2O

I2(aq) + 2S2O32-(aq) S4O62-(aq) + 2I-(aq)