JEE Main Previous year questions (2021-23): Redox Reactions PDF Download

Q.1. When Fe_0.93O is heated in presence of oxygen, it converts to Fe₂ O₃. The number of correct statement/s from the following is     (JEE Main 2023)
(a) The equivalent weight of Fe_0.93O is
(b) The number of moles of Fe²⁺ and Fe³⁺ in 1 mole of Fe_0.93O is 0.79 and 0.14 respectively
(c) Fe_0.93O is metal deficient with lattice comprising of cubic closed packed arrangement of O²⁻ ions
(d) The % composition of Fe²⁺ and Fe³⁺ in Fe_0.93 O is 85% and 15% respectively

Ans. d
Fe_0.93O
2x+(0.93-x)3 = 2
-x+3×0.93=2
x =0.79
0.79 = no. of Fe²⁺ ion
0.14 = no. of Fe³⁺ ion
nf = 0.79

Due to presence of Fe³⁺ in  FeO lattice, Metal deficiency occurs.

Q.2. In which of the following reactions the hydrogen peroxide acts as a reducing agent?     (JEE Main 2023)
(a) PbS + 4H₂O₂ → PbSO₄ + 4H₂O
(b) Mn²⁺ + H₂O₂ → Mn⁴⁺ + 2OH⁻ 
(c) HOCl + H₂O₂ → H₃O⁺ + Cl⁻ + O₂
(d) 2Fe²⁺ + H₂O₂ → 2Fe³⁺ + 2OH⁻ 

Ans. c

hydrogen peroxide acts as a reducing agent

Q.3. Consider the following sulphur based oxoacids.
H₂SO₃, H₂SO₄, H₂S₂O₈ and H₂S₂O₇.
Amongst these oxoacids, the number of those with peroxo (O−O) bond is ________.    (JEE Main 2022)

Ans. 1

Q.4. The disproportionation of  in acidic medium resulted in the formation of two manganese compounds A and B. If the oxidation state of Mn in B is smaller than that of A, then the spin-only magnetic moment (μ) value of B in BM is __________. (Nearest integer)    (JEE Main 2022)

Ans. 4

Mn → 4s²3d⁵
Mn⁺⁴ → 3d³
n = 3

= 3.87 ≈ 4 B.M

Q.5. The normality of H₂SO₄ in the solution obtained on mixing 100 mL of 0.1 MH₂SO₄ with
50 mL of 0.1MNaOH is _______________ × 10⁻¹ N. (Nearest Integer)    (JEE Main 2022)

Ans. 1
No. of equivalents of H₂SO₄ = 100 × 0.1 × 2 = 20
No. of equivalents of NaOH = 50 × 0.1 = 5
No. of equivalents of H₂SO₄ left = 20 − 5 = 15
⇒ 150 x x = 15
x = 1/10 = 0.1 N = 1 x 10^-1 N

Q.6. The difference in oxidation state of chromium in chromate and dichromate salts is ___________.    (JEE Main 2022)

Ans. 0
Chromate ion →, oxidation state of Cr = +6
Dichromate ion → , oxidation state of Cr = +6
∴ Difference in oxidation state = zero

Q.7. The neutralization occurs when 10 mL of 0.1M acid 'A' is allowed to react with 30 mL of 0.05 M base M(OH)₂. The basicity of the acid 'A' is __________.    (JEE Main 2022)
[M is a metal]

Ans. 3

Q.8. 0.01 M KMnO₄ solution was added to 20.0 mL of 0.05 M Mohr's salt solution through a burette. The initial reading of 50 mL burette is zero. The volume of KMnO₄ solution left in the burette after the end point is _____________ mL. (nearest integer)    (JEE Main 2022)

Ans. 30

Q.9. In neutral or faintly alkaline medium, KMnO₄ being a powerful oxidant can oxidize, thiosulphate almost quantitatively, to sulphate. In this reaction overall change in oxidation state of manganese will be:    (JEE Main 2022)
(a) 5
(b) 1
(c) 0
(d) 3

Ans. d
In neutral or Faintly alkaline medium, thiosulphate is oxidised almost quantitatively to sulphate ion according to reaction given below,

Here the Mn changes from Mn⁺⁷ to Mn⁺⁴ 
Thus overall change in its oxidation number would be of 3.

