The Estimation of Iron (II) and Iron (III) in a Mixture Containing Both


Fe2+ Fe3+ + e- (multiply by 5)

MnO4- + 5e- + 8H+ Mn2+ + 4H2O

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

Background Knowledge –

Iron, Fe, is a transition metal with the atomic number 26. Iron is a magnetic, malleable and metallic silver element. Iron has the ability of having multi valences, commonly Fe2+ or Fe3+. Fe2+ forms ferrous compounds, whereas Fe3+ tends to form ferric compounds. Iron can be used as a catalyst due to its multiple valances, e.g. it is used in the manufacturing of Ammonia in the Haber process.

An oxidising agent will convert Iron(II) into Iron(III), the iron will lose an electron. A reducing agent will convert Iron(III) into Iron(II), the iron will gain an electron.

Iron sulphate (FeSO4), also known as vitriol or copperas, is an important iron (II) compound as it forms pale green crystals containing seven molecules of water of hydration. It is mainly formed as a by – product in the pickling of iron.

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It can be used as a mordant in dyeing, tonics in medicines and also in the manufacturing of ink and pigments.

Iron (III) compounds also have many uses, such as iron (III) oxide being used as a pigment, known as iron red, or the compound is also used as a polishing abrasive, known as rouge. Iron (III) oxide is an amorphous red powder obtained by treating iron (III) salts with a base or by oxidizing pyrite.

In acid solution and under standard conditions, Fe2+(aq) is not oxidized to Fe3+(aq) by oxygen is the air or when dissolved in water, nor when it is dissolved in solution.

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2 Fe2+(aq) 2Fe3+(aq) + 2e- (-0.77V)

2H+(aq) + O2(g) + 2e- H2O2(aq) (+0.68V)

(Electrode Potentials)

Before I can begin the practical, I must calculate the concentration of KMnO4 that should be used for both titrations. In order to do this I must work out the concentration of iron using mole calculations:

Moles = Mass / RFM

= 1.3 / 55.8

= 0.023 moles

Concentration of Iron = Moles / Volume

= 0.023 / ( 200 / 1000 )

= 0.12 moldm-3

For the first titration I am assuming that the Fe2+ represents 70% of the concentration, therefore:

0.7 x 0.12 = 0.084 moldm-3

As the ratio from the equation above is 1 : 5 for KMnO4 to Fe2+, we must divide this concentration above by 5:

0.084 / 5 = 0.017 moldm-3

For the second titration, once again the ratio is 1 : 5, therefore the concentration is one fifth of that of the iron solution:

0.12 / 5 = 0.024 moldm-3

As both the concentrations are similar, I will use a concentration of KMnO4 to be 0.02M throughout the titrations.

Equipment, apparatus and chemicals:-

1. 100cm3 of Fe2+ and Fe3+ solution

2. 50 cm3 of Potassium permanganate at 0.02M

3. 1.0 moldm-3 Sulphuric Acid

4. Zinc granules

5. Clamp and Boss

6. Burette

7. Burette funnel – white markings so that it is clearly viewable through the purple KMnO4

8. 25 cm3 Pipette and safety pipette filler

9. White tile

10. Conical flasks and beakers

11. Filter paper

Titration 1

o The burette should be cleaned out with KMnO4, being allowed to run through the end of the burette, making sure there are no air bubbles, while cleaning out the tip.

o The titration must be set up now with the burette attached to the boss and clamp with a conical flask above a white tile below.

o The pipette must be cleaned out using a sample of the iron solution. Once cleaned, you must pipette 25cm3 of the iron solution into the conical flask. You must touch the end of the pipette on the conical flask to maintain accuracy throughout the experiment.

o Into the conical flask you must add roughly 20cm3 of sulphuric acid as acidified acidic solution is the requirements for which this reaction may take place.

o Fill the burette with KMnO4 using the burette filler, but be sure to remove the funnel as drops later on in the experiment may ruin results.

o Now you are setup to begin titrating. Allow the potassium permanganate to flow into the titre until a permanent colour change from colourless to pink has occurred. This is your rough titration. Record your value of your burette and fill in the table below.

o Repeat this procedure until you have concurrent titre volume readings i.e. your values must be within 0.1cm3 of each other. But, this time, allow the KMnO4 to flow into the solution until you are near the value you received for your rough titre. At this point you must drip the KMnO4 into the solution in the conical flask to increase the accuracy of your readings.

o Throughout the whole period of titrating, you must constantly be swirling the conical flask to make sure that the colour change is permanent.

o Keep all the conical flask solutions as they will be used in titration 2. ( Be sure to remember which one is rough and which ones are accurate.)

