Redox Titrations

This lesson covers: 

1. How redox titrations work
2. Redox titrations using potassium permanganate (KMnO₄)
3. Iodine-thiosulfate redox titrations
4. 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).

You need to know about two types of redox titration; permanganate (MnO₄^-) titrations and iodine/thiosulfate (I₂/S₂O₃^2-) titrations.

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

3. Hydrogen peroxide (H₂O₂)

• Half-equation:  H₂O₂ ➔ O₂ + 2H⁺ + 2e^-
• Full equation:  2MnO₄^- + 5H₂O₂ + 6H⁺ ➔ 2Mn²⁺ + 5O₂ + 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

Iodine-thiosulfate redox titrations

Iodine-thiosulfate redox titrations are used to determine the concentration of an oxidising agent. In these titrations, iodine (I₂) is reduced to iodide ions (I^-) by thiosulfate ions (S₂O₃^2-).

The relevant half-equations are:

I₂ + 2e^- ➔ 2I^-

2S₂O₃^2- ➔ S₄O₆^2- + 2e^-

And the full equation for the titration reaction is:

I₂ + 2S₂O₃^2- ➔ 2I^- + S₄O₆^2-

The process involves three main stages:

1. A known volume of the oxidising agent is reacted with an excess of acidified potassium iodide (KI) solution. This oxidises some of the iodide ions (I^-) to iodine (I₂).
2. The iodine produced is then titrated with a sodium thiosulfate (Na₂S₂O₃) solution of known concentration. The thiosulfate ions (S₂O₃^2-) reduce the iodine back to iodide ions.
3. The concentration of the original oxidising agent is calculated using the volume and concentration of the sodium thiosulfate solution used in the titration.

Iodine-thiosulfate titrations can be used to determine the concentration of various oxidising agents, including iodate ions (IO₃^-), hydrogen peroxide (H₂O₂), and chlorine (Cl₂).

The worked example below demonstrates how to apply the iodine-thiosulfate titration procedure to calculate the concentration of potassium iodate (KIO₃) in a solution.

Colour change at the end point of iodine-thiosulfate titrations

The end point of an iodine-thiosulfate titration is signalled by a distinct colour change caused by the presence of starch, which is added as an indicator near the end of the titration.

Here's how the colour change occurs:

1. During the titration, the dark brown colour of the iodine solution gradually fades as the iodine (I₂) is reduced to colourless iodide ions (I^-) by the thiosulfate ions (S₂O₃^2-) from the sodium thiosulfate solution.
2. When the majority of the iodine has been reduced and only a small amount remains, a few drops of starch solution are added to the titration mixture. Starch forms a deep blue-black complex with iodine, making it much easier to detect the presence of even trace amounts of iodine in the solution.
3. As more sodium thiosulfate is added, the blue-black colour persists until all the remaining iodine is reduced to iodide.
4. At the end point of the titration, the final drop of sodium thiosulfate reduces the last trace of iodine, causing the blue-black colour to disappear suddenly and completely, leaving a colourless solution.

This sharp colour change from blue-black to colourless is easily recognisable and indicates that all the iodine has been consumed, marking the end point of the titration.

Worked example 4 - Calculating the concentration of IO₃^- ions

25.0 cm³ of a solution containing an unknown concentration of IO₃^- ions is titrated with a 0.0800 mol dm^-3 sodium thiosulfate (Na₂S₂O₃) solution. The end point is reached after 30.0 cm³ of the sodium thiosulfate solution has been added.

Calculate the concentration of IO₃^- ions in the original solution.

Step 1: Write the balanced equations for the reactions

IO₃^- + 5I^- + 6H⁺ ➔ 3I₂ + 3H₂O

I₂ + 2S₂O₃^2- ➔ 2I^- + S₄O₆^2-

The combined equation is:

IO₃^- + 6H⁺ + 6S₂O₃^2- ➔  I^- + 3S₄O₆^2- + 3H₂O

Step 2: Conversion of cm³ into dm³

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

25.0 cm³ = 0.0250 dm³

30.0 cm³ = 0.0300 dm³

Step 3: Calculate number of moles of S₂O₃^2-

n = c × V =0.0800×0.0300=2.40×10−3 mol

Step 4: Calculate number of moles of IO₃^-

IO₃^- : S₂O₃^2- mole ratio = 1:6

Moles of IO₃^- = 6(2.40×10−3)​=4.00×10−4 mol

Step 5: Calculate concentration of IO₃^-

c =Vn​=0.0250(4.00×10−4)​=0.0160 mol dm−3