Redox Titration – Definition, Examples, Practice Problems and FAQ

Do you love butter? Who wouldn’t?

Do you know that the taste of butter drastically changes depending on whether it has less or more salt than is typical?

If the amount of salt in the butter is not regulated, the same butter we adore could end up making us queasy.

Have you ever considered how the salt content of the butter we eat is regulated?

Redox titration is a type of analytical method used to control the amount of salt in butter. Find out more about redox titration in this concept page article.

TABLE OF CONTENTS

• What is Titration?
• What is a Redox Titration?
• Redox Titration – Examples
• Equivalent Weight of Oxidising/Reducing Agents in Redox Titrations
• Stoichiometry of Redox Titrations
• Practice Problems
• Frequently Asked Questions – FAQ

What is Titration?

Titration is an analytical technique used to find the strength of an unknown solution (analyte) that is usually taken in a conical flask, by titrating it against a known solution (known as the titrant) that is usually taken in the burette. An indicator is added to the reaction mixture depending on the pH range at the endpoint to indicate the end of the reaction. Usually, an indicator acts by changing the colour or the intensity of the solution. The equivalence point is a point in the titration at which the amount of both the reactants is equal. The strength of the analyte can be calculated from the volume of the titrant used to reach the endpoint.

According to the law of volumetric analysis, V₁N₁ = V₂N₂

Where, V₁ is the volume of the titrant used to reach the endpoint of the reaction.

N₁ is the strength of the titrant used

V₂ is the volume of analyte taken

N₂ is the strength of the analyte that is to calculated.

If the values of V_1, N₁ and V₂ are known, then N₂ can be calculated as,
N₂ = V₁N₁/V₂

What is a Redox Titration?

Redox Titration is a laboratory method for determining the concentration of an analyte. It is based on a redox reaction between the analyte and the titrant, and it may occasionally include a potentiometer or a redox indicator as part of the process.

When we say that this titration is based on a redox reaction between the analyte and the titrant, we mean that the analyte and the titrant undergo a simultaneous oxidation-reduction reaction.

Redox titrations, as we know, are based on redox reactions, and a redox reaction is essentially an oxidation-reduction reaction.

How will we be able to tell if a substance has been oxidised?

If even one of the following four possibilities occurs, we can say that the substance has been oxidised.

1. If the substance's electrons are lost.
2. If the substance is exposed to oxygen.
3. If the given substance's hydrogen is removed.
4. When the oxidation state of the substance increases.

How will we know whether a substance has been reduced?

If even one of the following four options is true, we can say that a substance has been reduced.

1. If the substance accumulates electrons.
2. If the substance is treated with hydrogen.
3. If the given substance is depleted of oxygen.
4. When the oxidation state of the substance decreases.

Electron transfer occurs between the analyte and the titrant in redox titrations.

Redox Titration – Examples

Iodometric Titrations

The most common example of redox titration is when we use a starch indicator to detect the endpoint after treating an iodine solution with a reducing agent like thiosulfate to produce iodide ions. This is also referred to as iodimetric titration.

The reducing agents are titrated in iodimetry using an iodine solution because they can be quantitatively oxidised at the equivalence point. Only a small number of compounds can be oxidised by iodine because it is such a weak oxidising agent. One of the typical reducing agents, thiosulphate, needs to be determined iodimetrically.

Iodine (l₂) is a mild oxidising agent.

l₂ + 2e^- -> 2l^- (reduction)

Examples:

1. H₂S (aq) + I₂(aq) -> 2H⁺(aq) + S(aq) + 2I^-(aq)
2. H₂SO₃(aq) + I₂(aq) + H₂O(aq) -> SO₄^2-(aq) + 2I^-(aq) + 4H⁺(aq)
3. H3AsO3^-(aq) + I₂(aq) + H₂O(aq) -> H₂AsO₄^-(aq) + 2I^-(aq) + 2H⁺(aq)
4. Sn²⁺(aq) + I₂(aq) -> Sn⁴⁺(aq) + 2I^-(aq)
5. 2S₂O₃^2-(aq) + I₂(aq) -> S4O₆^2-(aq) + 2I^-(aq)

Iodometric Titrations

The strength or concentration of an oxidising agent is determined using iodometric titration. In this method, an oxidising agent is used to convert l^- into l₂. Titration with a standard thiosulphate solution yields a quantitative estimate of the liberated iodine. Almost all strong oxidising agents can be estimated using this method. In this case, starch is used as an indicator, producing a dark blue colour in the presence of l₂.