In this reaction it is only the Iron (II) that reacts with the KMnO4 allowing us to workout the mass of Iron (II). ( see later answers )

Titration 2

Essentially, this procedure is similar to titration 1. In this reaction zinc granules are added to the solutions from titration 1 as to reduce the remaining Fe3+ ions to Fe2+ ions:

2Fe3+(aq) + Zn(s) 2Fe2+(aq) + Zn2+(aq)

You should add zinc granules in excess to make sure that all the Fe3+ is converted into Fe2+.

* Once the conversion is finished you should filter the titre into separate conical flasks.

* Titrate using the same procedure as titration 1 using the same conical flasks i.e. rough for rough, 1 for 1, 2 for 2 and 3 for 3. Titrate until an end point is reached.

* Record the results on a tabular format as above.


Titration 1 Iron (II) only:

Moles of MnO4- : Volume x Concentration V = average titre volume (cm3)

= V / 1000 x 0.02

= 2.0 x 10-5.V moles

Using the 1 : 5 ratio, we can multiply this mole value by 5 to find the number of moles of Iron (II):

= (2.0 x 10-5).V x 5

= 1.0 x 10-4.V moles This value is for 25cm3, but we have to multiply up to 200cm3

This will be done by multiplying B by 200/25 = 8:

= (1.0 x 10-4).V x 8

= 8.0 x 10-4.V moles in 200cm3

Mass = Moles x RFM

= (8.0 x 10-4).V x 55.8

= 4.5 x 10-2.V g ( P )

Titration 2 Iron (II) and Iron (III):

Moles of MnO4- used = Volume x Concentration

= W / 1000 x 0.02

= 2.0 x 10-5.W moles

W = average titre volume for titration 2 (cm3)

Moles of Iron (II) (in 25cm3) is 5 times the MnO4- value because the ratio is 1 : 5. We can multiply this mole value by 5 to find the number of moles of Iron (II):

= (2.0 x 10-5).W x 5

= 1.0 x 10-4.W moles This value is for 25cm3, but we have to multiply up to 200cm3

This will be done by multiplying B by 200/25 = 8:

= (1.0 x 10-4).W x 8

= 8.0 x 10-4.W moles in 200cm3

Mass of Iron (II) in solution = Moles x RFM

= (8.0 x 10-4).W x 55.8

= 4.5 x 10-2.W g ( Q )

Overall, the mass of Iron (III) present is the difference between the mass of Iron (II) in titration 2 and titration 1:

= H – D

= ( 4.5 x 10-2 )( W – V ) g ( Z )

Percentage Composition:

Iron (II) : ( Mass of Iron (II) in titration 1 / Mass of Iron (II) in titration 2 ) x 100

= P / H x 100

Iron (III) : ( Mass of Iron (III) / Mass of Iron (II) in titration 2 ) x 100

= Q / H x 100


* Goggles must be worn at all times while working with chemicals.

* Lab coats must be worn at all times potassium permanganate stains clothes.

* Wear gloves when handling sulphuric acid as it is corrosive and toxic.

* Safety fillers must be used when using pipettes.

* If any chemical comes into contact with your skin, you must immediately wash the area thoroughly.

* Glass equipment must be handled carefully as breakages are possible. in the event of a cut, it should be seen to before continuing the practical.


* Chemistry in Context Fourth Edition – Graham Hill and John Holman

* Microsoft Encarta 2000 Encyclopedia

* Chemistry 2 textbook – OCR

* Compton’s Encyclopedia 2001 Edition

Jaimal B Amin A2 Practical Exp 9

Cite this page

The Estimation of Iron (II) and Iron (III) in a Mixture Containing Both. (2020, Jun 02). Retrieved from

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