JEE Redox Reactions PYQs With Video Solutions, Download PDF

JEE Redox Reactions PYQ

Redox Reactions is an important chapter in JEE Chemistry. It helps you understand how electrons move from one substance to another during a chemical reaction. In every redox reaction, oxidation and reduction happen at the same time. One substance loses electrons, while another substance gains them. In this chapter, you study oxidation numbers, oxidising agents, reducing agents, balancing redox equations, disproportionation reactions, equivalent mass, and electron transfer. These ideas are also useful in electrochemistry, metallurgy, and inorganic chemistry. That is why a clear understanding of this chapter is important for later topics as well. Most questions from Redox Reactions are based on oxidation-state calculations, identifying the oxidising or reducing agent, and balancing chemical equations. The calculations are usually simple, but a small mistake in sign, charge, or oxidation number can change the final answer. Solving previous-year questions helps you understand the common question patterns. It also improves your speed and makes it easier to apply the rules correctly.

On this page, you can revise the key concepts, download the PYQ PDF, learn important formulas, avoid common mistakes, and practise chapter-wise questions.

With regular practice, Redox Reactions can become an easy and scoring chapter in JEE Chemistry.

JEE Redox Reactions Previous-Year Questions

Previous-year questions help you understand which topics are asked most often. In this chapter, questions commonly come from oxidation numbers, oxidising and reducing agents, balancing equations, n-factor, equivalent mass, and disproportionation.

While solving JEE Mains PYQ, first identify the elements whose oxidation numbers are changing.

An increase in oxidation number means oxidation.

A decrease in oxidation number means reduction.

Once you identify these changes, it becomes easier to find the oxidising agent and reducing agent.

The substance that gets reduced acts as the oxidising agent. The substance that gets oxidised acts as the reducing agent.

For oxidation-number questions, follow the basic rules carefully.

The oxidation number of an element in its free state is zero. Oxygen is usually −2, while hydrogen is usually +1. The sum of the oxidation numbers of all atoms must be equal to the overall charge on the molecule or ion.

However, there are some important exceptions.

Oxygen has an oxidation number of −1 in peroxides. Hydrogen has an oxidation number of −1 in metal hydrides.

While balancing a redox equation, always check the reaction medium. The method may change depending on whether the reaction takes place in acidic, basic, or neutral medium.

Try to solve each question within a fixed time. After completing the set, check every wrong answer and find the exact reason behind the mistake.

Common reasons may include:

• You used the wrong oxidation number.
• You forgot an exception for oxygen or hydrogen.
• You confused the oxidising and reducing agents.
• You balanced the atoms but not the charge.
• You ignored the reaction medium.

Keeping a short list of oxidation-number rules can make these questions much easier.

JEE Redox Reactions PYQs PDF

You can download the JEE Redox Reactions PYQs PDF from the section below.

First, solve the PDF without looking at the answers. Try to complete it in one sitting so that you can understand your current speed and accuracy.

After solving the questions, divide them into three groups:

• Questions you solved easily
• Questions that took more time
• Questions you could not solve

Revise the rules and methods used in the difficult questions. Try those questions again after a few days.

While practising these JEE Questions, do not only compare the final answer with the solution.

Check the oxidation numbers, electron change, coefficients, atoms, and total charge.

Even if the equation looks balanced, count all atoms once again. Also make sure that the charge is the same on both sides.

For questions based on the ion-electron method, write the oxidation and reduction half-reactions separately.

First, balance the atoms other than oxygen and hydrogen. Then balance oxygen using H₂O. After that, balance hydrogen according to the reaction medium.

In acidic medium, H⁺ ions are used.

In basic medium, OH⁻ ions are used. You can first balance the reaction in acidic medium and then add OH⁻ ions to both sides to remove H⁺.

When you solve the PDF again, focus mainly on the questions that were incorrect or took too much time.

Important Formulas for JEE Redox Reactions PYQs

Redox Reactions does not have many long formulas, but it includes several important rules and relations.

You can add the following points to your JEE Chemistry Formula notes for quick revision.