Iodide (l^-) is a mild reducing agent.

2I^- -> I₂ + 2e^- (oxidation)

Examples:

1. 2HNO₂(aq) + 2I^-(aq) + 2H⁺(aq) -> 2NO(aq) + I₂(s) + 2H₂O(aq)
2. BrO3^-(aq) + 6I^-(aq) + 6H⁺(aq) -> Br^-(aq) + 3I₂(s) + 3H₂O(aq)
3. Cr₂O₇^2-(aq) + 6I^-(aq) + 14H⁺(aq) -> 2Cr³⁺(aq) + 3I₂(s) + 7H₂O(aq)
4. 2MnO4^-(aq) + 10I^-(aq) + 16H⁺(aq) -> 2Mn²⁺(aq) + 5I₂(s) + 8H₂O(aq)
5. IO₃^-(aq) + 5I^-(aq) + 6H⁺(aq) -> 3I₂(aq) + 3H₂O(aq)

Titration of Potassium Permanganate with Oxalic Acid (KMnO4 v/s C₂H₂O4)

Potassium permanganate is a strong oxidising agent that becomes even more potent in the presence of sulfuric acid. The oxidising ability of KMnO₄ in acidic medium is represented by the following equation.

In acidic media, MnO4^- + 8H⁺ + 5e^- -> Mn²⁺ + 4H₂O

Because solutions containing MnO₄^- ions are purple and solutions containing Mn²⁺ ions are colourless, permanganate solution is decolourised when added to a solution containing a reducing agent like oxalic acid. When there is an excess of potassium permanganate in the solution, it turns purple. As a result, KMnO₄ acts as a self-indicator in acidic solutions.

Potassium permanganate is tested for purity against pure oxalic acid. A redox reaction is involved. KMnO4 oxidises oxalic acid to carbon dioxide, which is then reduced to KMnO4. The following is how oxalic acid reacts with potassium permanganate.

Reduction Half Reaction

2KMnO4(aq) + 3H₂SO4(aq) -> K₂SO4(aq) + 2 MnSO4(aq) + 3H₂O(aq) + 5[O]

Oxidation Half Reaction

2(COOH)₂(aq) + 5[O] -> 5H₂O(aq) + 10CO₂(g)

Overall reaction

2KMnO4(aq) + 3H₂SO4(aq) + 2(COOH)₂(aq) -> K₂SO4(aq) + 2 MnSO4(aq) + 8H₂O(aq) + 10CO₂(g)

Equivalent Weight of Oxidising/Reducing Agents in Redox Titrations

To calculate the stoichiometry of any redox reaction, the stoichiometric ratio of the reacting species is required. This can be found from the balanced chemical equation of the redox reaction. By using the concept of chemical equivalence, this step can be avoided, since the number of gram equivalents of the oxidising and the reducing agents are equal.

The equivalent weight of an oxidising agent (OA) or a reducing agent (RA) can be calculated as follows.
Equivalent weight of OA/RA =Molar mass of OA/RA/Number of electrons gained or lost per formula unit of OA/RA

Example: In acidic media, MnO4^-+8H⁺+5e^- ->Mn²⁺+4H₂O

Number of electrons gained per formula unit = change in oxidation number per formula unit = 5

Therefore,

Irrespective of the reducing agent, in an acidic medium, permanganate ions acts as a strong oxidising agent and produce 5 electrons per formula unit.

In a faintly alkaline medium,  is reduced to  Mno₂ i,e, 3 electrons are consumed per formula unit.

Therefore, 

Stoichiometry of Redox Titrations

The normality of a solution of oxidising or reducing agent is defined as the number of gram equivalents of the oxidising or reducing agent present in one litre of its solution.

Relationship Between Normality and Molarity

Normality = Molarity ✕ n - factor
n-factor= Molar mass/Equivalent weight

Example:The n-factor of KMnO4 in an acidic medium is 5 since 5 electrons are produced per formula unit of KMnO4.

