Redox Reactions – JEE Advanced Previous Year Questions with Solutions

Redox Reactions JEE Advanced Previous Year Questions with Solutions provided here covers most of the questions that were asked in the entrance exam over the years. Thus, to prepare better for JEE Advanced the set of questions and solutions given here will be instrumental in students achieving maximum marks in their exams. The questions and solutions will further help students familiarize themselves with the format of the question paper and its types. Additionally, as JEE aspirants get extensive practice in answering the questions they will be able to develop adept problem-solving skills to tackle the questions in the upcoming exam.

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JEE Advanced Previous Year Questions on Redox Reaction

Question 1. To measure the quantity of MnCl₂ dissolved in an aqueous solution, it was completely converted to KMnO₄ using the reaction,

MnCl₂ + K₂S₂O₈ + H₂O → KMnO₄ + H₂SO₄ + HCl (equation not balanced).

Few drops of concentrated HCl were added to this solution and gently warmed. Further, oxalic acid (225 g) was added in portions till the colour of the permanganate ion disappeared. The quantity of MnCl2 (in mg) present in the initial solution is _____. (Atomic weights in g mol–1 : Mn = 55, Cl = 35.5)

Solution: (126)

MnCl₂ + K₂S₂O₈ + H₂O → KMnO₄ + H₂SO₄ + HCl

m_eq of C₂O₄^2- = m_eq of MnO₄^–

2 × 225 / 90 = a × 5

⇒ a = 1

w = 1 × [55+71]

= 126 mg

Question 2. The difference in the oxidation numbers of the two types of sulphur atoms in

Na₂S₄O₆ is:

Solution: (5)

The structure of Na₂S₄O₆ is;

The oxidation number of sulphur atom involved in coordinate bond formation is (+5) and that of the middle sulphur atom is zero. Hence, the difference in oxidation number of the two types of sulphur atoms will be (+5).

∴ The difference in the O.S of the two types of sulphur atoms is 5 – 0 = 5.

Question 3. For the reaction:

I^– + ClO^-3 + H₂SO₄ → Cl^– + HSO₄^– + I₂

The correct statement(s) in the balanced equation is/are:

A. Stoichiometric coefficient HSO₄^– of is 6

B. Iodide is oxidized

C. Sulphur is reduced

D. H₂O is one of the products

Solution: (A, B and D)

The balanced equation is:

6I^– + ClO^-3 + 6H₂SO₄ → Cl^– + 6HSO₄^– + 3I₂ + 3H₂O

Here you will see that iodine is oxidised and the stoichiometric coefficient of HSO₄^– is 6. Additionally, water is formed on the product side.

Question 4. The pair of compounds having metals in their highest oxidation state is:

A. MnO₂, FeCl₃

B. [MnO₄]^–, CrO₂Cl₂

C. [Fe(CN)₆]^3-, [Co(CN)₃]

D. [NiC1₄]^2-, [CoC1₄]^–

Solution: (B)

a. The oxidation state of Mn in MnO₂ is +4

The oxidation state of Fe in FeCl₃ is +3

b) The oxidation state of Mn in [MnO₄]^– is +7

The oxidation state of Cr in CrO₂Cl₂ is +6

c) The oxidation state of Fe in [Fe(CN)₆]^3- is +3

The oxidation state of Co in [Co(CN)₃] is +3

d) The oxidation state of Ni in [NiC1₄]^2- is +2

The oxidation state of Co in [CoC1₄]^– is +3

So the correct option is (B)

Question 5. Consider a titration of potassium dichromate solution with acidified Mohr’s salt solution using diphenylamine as an indicator. The number of moles of Mohr’s salt required per mole of dichromate is:

A. 3

B. 4

C. 5

D. 6

Solution: (D)

The redox reaction between potassiumdichromate and Mohr’s salt is:

6Fe²⁺ + Cr₂O₇^2- + 14H⁺ → 6Fe³⁺ + 2Cr³⁺ + 7H₂O

Mohr’s salt: Fe₂SO₄.(NH₄)₂SO₄.6H₂O

Dichromate will oxidise Fe²⁺ from Mohr’s salt to Fe³⁺.

Question 6. Which ordering of compounds is according to the decreasing order of the oxidation state of nitrogen?

A. HNO₃, NO, NH₄Cl, N₂

B. HNO₃, NO, N₂, NH₄Cl

C. HNO₃, NH₄Cl, NO, N₂

D. NO, HNO₃, NH₄Cl, N₂

Solution: (B)

Let the oxidation state of nitrogen = x

HNO₃: (+1) + X + 2(-3), x = +5

NO: x + (-2), x = +2

NH₄CI: X + 4(+1) + (-1), X = -3

N₂: 2x = 0, x = 0

So the correct order will be HNO₃, NO, N₂, NH₄Cl

Question 7. (NH₄)₂Cr₂O₇ on heating gives a gas which is also given by:

A. Heating NH₄NO₂

B. Heating NH₄NO₃

C. Mg₃N₂ + H₂O

D. Na(comp.) + H₂O₂

Solution: (A)

Ammonium dichromate on heating gives N₂ gas. This gas is also obtained by heating NH₄NO₂.

Question 8. The reaction 3ClO^– (aq) → ClO₃^– (aq) + 2Cl^– (aq) is an example of:

A. Oxidation

B. Reduction

C. Disproportionation

D. Decomposition reaction

Solution: (C)

In this reaction, chlorine is oxidized as well as reduced.

ClO^– is oxidised to ClO₃^–; ClO^– is also reduced to Cl^–.

The oxidation states of chlorine in ClO(aq)^–, ClO₃(aq)^– and Cl(aq)^– are +1,+5 and −1 respectively.

Question 9. Among the following identify the species with an atom in a +6 oxidation state.

A. MnO₄^–

B. Cr(CN)₆^3-

C. NiF₆^2-

D. CrO₂Cl₂

Solution: (D)

In MnO₄^–,Mn is in a +7 oxidation state as the oxidation state of oxygen is -2.

In Cr(CN)₆^3- , Cr is in +3 oxidation state as the oxidation state of cyanide is -1.

In NiF₆^2-, Ni is in a +4 oxidation state as the oxidation state of fluorine is -1.

What will be the oxidation number of Cr in CrO₂Cl₂?

We apply the rules for assigning oxidation numbers: x − 4 − 2 = 0

Solving, x = +6

In CrO₂Cl₂, Cr is in +6 oxidation state as the oxidation state of oxygen is -2 and the oxidation state of chlorine is -1.

Question 10. Oxidation number of sulphur in S₂, S₂F₂ and H₂S are:

(A) +2, 0, +2,

(B) -2, 0, +2

(C) 0, +1, -2

(D) 0, +1, +2

Solution: (C)

Since S₈ contains only one type of atom, so the oxidation number is 0. Therefore its oxidation state is 0.

In, S₂F₂, oxydation number of Fluorine is -1. Thus,

2S + 2F = 0

2S + 2(-1) = 0

2S -2 = 0

2S = +2

S = +1

Hence, the oxidation state of Sulphur in S₂F₂ is +1.

In H₂S, the oxidation number of hydrogen is 1. Thus,

2(H) + S = 0

2(1) + S = 0

2 + S = 0

S = -2

Hence, the oxidation state of Sulphur in H₂S is -2.