Do not memorise these rules without understanding the electron transfer.

Consider this reaction:

Zn + Cu²⁺ → Zn²⁺ + Cu

Zinc changes from 0 to +2. It loses two electrons, so it is oxidised.

Copper changes from +2 to 0. It gains two electrons, so it is reduced.

Therefore, zinc acts as the reducing agent, while Cu²⁺ acts as the oxidising agent.

Also remember that the n-factor of a substance is not always fixed. It depends on the reaction and the final oxidation state.

For example, the n-factor of KMnO₄ changes in acidic, neutral, and basic media because manganese forms different products in each medium.

Always identify the final product before using the n-factor, equivalent mass, or normality formula.

Common Mistakes to Avoid in JEE Redox Reactions PYQs

One common mistake is thinking that oxidation always means the addition of oxygen.

Oxidation can also mean loss of electrons, removal of hydrogen, or an increase in oxidation number.

Reduction means gain of electrons, addition of hydrogen, or a decrease in oxidation number.

Students also confuse the oxidising agent with the substance that is oxidised.

Remember that an oxidising agent accepts electrons and gets reduced. A reducing agent gives electrons and gets oxidised.

While solving JEE Redox Reactions Questions, do not find the oxidation number by looking at only one atom.

Use the total charge of the complete molecule or ion.

Another common mistake is taking the oxidation number of oxygen as −2 in every compound.

In peroxides, oxygen is −1. In superoxides, its average oxidation number is −1/2. In OF₂, oxygen has an oxidation number of +2.

Hydrogen is usually +1, but it becomes −1 in ionic metal hydrides such as NaH and CaH₂.

While balancing equations, students often balance the atoms but forget the charge.

A correctly balanced redox equation must have the same number of atoms and the same total charge on both sides.

Another mistake is using the same n-factor for a substance in every reaction.

The n-factor depends on the actual change in oxidation number, so it may change from one reaction to another.

Always check whether the reaction takes place in acidic, basic, or neutral medium. Using H⁺ in a basic-medium reaction or OH⁻ in an acidic-medium reaction can lead to the wrong equation.

Before finalising your answer, check the atoms, charge, reaction medium, and electron transfer once more.

List of JEE Redox Reactions PYQs

Use the questions given below as a short chapter test.

Try to solve them like a JEE Main Test. Set a time limit, avoid using notes, and check the answers only after completing the full set.

Question 1

For standardizing NaOH solution, which of the following is used as a primary standard?

A

Sodium tetra borate

B

Ferrous Ammonium sulphate

C

Oxalic acid

D

Dil. HCl

Show Answer Explanation

correct answer:- 3

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Question 2

When 10 mL of an aqueous solution of Fe$$^{2+}$$ ions was titrated in the presence of dil H$$_2$$SO$$_4$$ using diphenylamine indicator, 15 mL of 0.02 M solution of K$$_2$$Cr$$_2$$O$$_7$$ was required to get the end point. The molarity of the solution containing Fe$$^{2+}$$ ions is $$x \times 10^{-2}$$ M. The value of x is ___. (Nearest integer)

Show Answer Explanation Video Solution

correct answer:- 18

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Question 3

Which of the given reactions is not an example of disproportionation reaction?

A

$$2H_2O_2 \rightarrow 2H_2O + O_2$$

B

$$2NO_2 + H_2O \rightarrow HNO_3 + HNO_2$$

C

$$MnO_4^- + 4H^+ + 3e^- \rightarrow MnO_2 + 2H_2O$$

D

$$3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$$

Show Answer Explanation Video Solution

correct answer:- 3

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Question 4

$$H_2O_2$$ acts as a reducing agent in

A

$$2NaOCl + H_2O_2 \rightarrow 2NaCl + H_2O + O_2$$

B

$$2Fe^{2+} + 2H^+ + H_2O_2 \rightarrow 2Fe^{3+} + 2H_2O$$

C

$$Mn^{2+} + 2H_2O_2 \rightarrow MnO_2 + 2H_2O$$

D

$$Na_2S + 4H_2O_2 \rightarrow Na_2SO_4 + 4H_2O$$

Show Answer Explanation Video Solution

correct answer:- 1

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Question 5

Equal amounts (in moles) of potassium permanganate $$KMnO_4$$ are used to completely oxidize two different reducing agents in separate flasks.