The normality of a 0.2 M acidified solution of KMnO4 = 0.2 ✕ 5 = 1 N

Practice Problems

1. Find the n-factor of Mn in KMnO4 for the reaction

2KMnO4(aq) + 3H₂SO4(aq) + 2(COOH)₂(aq) -> K₂SO4(aq) + 2 MnSO4(aq) + 8H₂O(aq) + 10CO₂(g)

1. 5
2. 2
3. 3
4. 4

Answer:A

Solution:

Let the oxidation state of Mn in KMnO4 be 'x'

=> 1 + x - 8 = 0

=> x = + 7

Let the oxidation state of Mn in KMnO4 be 'y'

=> y+6-8=0

=> y= +2

n-factor of Mn = Number of Mn atom (O.S. of Mn in KMnO4-O.S. of Mn in MnSO4 )

n-factor of Mn= 1 |7 - 2| = 5

So, option A is the correct answer.

1. Pick out the correct statement for titration.

1. The equivalent point is a theoretical point at which the amounts of two reactants are simply equal.
2. Titrant is a known concentration solution placed in a burette.
3. Analyte is a solution of unknown concentration but known volume placed in a conical flask.
4. All of these

Answer:D

Solution:

• Titration is an analytical technique used to find the strength of an unknown solution (analyte) of a known volume that is usually taken in a conical flask, by titrating it against a known solution (known as the titrant) that is usually taken in the burette.

• An indicator is added to the reaction mixture depending on the pH range at the endpoint to indicate the end of the reaction. Usually, an indicator acts by changing the colour or the intensity of the solution.

• The equivalence point is a point in the titration at which the amount of both the reactants are equal.

• The strength of the analyte can be calculated from the volume of the titrant used to reach the endpoint.

Thus, the statements given in options A, B and C are correct.

So, option D is the correct answer.

1. The endpoint of a redox titration of KMnO4v/s C2H2O4 can be indicated by which observation?

1. Appearance of purple colour
2. Disappearance of brown colour
3. Disappearance of purple colour
4. Appearance of brown colour

Answer:A

Solution:

• When we first add potassium permanganate to an oxalic acid-containing conical flask, it discharges and the solution remains colourless.

• The endpoint is indicated by a purple colour due to an excess of unreacted potassium permanganate after complete consumption of oxalic acid ions (purple in colour).

• Solutions containing ions are purple and solutions containing ions are colourless, permanganate solution is decolourised when added to a solution containing a reducing agent like oxalic acid.

• When there is an excess of potassium permanganate in the solution, it turns purple. As a result, acts as a self-indicator in acidic solutions.

So, option A is the correct answer.

1. The endpoint of an iodometric titration can be indicated by which observation?

1. Appearance of pink colour
2. Disappearance of blue colour
3. Disappearance of pink colour
4. Appearance of blue colour

Answer:

Solution:The disappearance of blue colour indicates the endpoint of an iodometric titration. Starch, the indicator used in iodometric titrations, imparts a dark blue colour in the presence of iodine. The blue colour disappears at the endpoint with full iodine consumption.

So, option B is the correct answer.

Frequently Asked Questions – FAQ

Q1. What are the factors that affect a redox reaction?
Ans. pH is the only factor that affects redox titration as different species act differently in different pH. One of the most famous examples is potassium permanganate. It acts as an oxidising agent in alkaline and neutral media but acts as a strong oxidising agent in an acidic medium.

Q2. Why is potassium permanganate commonly used in redox reactions?
Ans. In redox titration, the endpoint of the titration is usually detected using an indicator. Potassium permanganate is commonly used in redox titrations because it acts as a self-indicator in the acidic medium. Solutions containing ions are purple and solutions containing ions are colourless, permanganate solution is decolourised when added to a solution containing a reducing agent like oxalic acid. When there is an excess of potassium permanganate in the solution, it turns purple. As a result, acts as a self-indicator in acidic solutions.

Q3. Can HCl be used as an acidifying agent in a redox titration involving permanganate?
Ans. HCl is not used in titrations involving permanganate because HCl reacts with permanganate ions and gets oxidised to chlorine gas. To avoid this reaction, sulphuric acid is preferred to hydrochloric acid.

Q4. Why is the permanganate titration end point not permanent?
Ans. The excess permanganate ions react slowly with the relatively high concentration of manganese (II) ions present at the endpoint. This is why the permanganate endpoint is not permanent.