Redox Titrations

This lesson covers: 

1. How redox titrations work
2. Redox titrations using potassium permanganate (KMnO₄)
3. Calculations involving redox titrations

Redox titrations determine oxidising or reducing agent concentrations

Redox titrations are used to find the concentration of an oxidising agent or reducing agent in solution.

While they share some similarities with acid-base titrations, there are key differences:

• In many cases, no additional indicator is needed as the reactants themselves undergo colour changes that signal the end point. This is known as a self-indicating titration.
• The titration determines the amount of oxidising agent needed to exactly react with a given quantity of reducing agent (or vice versa).

Transition elements are versatile redox reagents

Transition metals are particularly well-suited for use in redox titrations due to their unique properties:

• They can readily change oxidation states by accepting or donating electrons, making them effective oxidising or reducing agents.
• These changes in oxidation state are often accompanied by distinct colour changes, which serve as built-in visual indicators.
• The colour changes make it easy to identify the end point of the titration without the need for an additional indicator in many cases.

One of the most commonly used transition metal compounds in redox titrations is potassium permanganate (KMnO₄).

Redox titrations with potassium permanganate (KMnO₄)

Potassium permanganate (KMnO₄) is a powerful oxidising agent commonly used in redox titrations. The permanganate ion (MnO₄^-) is reduced to manganese(II) ions (Mn²⁺) during these reactions.

The half-equation for this reduction is:

MnO₄^- + 8H⁺ + 5e^- ➔ Mn²⁺ + 4H₂O

The end point of a permanganate titration is signalled by a sharp colour change from the deep purple of MnO₄^- to a faint pink or colourless solution, indicating that all the permanganate ions have been reduced to manganese(II) ions.

MnO₄^- can oxidise a variety of reducing agents, including:

1. Iron(II) ions (Fe²⁺)

• Half-equation:  Fe²⁺ ➔ Fe³⁺ + e^-
• Full equation:  MnO₄^- + 5Fe²⁺ + 8H⁺ ➔ Mn²⁺ + 5Fe³⁺ + 4H₂O

2. Ethanedioate ions (C₂O₄^2-)

• Half-equation:  C₂O₄^2- ➔ 2CO₂ + 2e^-
• Full equation:  2MnO₄^- + 5C₂O₄^2- + 16H⁺ ➔ 2Mn²⁺ + 10CO₂ + 8H₂O

Titration procedure for determining MnO₄^- concentration

To determine the concentration of MnO₄^- needed to react with a reducing agent:

1. Measure a known volume of the reducing agent solution (e.g., Fe²⁺) using a pipette into a conical flask. Add an excess of dilute sulfuric acid to ensure sufficient H⁺ ions for MnO₄^- reduction.
2. Gradually add MnO₄^- solution from a burette to the flask, swirling constantly. Stop when the mixture becomes faintly tinted with the purple MnO₄^- colour, which indicates the end point. Record the volume of MnO₄^- added.
3. Repeat the titration until concordant results are obtained (titre volumes that differ by no more than 0.10 cm³).
4. Calculate the mean volume of MnO₄^- added from the concordant titres.

The following worked examples demonstrate how to apply this titration procedure to calculate the concentration of various analytes and determine the percentage composition of a sample.

Worked example 1 - Calculating the concentration of Fe²⁺ ions

30.0 cm³ of an acidified solution containing Fe²⁺ ions is titrated with a 0.0150 mol dm^-3 KMnO₄ solution. The end point is reached after 28.0 cm³ of KMnO₄ has been added.

Calculate the concentration of Fe²⁺ in the original solution.

Step 1: Write the balanced equation for the reaction

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

Step 2: Conversion of cm³ into dm³

To convert from cm³ into dm³, divide by 1,000

30.0 cm³ = 0.0300 dm³

28.0 cm³ = 0.0280 dm³

Step 3: Calculate number of moles of MnO₄^- added

n = c × V =0.0150×0.0280=4.20×10−4 mol

Step 4: Calculate number of moles of Fe²⁺

Fe²⁺ : MnO₄^- mole ratio = 5:1

Moles of Fe²⁺ = (4.20 x 10^-4) x 5 = 2.10 x 10^-3 mol

Step 5: Calculate concentration of Fe²⁺

c =Vn​=0.0300(2.10×10−3)​=0.070 mol dm−3

Worked example 2 - Estimating the percentage of iron in iron tablets

A 3.00 g iron tablet was dissolved in dilute sulfuric acid to give 300 cm³ of solution. 30.0 cm³ of this solution was found to react with 15.0 cm³ of 0.0300 mol dm^-3 potassium manganate(VII) solution.

Calculate the percentage of iron in the tablet.

Step 1: Write the balanced equation for the reaction

MnO₄^-_(aq) + 8H⁺_(aq) + 5Fe²⁺_(aq) ➔ Mn²⁺_(aq) + 4H₂O_(l) + 5Fe³⁺_(aq)

Step 2: Conversion of cm³ into dm³

To convert from cm³ into dm³, divide by 1,000

300 cm³ = 0.300 dm³

15.0 cm³ = 0.0150 dm³

Step 3: Calculate the number of moles of MnO₄^- 

n = c × V =0.0300×0.0150=4.50×10−4 mol

Step 4: Calculate number of moles of Fe²⁺ in 30.0 cm³ portion

Fe²⁺ : MnO₄^- mole ratio = 5:1

Moles of Fe²⁺ = (4.50 x 10^-4) x 5 = 2.25 x 10^-3 mol

Step 5: Calculate number of moles of Fe²⁺ in 300 cm³ solution

Moles of Fe2+ = (2.25 x 10^-3) x 10 = 2.25 x 10^-2 mol$

Step 6: Calculate mass of iron in the tablet

m = n x A_r = (2.25 x 10^-2) x 55.8 = 1.26 g

Step 7: Calculate percentage of iron in the tablet

% of iron =3.001.26​×100=41.9%

Worked example 3 - Calculating the concentration of C₂O₄^2- ions

20.0 cm³ of a solution containing an unknown concentration of C₂O₄^2- ions is titrated with a 0.0600 mol dm^-3 potassium permanganate (KMnO₄) solution. The end point is reached after 22.0 cm³ of the potassium permanganate solution has been added.

Calculate the concentration of C₂O₄^2- ions in the original solution.

Step 1: Write the balanced equation for the reaction

2MnO₄^- + 5C₂O₄^2- + 16H⁺ ➔ 2Mn²⁺ + 10CO₂ + 8H₂O

Step 2: Conversion of cm³ into dm³

To convert from cm³ into dm³, divide by 1,000

20.0 cm³ = 0.0200 dm³

22.0 cm³ = 0.0220 dm³

Step 3: Calculate number of moles of MnO₄^-

n = c × V =0.0600×0.0220=1.32×10−3 mol

Step 4: Calculate number of moles of C₂O₄^2-

C₂O₄^2- : MnO₄^- mole ratio = 5:2

Moles of C₂O₄^2- = 1.32×10−3×(25​)=3.30×10−3 mol

Step 5: Calculate concentration of C₂O₄^2-

c =Vn​=0.0200(3.30×10−3)​=0.165 mol dm−3