In Flask A, the $$KMnO_4$$ reacts with ferrous oxalate $$FeC_2O_4$$ in a strongly acidic medium.

In Flask B, the $$KMnO_4$$ reacts with potassium iodide $$KI$$ in a faintly alkaline (neutral) medium.

What is the ratio of the number of moles of $$FeC_2O_4$$ oxidized in Flask A to the number of moles of $$KI$$ oxidized in Flask B?

A

$$5 : 3$$

B

$$10 : 3$$

C

$$5 : 6$$

D

$$3 : 1$$

Show Answer Explanation

Solution

Key Principle: Law of Equivalence

According to the law of chemical equivalence, the total number of equivalents of an oxidizing agent completely reacts with an equal number of equivalents of a reducing agent:

$$\text{Equivalents of Oxidizing Agent} = \text{Equivalents of Reducing Agent}$$

$$\text{Moles} \times \text{n-factor} = \text{Moles} \times \text{n-factor}$$

Let the number of moles of $$\text{KMnO}_4$$ used in each flask be $$x$$.

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Analysis of Flask A (Acidic Medium)

• Reaction of $$\text{KMnO}_4$$:

  In a strongly acidic medium, the permanganate ion ($$\text{MnO}_4^-$$) is reduced to manganese(II) ions ($$\text{Mn}^{2+}$$):

  $$\overset{+7}{\text{Mn}}\text{O}_4^- \rightarrow \overset{+2}{\text{Mn}}^{2+}$$

  The change in oxidation state per manganese atom is $$7 - 2 = 5$$. Thus, the $$\text{n-factor}$$ of $$\text{KMnO}_4$$ in an acidic medium is 5 .

• Reaction of Ferrous Oxalate ($$\text{FeC}_2\text{O}_4$$):

  During oxidation, both the cation and the anion in ferrous oxalate undergo an increase in their oxidation states:

  1. $$\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \quad (\text{loses } 1 \text{ electron})$$
  2. $$\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2e^- \quad (\text{loses } 2 \text{ electrons})$$

  The total number of electrons lost per molecule of $$\text{FeC}_2\text{O}_4$$ is $$1 + 2 = 3$$. Thus, its $$\text{n-factor}$$ is 3 .

Applying the equivalence formula for Flask A:

$$\text{Moles of KMnO}_4 \times 5 = \text{Moles of FeC}_2\text{O}_4 \times 3$$

$$x \times 5 = \text{Moles of FeC}_2\text{O}_4 \times 3 \implies \text{Moles of FeC}_2\text{O}_4 = \frac{5x}{3}$$

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Analysis of Flask B (Faintly Alkaline / Neutral Medium)

• Reaction of $$\text{KMnO}_4$$:

  In a neutral or faintly alkaline medium, the permanganate ion ($$\text{MnO}_4^-$$) is reduced to manganese dioxide ($$\text{MnO}_2$$):

  $$\overset{+7}{\text{Mn}}\text{O}_4^- \rightarrow \overset{+4}{\text{Mn}}\text{O}_2$$

  The change in oxidation state per manganese atom is $$7 - 4 = 3$$. Thus, the $$\text{n-factor}$$ of $$\text{KMnO}_4$$ in a neutral/faintly alkaline medium is 3 .

• Reaction of Potassium Iodide ($$\text{KI}$$):

  In a faintly alkaline medium, the iodide ion ($$\text{I}^-$$) is oxidized to the iodate ion ($$\text{IO}_3^-$$):

  $$\overset{-1}{\text{I}}^- \rightarrow \overset{+5}{\text{IO}}_3^-$$

  The change in oxidation state per iodine atom is $$+5 - (-1) = 6$$. Thus, the $$\text{n-factor}$$ of $$\text{KI}$$ is 6 .