Q.10. The dark purple colour of KMnO₄ disappears in the titration with oxalic acid in acidic medium. The overall change in the oxidation number of manganese in the reaction is :    (JEE Main 2022)
(a) 5
(b) 1
(c)7
(d) 2

Ans. 5

Q.11. Which of the given reactions is not an example of disproportionation reaction?    (JEE Main 2022)
(a) 2H₂O₂ → 2H₂O + O₂
(b) 2NO₂ + H₂O → HNO₃ + HNO₂
(c)  + 4H⁺ +3e⁻ → MnO₂ + 2H₂O
(d) + 4H⁺ → + MnO₂ + 2H₂O

Ans. c

Q.12. Which one of the following is an example of disproportionation reaction ?    (JEE Main 2022)
(a) + 4H⁺ →  + MnO₂ + 2H₂O
(b)  + 4H⁺ + 4e⁻ → MnO₂ + 2H₂O
(c) 10I⁻ +  + 16H⁺ → 2Mn²⁺ + 8H₂O + 5I₂
(d) + H₂O → 8MnO₂ +  + 2OH⁻

Ans. a

is an intermediate oxidation state and is converted into compounds having higher and lower oxidation states. 

Q.13. Among the following, basic oxide is :    (JEE Main 2022)
(a) SO₃
(b) SiO₂
(c) CaO
(d) Al₂O₃

Ans. c
Since, oxides of metals are basic in nature. Hence CaO is a basic oxide.
SO₃ and SiO₂ are acidic oxides and Al₂O₃ is a amphoteric oxide.

Q.14. Ozonolysis of ClO₂ produces an oxide of chlorine. The average oxidation state of chlorine in this oxide is __________.    (JEE Advanced 2021)

Ans. 6

Oxidation state of Cl in Cl₂O₆ = 2x + 6(−2) = 0
2x − 12 = 0
2x = 12 ⇒ x = + 6
Average oxidation state of Cl in Cl₂O₆ is 6.

Q.15. A sample (5.6 g) containing iron is completely dissolved in cold dilute HCl to prepare a 250 mL of solution. Titration of 25.0 mL of this solution requires 12.5 mL of 0.03 M KMnO₄ solution to reach the end point. Number of moles of Fe²⁺ present in 250 mL solution is x × 10⁻² (consider complete dissolution of FeCl₂). The amount of iron present in the sample is y% by weight
(Assume : KMnO₄ reacts only with Fe²⁺ in the solution
Use : Molar mass of iron as 56 g mol⁻¹)    (JEE Advanced 2021)

Ans. 18.75
The value of y is ______.
Mass of sample = 5.6 g
Number of moles of Fe = 1.88 × 10⁻² mol
Molar mass of Fe = 56 g/mol
Mass of Fe = Moles of Fe × Molar mass
= 1.88 × 10⁻² × 56 = 1.05 g

⇒ y = 18.75%
The value of y is 18.75.

Q.16. A sample (5.6 g) containing iron is completely dissolved in cold dilute HCl to prepare a 250 mL of solution. Titration of 25.0 mL of this solution requires 12.5 mL of 0.03 M KMnO₄ solution to reach the end point. Number of moles of Fe²⁺ present in 250 mL solution is x × 10⁻² (consider complete dissolution of FeCl₂). The amount of iron present in the sample is y% by weight.(Assume : KMnO₄ reacts only with Fe²⁺ in the solution
Use : Molar mass of iron as 56 g mol⁻¹)
The value of x is ______.    (JEE Advanced 2021)

Ans. 1.875

Concentration of Fe²⁺, M₁ =

Volume of Fe²⁺ solution titrated, V₁ = 25.0 mL
Concentration of , M2 = 0.03 M
Volume of used, V2 = 12.5 mL

Using relation,

n₂M₁V₁ = n₁M₂V₂ [n₁ and n₂ are number of moles of Fe²⁺ and  reacting]
1 × x/25 × 25 = 5 × 0.03 × 12.5 ⇒ x = 1.88
The volume of x is 1.88.

