JEE Chemistry Redox Reactions MCQs Set 13

Download JEE MCQs for JEE Chemistry: Redox Reactions

Access targeted multiple-choice questions for Redox Reactions designed to align with the latest JEE academic syllabus for JEE Chemistry. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Chapter-wise Objective Questions: Redox Reactions

Navigate directly to the 50 objective questions for Redox Reactions using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.

Multiple Choice Questions

Question 1: The oxidation number of nitrogen in \( \ce{N2H5+} \) is
(a) -3
(b) -2
(c) -1
(d) +2
Show Answer & Explanation

Answer: (b) -2

Explanation:
1. Let N = x: 2x + 5(+1) = +1.
2. 2x = -4, so x = -2.

Teacher's Note:
1. Assign oxidation numbers using the rules: O is -2 (except in peroxides and superoxides), H is +1 (except in metal hydrides), and the sum equals the overall charge.
2. Write the sum of oxidation numbers as an equation and solve for the unknown atom.

Question 2: In which of the following compounds is the oxidation number of carbon maximum?
(a) \( \ce{HCHO} \)
(b) \( \ce{CHCl3} \)
(c) \( \ce{CH3OH} \)
(d) \( \ce{C12H22O11} \)
Show Answer & Explanation

Answer: (b) \( \ce{CHCl3} \)

Explanation:
1. HCHO: 0; CHCl3: +2; CH3OH: -2; sucrose: 0 (average).
2. Carbon is highest (+2) in CHCl3.

Teacher's Note:
1. Assign oxidation numbers using the rules: O is -2 (except in peroxides and superoxides), H is +1 (except in metal hydrides), and the sum equals the overall charge.
2. Write the sum of oxidation numbers as an equation and solve for the unknown atom.

Question 3: The oxidation state of iodine in \( \ce{H4IO6-} \) is
(a) +7
(b) +5
(c) +1
(d) -1
Show Answer & Explanation

Answer: (a) +7

Explanation:
1. 4(+1) + x + 6(-2) = -1.
2. x = +7.

Teacher's Note:
1. Assign oxidation numbers using the rules: O is -2 (except in peroxides and superoxides), H is +1 (except in metal hydrides), and the sum equals the overall charge.
2. Write the sum of oxidation numbers as an equation and solve for the unknown atom.

Question 4: In which one of the following changes are five electrons transferred?
(a) \( \ce{MnO4- -> Mn^{2+}} \)
(b) \( \ce{CrO4^{2-} -> Cr^{3+}} \)
(c) \( \ce{MnO4^{2-} -> MnO2} \)
(d) \( \ce{Cr2O7^{2-} -> 2Cr^{3+}} \)
Show Answer & Explanation

Answer: (a) \( \ce{MnO4- -> Mn^{2+}} \)

Explanation:
1. Mn: +7 to +2 gains 5 electrons.
2. The others involve 3 (Cr +6 to +3), 2 (Mn +6 to +4) and 6 electrons respectively.

Teacher's Note:
1. Number of electrons transferred = change in oxidation number multiplied by the number of atoms changing.
2. Equivalent weight of an oxidising or reducing agent = molar mass / electrons gained or lost per formula unit.

Question 5: In which of the following compounds does iron have the lowest oxidation state?
(a) \( \ce{FeSO4.(NH4)2SO4.6H2O} \)
(b) \( \ce{K4Fe(CN)6} \)
(c) \( \ce{Fe(CO)5} \)
(d) \( \ce{Fe2O3} \)
Show Answer & Explanation

Answer: (c) \( \ce{Fe(CO)5} \)

Explanation:
1. The two double/complex salts in (a) and (b) have Fe(II); (d) has Fe(III).
2. In Fe(CO)5 the carbonyl ligands are neutral, so Fe is 0, the lowest.

Teacher's Note:
1. Assign oxidation numbers using the rules: O is -2 (except in peroxides and superoxides), H is +1 (except in metal hydrides), and the sum equals the overall charge.
2. Write the sum of oxidation numbers as an equation and solve for the unknown atom.

Question 6: When SO2 is passed through an acidic solution of potassium dichromate, chromium sulphate is formed. The change in valency of chromium is
(a) +4 to +2
(b) +5 to +3
(c) +6 to +3
(d) +7 to +2
Show Answer & Explanation

Answer: (c) +6 to +3

Explanation:
1. Dichromate has Cr in the +6 state; Cr\( _2(SO_4)_3 \) has Cr in the +3 state.
2. SO2 reduces Cr from +6 to +3.

Teacher's Note:
1. Assign oxidation numbers using the rules: O is -2 (except in peroxides and superoxides), H is +1 (except in metal hydrides), and the sum equals the overall charge.
2. Write the sum of oxidation numbers as an equation and solve for the unknown atom.

Question 7: BaO2 reacts with dilute H2SO4 to give BaSO4 and H2O2. The oxidation states of the most electronegative element (oxygen) in the products are
(a) 0 and -2
(b) -1 and -2
(c) -2 and 0
(d) -2 and +1
Show Answer & Explanation

Answer: (b) -1 and -2

Explanation:
1. In BaSO4 oxygen is -2.
2. In H2O2 (a peroxide) oxygen is -1.