Applying the equivalence formula for Flask B:

$$\text{Moles of KMnO}_4 \times 3 = \text{Moles of KI} \times 6$$

$$x \times 3 = \text{Moles of KI} \times 6 \implies \text{Moles of KI} = \frac{3x}{6} = \frac{x}{2}$$

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Calculating the Ratio

Now, we compute the ratio of the moles of $$\text{FeC}_2\text{O}_4$$ oxidized to the moles of $$\text{KI}$$ oxidized:

$$\text{Ratio} = \frac{\text{Moles of FeC}_2\text{O}_4}{\text{Moles of KI}} = \frac{\frac{5x}{3}}{\frac{x}{2}}$$

$$\text{Ratio} = \frac{5}{3} \times \frac{2}{1} = \frac{10}{3}$$

Therefore, the required ratio is 10 : 3 .

Answer: Option B — 10 : 3

correct answer:- 2

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Question 6

The reaction of $$H_2O_2$$ with potassium permanganate in acidic medium leads to the formation of mainly

A

$$Mn^{2+}$$

B

$$Mn^{4+}$$

C

$$Mn^{3+}$$

D

$$Mn^{6+}$$

Show Answer Explanation Video Solution

correct answer:- 1

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Question 7

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R
Assertion A: Permanganate titrations are not performed in presence of hydrochloric acid.
Reason R: Chlorine is formed as a consequence of oxidation of hydrochloric acid.
In the light of the above statements, choose the correct answer from the options given below

A

Both A and R are true and R is the correct explanation of A

B

Both A and R are true but R is NOT the correct explanation of A

C

A is true but R is false

D

A is false but R is true

Show Answer Explanation

correct answer:- 1

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Question 8

Given below are two statements:

Statement I: Hydrogen peroxide can act as an oxidizing agent in both acidic and basic conditions.
Statement II: Density of hydrogen peroxide at $$298 \text{ K}$$ is lower than that of $$D_2O$$.
In the light of the above statements. Choose the correct answer from the options

A

Both Statement I and Statement II are true

B

Both Statement I and Statement II are false

C

Statement I is true but Statement II is false

D

Statement I is false but Statement II is true

Show Answer Explanation

correct answer:- 3

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Question 9

High purity ($$> 99.95\%$$) dihydrogen is obtained by

A

reaction of zinc with aqueous alkali

B

electrolysis of warm aqueous barium hydroxide solution between nickel electrodes

C

electrolysis of acidified water using platinum electrodes

D

reaction of zinc with dilute acid

Show Answer Explanation Video Solution

correct answer:- 2

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Question 10

$$\text{MnO}_4^{2-}$$, in acidic medium, disproportionates to :

A

$$\text{MnO}_4^{-}  \text{and}  \text{MnO}_{2}$$

B

$$\text{Mn}_2 \text{O}_7  \text{and}  \text{MnO}$$

C

$$\text{Mn}_2 \text{O}_7  \text{and}  \text{MnO}_2$$

D

$$\text{MnO}_4^{-}  \text{and}  \text{MnO}$$

Show Answer Explanation Video Solution

correct answer:- 1

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Question 11

During water-gas shift reaction

A

Carbon monoxide is oxidized to carbon dioxide.

B

Water is evaporated in presence of catalyst.

C

Carbon is oxidized to carbon monoxide.

D

Carbon dioxide is reduced to carbon monoxide.

Show Answer Explanation Video Solution

correct answer:- 1

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Question 12

See the following chemical reaction:
Cr$$_2$$O$$_7^{2-}$$ + XH$$^+$$ + 6Fe$$^{2+}$$ $$\to$$ YCr$$^{3+}$$ + 6Fe$$^{3+}$$ + ZH$$_2$$O
The sum of X, Y and Z is _____.

