Transition Metals - Carrying Titrations with Potassium Permanganate (A-Level Chemistry)

Carrying out a Redox Titration with Potassium Manganate (VII)

Redox Titrations with Potassium Manganate (VII)

Carrying out a Redox Titration with Potassium Manganate (VII)

Acidified potassium manganate (VII) is used as it is a strong oxidising agent.

When is acts as an oxidising agent, the colour changes from purple (MnO₄⁻ ) to colourless (Mn²⁺)

MnO₄⁻ + 8H⁺ +5e⁻ → Mn²⁺ + 4H₂O

The choice and amount of acid is important and justified by considering the half equation and redox potentials.

MnO₄⁻ + 8H⁺ +5e⁻ ⇌ Mn²⁺ + 4H₂O E⦵ = +1.51V

From the half equation, 8 moles of acid are needed for each mole of manganate (VII).

Insufficient volume of acid will mean that the solution will not be acidic enough and a brown solid, MnO₂, will be produced instead of Mn²⁺ ions.

MnO₄⁻ (aq) + 4H⁺(aq) + 3e⁻ → MnO₂ (s) + 2H₂O(l)

The brown solid will mask the colour change and lead to larger volumes of manganate ions used than actually required.

Why can ethanoic acid not be used?

A weak acid, like ethanoic acid, cannot be used for the same reason. There are insufficient hydrogen ions produced.

Why can concentrated hydrochloric acid not be used?

Concentrated hydrochloric acid cannot be used as the manganate (VII) ions will oxidise the chloride ions to chlorine. The E⦵MnO₄⁻ / Mn²⁺＞ E⦵Cl₂/Cl⁻ :

MnO₄⁻ + 8H⁺ +5e⁻ ⇌ Mn²⁺ + 4H₂O E⦵ = +1.51V

Cl₂(aq) + 2e⁻ ⇌ 2Cl⁻ (aq) E⦵ = +1.36V

As manganate ions will be used up in the reaction with chloride ions, it will lead to a greater volume of manganate being used in the titration than required and an inaccurate reading.

Redox Titration of Potassium Manganate (VII) with Iron (II) Ions

In the redox titration between potassium manganate (VII) and Fe²⁺ ions the following reactions occur:

Fe³⁺ +e⁻ ⇌ Fe²⁺                                      E⦵ = +0.77V

MnO₄⁻ + 8H⁺ +5e⁻ ⇌ Mn²⁺ + 4H₂O E⦵ = +1.51V
Purple                            Colourless

The more positive electrode potential will be reduced so the reactions that will actually occur are:

MnO₄⁻ + 8H⁺ +5e⁻ ⇌ Mn²⁺ + 4H₂O

Fe²⁺ → Fe³⁺ + e

Balancing the electrons and combining the two equations gives the overall reaction:

MnO₄⁻ + 8H⁺ + 5Fe²⁺→ 5Fe³⁺ + Mn²⁺ + 4H₂O

Why can concentrated nitric acid not be used?

Dilute sulfuric acid can be used in this titration.

However concentrated nitric acid cannot be used, concentrated nitric acid oxidises Fe²⁺ to Fe³⁺. This would lead to a smaller volume of manganate (VII) ions being used than required.

Worked example: A student dissolved 1550 mg of iron tablets in an excess of dilute sulfuric acid. The solution was titrated with 0.0200 mol dm⁻³ potassium manganate (VII) solution. 27.50 cm³ of potassium manganate (VII) was required to reach the end point in the titration. Calculate the percentage of iron in the sample of tablets. (4 marks)

Answer

Step 1: Work out moles of manganate (VII)

Amount of manganate (VII) = 27.5/1000 x 0.02 = 5.5 x 10⁻⁴ mol (1)
MnO₄⁻ + 8H⁺ + 5Fe²⁺→ 5Fe³⁺ + Mn²⁺ + 4H₂O

Step 2:- use the mole ratios to work out moles of Fe²⁺

Mole ratio of MnO₄⁻ : Fe²⁺ = 1:5
So moles of Fe²⁺ = 5 x = 2.75 x 10⁻³ mol (1)

Step 3: work out the mass from the moles

Mass of iron = moles/ Ar = 2.75 x 10⁻³ x 55.8 = 0.15345 g = 153.5 mg (1)

Step 4: Find the percentage of mass from the mass reacted.

Percentage of Fe = 153.5/1550 x 100 = 9.90% (1)

Redox Titration of Potassium Manganate (VII) with Ethanedioate ions

In the redox titration between potassium manganate (VII) and ethanedioate ions, the following reactions occur:

C₂O₄²⁻ → 2CO₂ + 2e⁻
MnO₄⁻ + 8H⁺ +5e⁻ → Mn²⁺ + 4H₂O

Balancing the electrons and combining the half equations, gives the overall reaction:

2MnO₄⁻ + 16H⁺ + 5C₂O₄²⁻ → 10CO₂ + 2Mn²⁺ + 8H₂O

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What reaction conditions are needed?

Again dilute sulfuric acid can be used in this reaction. In this reaction there are two negative ions reacting together (MnO₄⁻ ) and (C₂O₄²⁻). This results in a high activation energy and the solution needs to be heated to about 60°C to speed up the initial reaction.

Worked example: 4.56g of impure iron (II) ethanedioate (FeC₂O₄.2H₂O) is dissolved in excess sulfuric acid and made up to 250 cm³ standard solution. 25cm³ of this solution is titrated against 0.05 mol dm⁻³ potassium manganate (VII). 24.55 cm³ of the manganate (VII) ions were needed until the solution decolourised. What is the percentage of iron(II) ethanedioate in the original sample?

Answer

Step 1: Determine the overall reaction

Both the Fe²⁺ and C₂O₄²⁻ ions react with the MnO4- ions
2MnO₄⁻ + 16H⁺ + 5C₂O₄²⁻ → 10CO₂ + 2Mn²⁺ + 8H₂O
MnO₄⁻ + 8H⁺ + 5Fe²⁺→ 5Fe³⁺ + Mn²⁺ + 4H₂O
So overall
3MnO₄⁻ + 24H⁺ + 5FeC₂O₄ → 10CO₂ + 3Mn²⁺+ 12H₂O + 5Fe³⁺
So the mole ratio of MnO₄⁻ :FeC₂O₄ = 3:5

Step 2: work out the moles of manganate (VII) reacted

Moles of manganate (VII) = 24.55/1000 x 0.05 = 1.2275 x10⁻³ mol

Step 3 use the mole ratio to find the moles of FeC2O4.2H2O in 25 cm³.

Moles of MnO₄⁻ x 5/3 = 1.2275 x 10⁻³ x 5/3 = 2.046 x 10⁻³ mol

Step 4 Find the moles of FeC₂O₄.2H₂O in 250 cm³

2.046 x 10⁻³ x 10 = 2.046 x 10⁻² mol

Step 5 find the mass of FeC₂O₄.2H₂O

Mass = moles x Mr = 2.046 x 10⁻² x 179.8 = 3.678 g

Step 6 Find the % mass

% mass = 3.678/ 4.56 x 100 = 80.67%

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