Teacher's Note:
1. Assign oxidation numbers using the rules: O is -2 (except in peroxides and superoxides), H is +1 (except in metal hydrides), and the sum equals the overall charge.
2. Write the sum of oxidation numbers as an equation and solve for the unknown atom.

Question 8: The oxidation number of sulphur in \( \ce{Na2S4O6} \) is
(a) \( \frac23 \)
(b) \( \frac32 \)
(c) \( \frac35 \)
(d) \( \frac52 \)
Show Answer & Explanation

Answer: (d) \( \frac52 \)

Explanation:
1. 2(+1) + 4x + 6(-2) = 0.
2. 4x = 10, so x = +5/2 (average oxidation state).

Teacher's Note:
1. Assign oxidation numbers using the rules: O is -2 (except in peroxides and superoxides), H is +1 (except in metal hydrides), and the sum equals the overall charge.
2. Write the sum of oxidation numbers as an equation and solve for the unknown atom.

Question 9: The oxidation state of iron in \( \ce{Fe3O4} \) is
(a) \( \frac32 \)
(b) \( \frac45 \)
(c) \( \frac54 \)
(d) \( \frac83 \)
Show Answer & Explanation

Answer: (d) \( \frac83 \)

Explanation:
1. 3x + 4(-2) = 0.
2. x = +8/3 (average; two Fe(III) and one Fe(II)).

Teacher's Note:
1. Assign oxidation numbers using the rules: O is -2 (except in peroxides and superoxides), H is +1 (except in metal hydrides), and the sum equals the overall charge.
2. Write the sum of oxidation numbers as an equation and solve for the unknown atom.

Question 10: The oxidation number of sulphur in \( \ce{S2Cl2} \) is
(a) +1
(b) -1
(c) +6
(d) 0
Show Answer & Explanation

Answer: (a) +1

Explanation:
1. 2x + 2(-1) = 0, so x = +1.

Teacher's Note:
1. Assign oxidation numbers using the rules: O is -2 (except in peroxides and superoxides), H is +1 (except in metal hydrides), and the sum equals the overall charge.
2. Write the sum of oxidation numbers as an equation and solve for the unknown atom.

Question 11: When CuSO4 reacts with KI to give CuI and iodine, the oxidation number of copper changes by
(a) 0
(b) -1
(c) +1
(d) -2
Show Answer & Explanation

Answer: (b) -1

Explanation:
1. Cu goes from +2 in CuSO4 to +1 in CuI.
2. The change is -1 (reduction).

Teacher's Note:
1. Number of electrons transferred = change in oxidation number multiplied by the number of atoms changing.
2. Equivalent weight of an oxidising or reducing agent = molar mass / electrons gained or lost per formula unit.

Question 12: The oxidation number of manganese in K2MnO4 and MnSO4 respectively are
(a) +7, +2
(b) +6, +2
(c) +5, +2
(d) +2, +6
Show Answer & Explanation

Answer: (b) +6, +2

Explanation:
1. K2MnO4: 2(+1) + x + 4(-2) = 0, so x = +6.
2. MnSO4: x + (+6) + (-8) = 0, so x = +2.

Teacher's Note:
1. Assign oxidation numbers using the rules: O is -2 (except in peroxides and superoxides), H is +1 (except in metal hydrides), and the sum equals the overall charge.
2. Write the sum of oxidation numbers as an equation and solve for the unknown atom.

Question 13: The oxidation number of fluorine in \( \ce{F2O} \) is
(a) -1
(b) +1
(c) +2
(d) -2
Show Answer & Explanation

Answer: (a) -1

Explanation:
1. Fluorine is the most electronegative element and is always -1 in compounds.
2. In OF2 oxygen is therefore +2.

Teacher's Note:
1. Assign oxidation numbers using the rules: O is -2 (except in peroxides and superoxides), H is +1 (except in metal hydrides), and the sum equals the overall charge.
2. Write the sum of oxidation numbers as an equation and solve for the unknown atom.

Question 14: The oxidation number of nickel in \( \ce{K4[Ni(CN)4]} \) is
(a) -2
(b) -1
(c) 0
(d) +2
Show Answer & Explanation

Answer: (c) 0

Explanation:
1. 4(+1) + x + 4(-1) = 0.
2. x = 0.

Teacher's Note:
1. Assign oxidation numbers using the rules: O is -2 (except in peroxides and superoxides), H is +1 (except in metal hydrides), and the sum equals the overall charge.
2. Write the sum of oxidation numbers as an equation and solve for the unknown atom.

Question 15: The oxidation state shown by silicon when it combines with strongly electropositive metals is
(a) -2
(b) -4
(c) +4
(d) +2
Show Answer & Explanation

Answer: (b) -4

Explanation:
1. With metals silicon is more electronegative and forms silicides, e.g. \( \ce{Mg2Si} \).
2. Its oxidation state is -4.

Teacher's Note:
1. Assign oxidation numbers using the rules: O is -2 (except in peroxides and superoxides), H is +1 (except in metal hydrides), and the sum equals the overall charge.
2. Write the sum of oxidation numbers as an equation and solve for the unknown atom.