Show Answer Explanation Video Solution

correct answer:- 23

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Question 13

KMnO$$_4$$ oxidises I$$^-$$ in acidic and neutral/faintly alkaline solution, respectively to

A

I$$_2$$ and IO$$_3^-$$

B

IO$$_3^-$$ and I$$_2$$

C

IO$$_3^-$$ and IO$$_3^-$$

D

I$$_2$$ and I$$_2$$

Show Answer Explanation

correct answer:- 3

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Question 14

Which of the following reactions are disproportionation reactions? (1) $$Cu^+ \rightarrow Cu^{2+} + Cu$$ (2) $$3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$$ (3) $$2KMnO_4 \rightarrow K_2MnO_4 + MnO_2 + O_2$$ (4) $$2MnO_4^- + 3Mn^{2+} + 2H_2O \rightarrow 5MnO_2 + 4H^+$$. Choose the correct answer from the options given below:

A

1, 2

B

2, 3, 4

C

1, 2, 4

D

1, 4

Show Answer Explanation

correct answer:- 1

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Question 15

Which one of the following reactions indicates the reducing ability of hydrogen peroxide in basic medium?

A

$$HOCl + H_2O_2 \rightarrow H_3O^+ + Cl^- + O_2$$

B

$$PbS + 4H_2O_2 \rightarrow PbSO_4 + 4H_2O$$

C

$$2MnO_4^- + 3H_2O_2 \rightarrow 2MnO_2 + 3O_2 + 2H_2O + 2OH^-$$

D

$$Mn^{2+} + H_2O_2 \rightarrow Mn^{4+} + 2OH^-$$

Show Answer Explanation

correct answer:- 3

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Question 16

The dark purple colour of $$KMnO_4$$ disappears in the titration with oxalic acid in acidic medium. The overall change in the oxidation number of manganese in the reaction is

A

$$5$$

B

$$1$$

C

$$7$$

D

$$2$$

Show Answer Explanation Video Solution

correct answer:- 1

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Question 17

The products obtained from a reaction of hydrogen peroxide and acidified potassium permanganate are

A

$$Mn^{4+}$$, $$H_2O$$ only

B

$$Mn^{2+}$$, $$H_2O$$ only

C

$$Mn^{4+}$$, $$H_2O$$, $$O_2$$ only

D

$$Mn^{2+}$$, $$H_2O$$, $$O_2$$ only

Show Answer Explanation

correct answer:- 4

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Question 18

Which respect to an ore, Ellingham diagram helps to predict the feasibility of its:

A

Electrolysis

B

Zone refining

C

Vapour phase refining

D

Thermal reduction

Show Answer Explanation Video Solution

correct answer:- 4

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Question 19

Identify the process in which change in the oxidation state is five:

A

$$Cr_2O_7^{2-} \rightarrow 2Cr^{3+}$$

B

$$MnO_4^{-} \rightarrow Mn^{2+}$$

C

$$CrO_4^{2-} \rightarrow Cr^{3+}$$

D

$$C_2O_4^{2-} \rightarrow 2CO_2$$

Show Answer Explanation Video Solution

correct answer:- 2

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Question 20

The INCORRECT statement(s) about heavy water is (are):
(A) used as a moderator in nuclear reactor
(B) obtained as a by-product in fertilizer industry.
(C) used for the study of reaction mechanism
(D) has a higher dielectric constant than water
Choose the correct answer from the options given below:

A

(B) only

B

(C) only

C

(D) only

D

(B) and (D) only

Show Answer Explanation

correct answer:- 3

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Question 21

10.0 mL of 0.05 M KMnO$$_4$$ solution was consumed in a titration with 10.0 mL of given oxalic acid dihydrate solution. The strength of given oxalic acid solution is _________ $$\times 10^{-2}$$ g/L. (Round off to the nearest integer)

Show Answer Explanation

correct answer:- 1575

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Question 22

The oxidation states of nitrogen in NO, NO$$_2$$, N$$_2$$O and NO$$_3^-$$ are in the order of :