Q.17. In the given chemical reaction, colors of the Fe²⁺ and Fe³⁺ ions, are respectively :
5Fe²⁺ +  + 8H⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺    (JEE Main 2021)
(a) Yellow, Orange
(b) Yellow, Green
(c) Green, Orange
(d) Green, Yellow

Ans. d
Colour of Fe²⁺ is observed green and Fe³⁺ is yellow.

Q.18. Experimentally reducing a functional group cannot be done by which one of the following reagents?    (JEE Main 2021)
(a) Pt-C/H₂
(b) Na/H₂
(c) Pd-C/H₂
(d) Zn/H₂O

Ans. b
Na in presence of H₂, will not release electron which are required for reduction.
H₂ gas also not get adsorbed on Na. Hence Na/H₂ cannot be used as a reducing agent.

Q.19. In which one of the following sets all species show disproportionation reaction?    (JEE Main 2021)
(a)
(b)
(c)
(d)

Ans. c
No option contains all species that show disproportionation reaction. So question is bonus.
- Cl, Mn, Cr in these anions are present in highest oxidation state. These will not undergo disproportionation.

Q.20. Potassium permanganate on heating at 513 K gives a product which is :    (JEE Main 2021)
(a) paramagnetic and colourless
(b) diamagnetic and green
(c) diamagnetic and colourless
(d) paramagnetic and green

Ans. d

In K₂MnO₄, manganese oxidation state is +6 and hence it has one unpaired e⁻.

Q.21. The nature of oxides V₂O₃ and CrO is indexed as 'X' and 'Y' type respectively. The correct set of X and Y is :    (JEE Main 2021)
(a) X = basic, Y = amphoteric
(b) X = amphoteric, Y = basic
(c) X = acidic, Y = acidic
(d) X = basic, Y = basic

Ans. d
V₂O₃ basic
CrO basic

Q.22. The incorrect statement is :    (JEE Main 2021)
(a) Cl₂ is more reactive than ClF.
(b) F₂ is more reactive than ClF.
(c) On hydrolysis ClF forms HOCl and HF.
(d) F₂ is a stronger oxidizing agent than Cl₂ in aqueous solution.

Ans. a

(i) Reactivity order :
F₂ > ClF (inter halogen) > Cl₂
(ii) ClF + H₂O → HOCl + HF
(iii) Oxidizing power is aqueous solution
F₂ > Cl₂ > Br₂ > I₂

Q.23. Match List - I with List - II :

Choose the most appropriate answer from the options given below    (JEE Main 2021)
(a) (a)-(ii), (b)-(ii), (c)-(iii), (d)-(ii), (e)-(iii)
(b) (a)-(ii), (b)-(iii), (c)-(ii), (d)-(i), (e)-(iii)
(c) (a)-(ii), (b)-(ii), (c)-(iii), (d)-(i), (e)-(iii)
(d) (a)-(ii), (b)-(i), (c)-(ii), (d)-(iii), (e)-(iii)

Ans. b
NaOH → Basic
Be(OH)₂ → Amphoteric
Ca(OH)₂ → Basic
B(OH)₃ → Acidic
Al(OH)₃ → Amphoteric

Q.24. The oxidation states of 'P' in H₄P₂O₇, H₄P₂O₅ and H₄P₂O₆, respectively, are :    (JEE Main 2021)
(a) 7, 5 and 6
(b) 5, 4 and 3
(c) 5, 3 and 4
(d) 6, 4 and 5

Ans. c
Oxidation state of P in H₄P₂O₇, H₄P₂O₅ and H₄P₂O₆ is 5, 3 & 4 respectively
H₄P₂O₇
2x + 4(+ 1) + 7 (−2) = 0
x = + 5

2x + 4(+ 1) + 5 (−2) = 0
x = + 3

2x + 4 (+ 1) + 6(−2) = 0
x = +4

Q.25. Identify the process in which change in the oxidation state is five :    (JEE Main 2021)
(a)
(b)
(c)
(d)