Question 16: The oxidation state of chromium in the final product formed by the reaction between KI and acidified potassium dichromate solution is
(a) +4
(b) +6
(c) +2
(d) +3
Show Answer & Explanation

Answer: (d) +3

Explanation:
1. Dichromate oxidises iodide to iodine in acid medium.
2. Cr(VI) is reduced to Cr\( ^{3+} \), i.e. +3.

Teacher's Note:
1. Assign oxidation numbers using the rules: O is -2 (except in peroxides and superoxides), H is +1 (except in metal hydrides), and the sum equals the overall charge.
2. Write the sum of oxidation numbers as an equation and solve for the unknown atom.

Question 17: The value of x in the partial redox equation \( \ce{MnO4- + 8H+ + x e- <=> Mn^{2+} + 4H2O} \) is
(a) 5
(b) 3
(c) 1
(d) 0
Show Answer & Explanation

Answer: (a) 5

Explanation:
1. Mn goes from +7 to +2.
2. So 5 electrons are gained, x = 5.

Teacher's Note:
1. Number of electrons transferred = change in oxidation number multiplied by the number of atoms changing.
2. Equivalent weight of an oxidising or reducing agent = molar mass / electrons gained or lost per formula unit.

Question 18: In the equation \( \ce{C2H6(g) + nO2 -> CO2(g) + H2O(l)} \) (balanced with whole numbers), the ratio of the coefficients of CO2 and H2O is
(a) 1 : 1
(b) 2 : 3
(c) 3 : 2
(d) 1 : 3
Show Answer & Explanation

Answer: (b) 2 : 3

Explanation:
1. Balanced: \( \ce{2C2H6 + 7O2 -> 4CO2 + 6H2O} \).
2. CO2 : H2O = 4 : 6 = 2 : 3.

Teacher's Note:
1. Number of electrons transferred = change in oxidation number multiplied by the number of atoms changing.
2. Equivalent weight of an oxidising or reducing agent = molar mass / electrons gained or lost per formula unit.

Question 19: For decolourisation of 1 mole of KMnO4 (acidic medium), the moles of H2O2 required is
(a) 1/2
(b) 3/2
(c) 5/2
(d) 7/2
Show Answer & Explanation

Answer: (c) 5/2

Explanation:
1. \( \ce{2MnO4- + 5H2O2 + 6H+ -> 2Mn^{2+} + 5O2 + 8H2O} \).
2. 2 mol KMnO4 need 5 mol H2O2, so 1 mol needs 5/2 mol.

Teacher's Note:
1. Number of electrons transferred = change in oxidation number multiplied by the number of atoms changing.
2. Equivalent weight of an oxidising or reducing agent = molar mass / electrons gained or lost per formula unit.

Question 20: In the reaction \( \ce{2Cr(OH)3 + 4OH- + KIO3 -> 2CrO4^{2-} + 5H2O + KI} \), the equivalent weight of KIO3 is
(a) mol. wt.
(b) mol. wt. / 6
(c) mol. wt. / 2
(d) mol. wt. / 3
Show Answer & Explanation

Answer: (b) mol. wt. / 6

Explanation:
1. Iodine goes from +5 in KIO3 to -1 in KI, a gain of 6 electrons.
2. Equivalent weight = mol. wt. / 6.

Teacher's Note:
1. Number of electrons transferred = change in oxidation number multiplied by the number of atoms changing.
2. Equivalent weight of an oxidising or reducing agent = molar mass / electrons gained or lost per formula unit.

Question 21: In alkaline medium, ClO2 oxidises H2O2 to O2 and is itself reduced to Cl\( ^- \). How many moles of H2O2 will be oxidised by one mole of ClO2?
(a) 1.0
(b) 1.5
(c) 2.5
(d) 3.5
Show Answer & Explanation

Answer: (c) 2.5

Explanation:
1. ClO2 to Cl\( ^- \): +4 to -1, gains 5 electrons.
2. H2O2 to O2 loses 2 electrons per mole, so moles of H2O2 = 5/2 = 2.5.

Teacher's Note:
1. Number of electrons transferred = change in oxidation number multiplied by the number of atoms changing.
2. Equivalent weight of an oxidising or reducing agent = molar mass / electrons gained or lost per formula unit.

Question 22: One mole of N2H4 loses 10 mol of electrons to form a new compound Y. Assuming all nitrogen appears in Y and the oxidation state of hydrogen is unchanged, the oxidation state of nitrogen in Y is
(a) +3
(b) -3
(c) -1
(d) +5
Show Answer & Explanation

Answer: (a) +3

Explanation:
1. In N2H4 nitrogen is -2, so the two N atoms total -4.
2. After losing 10 electrons the total is +6, so each N is +3.

Teacher's Note:
1. Assign oxidation numbers using the rules: O is -2 (except in peroxides and superoxides), H is +1 (except in metal hydrides), and the sum equals the overall charge.
2. Write the sum of oxidation numbers as an equation and solve for the unknown atom.

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Redox Reactions Objective Questions & Solutions for JEE Chemistry

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