A

$$NO_3^- > NO_2 > NO > N_2O$$

B

$$NO_2 > NO_3^- > NO > N_2O$$

C

$$N_2O > NO_2 > NO > NO_3^-$$

D

$$NO > NO_2 > N_2O > NO_3^-$$

Show Answer Explanation Video Solution

correct answer:- 1

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Question 23

The redox reaction among the following is

A

formation of ozone from atmospheric oxygen in the presence of sunlight

B

reaction of [Co(H$$_2$$O)$$_6$$]Cl$$_3$$ with AgNO$$_3$$

C

reaction of H$$_2$$SO$$_4$$ with NaOH

D

combination of dinitrogen with dioxygen at 2000K

Show Answer Explanation Video Solution

correct answer:- 4

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Question 24

According to the following diagram, A reduces $$BO_2$$ when the temperature is:

A

$$< 1400°C$$

B

$$> 1400°C$$

C

$$> 1200°C$$ but $$< 1400°C$$

D

$$< 1200°C$$

Show Answer Explanation Video Solution

correct answer:- 2

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Question 25

Chlorine undergoes disproportionation in alkaline medium as shown below :
$$aCl_2(g) + bOH^-(aq) \rightarrow cClO^-(aq) + dCl^-(aq) + eH_2O(l)$$
The values of $$a, b, c$$ and $$d$$ in a balanced redox reaction are respectively :

A

$$1, 2, 1$$ and $$1$$

B

$$2, 2, 1$$ and $$3$$

C

$$3, 4, 4$$ and $$2$$

D

$$2, 4, 1$$ and $$3$$

Show Answer Explanation

correct answer:- 1

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Question 26

In basic medium $$CrO_4^{2-}$$ oxidises $$S_2O_3^{2-}$$ to form $$SO_4^{2-}$$ and itself changes into $$Cr(OH)_4^-$$. The volume of 0.154 M $$CrO_4^{2-}$$ required to react with 40 mL of 0.25 M $$S_2O_3^{2-}$$ is ______ mL. (Rounded-off to the nearest integer)

Show Answer Explanation

correct answer:- 173

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Question 27

Which of the following equation depicts the oxidizing nature of $$H_2O_2$$?

A

$$2I^- + H_2O_2 + 2H^+ \to I_2 + 2H_2O$$

B

$$I_2 + H_2O_2 + 2OH^- \to 2I^- + 2H_2O + O_2$$

C

$$Cl_2 + H_2O_2 \to 2HCl + O_2$$

D

$$KIO_4 + H_2O_2 \to KIO_3 + H_2O + O_2$$

Show Answer Explanation

correct answer:- 1

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Question 28

The oxidation states of $$P$$ in H$$_4$$P$$_2$$O$$_7$$, H$$_4$$P$$_2$$O$$_5$$ and H$$_4$$P$$_2$$O$$_6$$, respectively, are:

A

7, 5 and 6

B

5, 4 and 3

C

5, 3 and 4

D

6, 4 and 5

Show Answer Explanation

correct answer:- 3

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Question 29

In neutral or faintly alkaline medium, $$KMnO_4$$ being a powerful oxidant can oxidise, thiosulphate almost quantitatively, to sulphate. In this reaction overall change in oxidation state of manganese will be

A

5

B

1

C

0

D

3

Show Answer Explanation Video Solution

correct answer:- 4

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Question 30

Water does not produce CO on reacting with:

A

$$CO_2$$

B

$$CH_4$$

C

$$C_3H_8$$

D

C

Show Answer Explanation

correct answer:- 1

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Question 31

Which one of the following is an example of disproportionation reaction?

A

$$3MnO_4^{2-} + 4H^+ \to 2MnO_4^- + MnO_2 + 2H_2O$$

B

$$MnO_4^- + 4H^+ + 4e^- \to MnO_2 + 2H_2O$$

C

$$10I^- + 2MnO_4^- + 16H^+ \to 2Mn^{2+} + 8H_2O + 6I_2$$

D

$$8MnO_4^- + 3S_2O_3^{2-} + H_2O \to 8MnO_2 + 6SO_4^{2-} + 2OH^-$$

Show Answer Explanation Video Solution

correct answer:- 1