Ans. b

Q.26. The oxidation states of nitrogen in NO, NO₂, N₂O and  are in the order of :    (JEE Main 2021)
(a) N₂O > NO₂ > NO >
(b) NO > NO₂ > N2O >
(c) NO₂ >  > NO > N₂O
(d)  > NO₂ > NO > N₂O

Ans. d
The oxidation states of Nitrogen in following molecules are as follows
→ +5
NO₂ → +4
NO → +2
N₂O → +1

Q.27. The set that represents the pair of neutral oxides of nitrogen is :    (JEE Main 2021)
(a) NO and N₂O
(b) N₂O and NO₂
(c) NO and NO₂
(d) N₂O and N₂O₃

Ans. a
NO and N₂O are neutral oxides and N₂O₃ , NO₂ and N₂O₅ are acidic oxides.

Q.28. Statement I : Sodium hydride can be used as an oxidising agent.
Statement II : The lone pair of electrons on nitrogen in pyridine makes it basic.
Choose the CORRECT answer from the options given below :    (JEE Main 2021)
(a) Statement I is false but statement II is true
(b) Both statement I and statement II are true
(c) Statement I is true but statement II is false
(d) Both statement I and statement II are false

Ans. a
NaH is a strong H^– (hydride) donor. Hence cannot be used as an oxidising agent.

In Pyridine, the lone pair of ‘N’ is localised, makes it basic.

Q.29. Compound A used as a strong oxidizing agent is amphoteric in nature. It is the part of lead storage batteries. Compound A is :    (JEE Main 2021)
(a) PbO
(b) PbO₂
(c) PbSO₄
(d) Pb₃O₄

Ans. b
PbO₂ is strong oxidizing agent because Pb+4 is not stable and can be easily reduced to Pb⁺².
PbO₂ is used in lead storage batteries. It is also amphoteric in nature.

Q.30. Which of the following equation depicts the oxidizing nature of H₂O₂?    (JEE Main 2021)
(a) CL₂ + H₂O₂ → 2HCl + O₂
(b) KIO₄ + H₂O₂ → KIO₃ + H₂O + O₂
(c) 2I− + H₂O₂ + 2H+ → I₂ + 2H₂O
(d) I₂ + H₂O₂ + 2OH⁻ → 2I− + 2H₂O + O₂

Ans. c
2I− + H₂O₂ + 2H+ → I₂ + 2H₂O
Oxygen reduces from –1 to –2.
So, its reduction will takes place. Hence it will behave as oxidising agent or it shows oxidising nature.
While in other option it change from (–1) to 0.

Q.31. The incorrect statement among the following is :    (JEE Main 2021)
(a) RuO₄ is an oxidizing agent
(b) Cr₂O₃ is an amphoteric oxide.
(c) VOSO₄ is a reducing agent
(d) Red colour of ruby is due to the presence of Co³⁺

Ans. d
Red colour of ruby is due to presence of Cr³⁺ ions in Al₂O₃. Chromium is the trace element that causes ruby’s red colour, which ranges from an orange red to a publish red. The strength of ruby’s red depends on how much chromium is present.

Q.32. (A) HOCI + H₂O₂ → H₃O+ + Cl^- + O₂
(B) I₂ + H₂O₂ + 2OH^- → 2I^- + 2H₂O + O₂
Choose the correct option.    (JEE Main 2021)
(a) H₂O₂ acts as reducing and oxidising agent respectively in equations (A) and (B).
(b) H₂O₂ act as oxidizing and reducing agent respectively in equations (A) and (B).
(c) H₂O₂ acts as oxidising agent in equations (A) and (B).
(d) H₂O₂ acts as reducing agent in equations (A) and (B).

Ans. d
In equation (A), HOCl undergoes reduction in presence of H₂O₂.

Here, oxidation state of Cl changes from +1 to –1 (i.e. reduces)
∴ H₂O₂ act as reducing agent I₂ reduces to I^- in presence of H₂O₂.
In equation (B),

Here oxidation state of iodine decreases (from 0 to –1)
∴ H₂O₂ act as reducing agent in both the equations.