A Redox Titration between Manganate (VII) and Iron (II) Essay Sample

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Objective: To determine the x in the formula Fe (NH4)2(SO4)2�xH2O by titration against a standard solution of potassium manganate (VII) (permanganate). Theory: The experiment involves a redox reaction between potassium manganate (VII) and ammonium Potassium manganate (VII) is used in the experiment as it reacts completely and it is its own indicator. Potassium manganate (VII) solution is a strong oxidizing agent. In an acidic medium, manganate(VII) ion undergoes reduction as shown below. MnO4- (aq) +

medium, manganate(VII) ion undergoes reduction as shown below.

MnO4- (aq) + 8H+ (aq) + 5e- –> 4H2O (l) + Mn2+ (aq)

Ammonium Iron (II) sulphate is a strong reducing agent because of the presence of Iron (II) ions. Iron (II) ions can be oxidized to form Fe3+ which is yellow in color.

Manganate (VII) ions are purple in color, when the MnO4- are reduced to Mn 2+, the color of the solution becomes pale pink, thus a self indicating titration can occur

According to the overall reaction:

MnO4- (aq) + 5Fe 2+ (aq) +8H+(aq) –> Mn2+(aq) +5Fe3+(aq)+4H2O(l)

Between the titration of potassium manganate (VII) solution and ammonium iron (II) sulphate solution, The molarity of ammonium iron (II)


Safety spectacles

25cm3 pipette

Pipette filler

4 conical flasks

50cm3 burette

Small funnel

White tile

Wash-bottle of distilled water


Solution A: ammonium Iron(II) sulphate solution

1M dilute sulphuric acid

Solution B: potassium manganate (VII)


1. The burette was filled with potassium manganate (VII) solution B

2. 25 cm3of the ammonium iron (II) sulphate solution A into was pipette into a conical flask.

3. 25 cm3of the dilute sulphuric acid was added into the conical flask filled with 25 cm3of ammonium iron (II) sulphate solution A

4. The conical flask was titrated with potassium permanganate solution B

5. The titration was stopped when a pale pink colour was formed in the conical flask

6. The titration was repeated several times to obtain consistent results



Concentration of potassium manganate (VII) solution, B: 0.0189 ( mol dm3)

Concentration of ammonium iron (II) sulphate solution, A: 39.214 (g dm3)






Final Burette reading(cm3)






Initial Burette reading (cm3)






Volume used (cm3)






Average volume used :

26.1 + 26.1 + 25.9 + 26.1 = 26.05 cm3



MnO4- (aq) + 5Fe 2+ (aq) +8H+(aq) –> Mn2+(aq) +5Fe3+(aq)+4H2O(l)

We can find the no. of moles of MnO4- by:

0.0189 � 26.05 = 4.293 � 10-4


According to the equation above, 1 mol of MnO4- react with 5 moles of Fe 2+

So, the no of moles of Fe 2+ is :

4.293 � 10-3� 5 = 2.46� 10-3

By the finding the no of moles of Fe 2+, we can find the Molarity by using the formula,

Molarity = No of moles


Molarity of Fe 2+ = 2.46� 10-3


= 0.0985 mol dm-3

By finding the Molarity we can find the mass of anhydrous Fe (NH4)2(SO4)2 in one dm3.

Molar mass of Fe (NH4)2(SO4)2 = 55.8+ (14.0+4) + (32.1 + 16�4) �2

= 55.8+ 36+192.2

= 284 g mol

Molarity of Fe 2+ = molarity of Fe (NH4)2(SO4)2

Therefore, Molarity of Fe (NH4)2(SO4)2 = 0.0985 mol dm-3

The No of moles of Fe (NH4)2(SO4)2 can be found by the formula:

Molarity � Volume = no of moles

0.0985 � 1 dm-3 = 0.0985 moles

Therefore the no of moles of Fe (NH4)2(SO4)2 is 0.0985 moles.

The mass of anhydrous Fe (NH4)2(SO4)2 can be found by:

0.0985 mol � 284 g mol = 27.97g

Therefore the mass of anhydrous Fe (NH4)2(SO4)2 in 1 dm-3 is 28 g.

Using the concentration of ammonium iron (II) sulphate solution and the mass of anhydrous Fe (NH4)2(SO4)2 we can obtain volume of the Fe (NH4)2(SO4)2�xH2O

Let the volume of the solution of ammonium iron (II) sulphate be V:

39.214 g dm-3 = 27.97g


39.214 g dm-3 V= 27.97

V= 0.7132 dm-3

The no of moles of Fe (NH4)2(SO4)2�xH2O can be found by the formula:

Molarity � Volume = no of moles

The Molarity of Fe (NH4)2(SO4)2�xH2O= Molarity of Fe 2+

0.0985mol dm3� 0.7132 dm3 = 0.703mol

Therefore the no of moles of Fe (NH4)2(SO4)2�xH2O is 0.0703mol

The molar mass of Fe (NH4)2(SO4)2�xH2O Can be found by:

Let the molar mass of Fe (NH4)2(SO4)2�xH2O be MM

27.97 = 0.0703



MM= 398.1

Therefore the total molar mass Fe (NH4)2(SO4)2�xH2O is 398.

The Molar mass of Fe (NH4)2(SO4)2�xH2O is :


= 284 + 18x

By using the total molar mass:

284 + 18x = 398

18x= 398-284

18x= 114



Therefore the value of x is 6.

Therefore the formula is Fe (NH4)2(SO4)2�6H2O.


A redox titration involves a reducing agent and an oxidizing agent. In this experiment, ammonium iron (II) sulphate was the reducing agent and potassium manganate was the oxidizing agent.


The solution in the conical flask was titrated with potassium permanganate, the solution in the flask first turn into a yellowish colour; this was because of the formation of Iron (III) ions formed during the redox reaction.

According to the following equation:

Fe 2+ (aq)+ e—> Fe3+(aq)

The formation of Fe3+ caused the solution to turn yellow when it was titrated against the potassium manganate (VII).

Solution gradually turned into pale pink because of the formation of Mn2+

The excess Mn2+ caused the solution to turn to pale pink. According to the eqation.

MnO4- (aq) + 8H+ (aq) + 5e- –> 4H2O (l) + Mn2+ (aq)


1. Explain why it is unnecessary to use any indicator in this experiment.

–>This is because potassium permanganate can be its own self indicator because it can form a pale pink colour due to the formation of Mn2+ when it is being reduced

2. Explain why dilute sulphuric acid should be added to ammonium iron (II) sulphate solution before titration.

–> It prevents premature oxidation of iron (II) ions as hydrated ammonium iron (II) ions may react with air to form Iron (III) ions.

It also ensures the complete reaction for manganate from manganate (VII) to Mn2+

2. Suggest one possible error in the experiment

–> The possible error of this experiment is that maybe the pipette was not washed properly as the same pipette was used to pipette 25cm3 of ammonium iron (II) sulphate and 25cm3 of sulphuric acid. The solution could have already reacted when pipette.


The value of x in the formula of Fe (NH4)2(SO4)2�xH2O is 6.

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