JEE Chemistry Redox Reactions MCQs Set 05

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Chapter-wise Objective Questions: Redox Reactions

Access the complete set of multiple-choice questions for Redox Reactions below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: A sulphur-containing species which cannot act as a reducing agent is
(a) \( \ce{SO2} \)
(b) \( \ce{SO3^2-} \)
(c) \( \ce{H2SO4} \)
(d) \( \ce{S2O3^2-} \)
Show Answer & Explanation

Answer: (c) \( \ce{H2SO4} \)

Explanation:
1. Sulphur in H2SO4 is already at its maximum state, +6.
2. The others have S in a lower state and can be oxidised.

Teacher's Note:
1. The maximum oxidation state of S is +6.

Question 2: When \( \ce{H2} \) reacts with Na it acts as
(a) an oxidising agent
(b) a reducing agent
(c) both
(d) cannot be predicted
Show Answer & Explanation

Answer: (a) an oxidising agent

Explanation:
1. \( \ce{2Na + H2 -> 2NaH} \): H goes from 0 to -1.
2. Gain of electrons means H2 is the oxidant.

Teacher's Note:
1. With metals hydrogen forms hydrides (H-), with non-metals it is oxidised.

Question 3: A compound contains atoms A, B and C with oxidation numbers +2, +5 and -2 respectively. The possible formula is
(a) \( \ce{ABC2} \)
(b) \( \ce{B2(AC3)2} \)
(c) \( \ce{A3(BC4)2} \)
(d) \( \ce{A3(B4C)2} \)
Show Answer & Explanation

Answer: (c) \( \ce{A3(BC4)2} \)

Explanation:
1. The total must be zero.
2. A3(BC4)2: 3(2) + 2(5) + 8(-2) = 6 + 10 - 16 = 0.

Teacher's Note:
1. Test each formula by multiplying by the subscripts.

Question 4: The equivalent weight of \( \ce{N2} \) in the change \( \ce{N2 -> NH3} \) is
(a) \( \frac{28}{6} \)
(b) 28
(c) \( \frac{28}{2} \)
(d) \( \frac{28}{3} \)
Show Answer & Explanation

Answer: (a) \( \frac{28}{6} \)

Explanation:
1. N2 (0) forms 2 N(-3): 6 electrons gained per mole of N2.
2. E = 28/6.

Teacher's Note:
1. Equivalent weight = molar mass divided by electrons exchanged per mole.

Question 5: The equivalent weight of \( \ce{NH3} \) in the change \( \ce{N2 -> NH3} \) is
(a) \( \frac{17}{6} \)
(b) 17
(c) \( \frac{17}{2} \)
(d) \( \frac{17}{3} \)
Show Answer & Explanation

Answer: (d) \( \frac{17}{3} \)

Explanation:
1. Per mole of NH3 formed, one N goes from 0 to -3: 3 electrons.
2. E = 17/3.

Teacher's Note:
1. Count electrons per formula unit of the species whose equivalent weight is asked.

Question 6: In the reaction \( \ce{VO + Fe2O3 -> FeO + V2O5} \), the equivalent weight of \( \ce{V2O5} \) is
(a) M/8
(b) M/2
(c) M/6
(d) M/4
Show Answer & Explanation

Answer: (c) M/6

Explanation:
1. V changes from +2 to +5: 3 electrons per V atom.
2. Two V atoms in V2O5 means 6 electrons per mole, so E = M/6.

Teacher's Note:
1. Multiply the change per atom by the number of such atoms in the formula.

Question 7: The equivalent weight of iodine in \( \ce{I2 + 2S2O3^2- -> 2I^- + S4O6^2-} \) is
(a) its molecular weight
(b) molecular weight/2
(c) molecular weight/4
(d) none of these
Show Answer & Explanation

Answer: (b) molecular weight/2

Explanation:
1. I2 gains 2 electrons to give 2 I^-.
2. E = M/2.

Teacher's Note:
1. Halogen molecules have n-factor 2 when they form halide ions.

Question 8: In the reaction \( \ce{A^{n2+} + x e^- -> A^{n1+}} \), the value of x is
(a) n1 + n2
(b) n2 - n1
(c) n1 - n2
(d) n1 x n2
Show Answer & Explanation

Answer: (b) n2 - n1

Explanation:
1. The change in oxidation number equals the electrons gained: x = n2 - n1.

Teacher's Note:
1. The electrons gained equal the drop in oxidation number.

Question 9: What is the equivalent weight of the reductant in \( \ce{[Fe(CN)6]^3- + H2O2 + 2OH^- -> 2[Fe(CN)6]^4- + 2H2O + O2} \)? (Fe = 56, C = 12, N = 14, O = 16, H = 1)
(a) 17
(b) 212
(c) 34
(d) 32
Show Answer & Explanation

Answer: (a) 17

Explanation:
1. Fe is reduced (+3 to +2), so the reductant is H2O2.
2. H2O2 gives O2: 2 electrons per mole. E = 34/2 = 17.

Teacher's Note:
1. Identify first which species loses the electrons.

Question 10: The equivalent weight of \( \ce{FeC2O4} \) in the change \( \ce{FeC2O4 -> Fe^3+ + CO2} \) is
(a) M/3
(b) M/6
(c) M/2
(d) M/1
Show Answer & Explanation

Answer: (a) M/3

Explanation:
1. Fe: +2 to +3 gives 1 electron.
2. C2O4^2- (C: +3 to +4, two C atoms) gives 2 electrons.
3. Total 3, so E = M/3.

Teacher's Note:
1. Ferrous oxalate has two oxidisable parts.

Question 11: The number of moles of oxalate ions oxidised by one mole of \( \ce{MnO4^-} \) in acidic medium is
(a) \( \frac{1}{5} \)
(b) \( \frac{2}{5} \)
(c) \( \frac{5}{2} \)
(d) 5
Show Answer & Explanation

Answer: (c) \( \frac{5}{2} \)

Explanation:
1. MnO4^- takes 5 electrons; each oxalate gives 2.
2. n(oxalate) x 2 = 1 x 5, so n = 5/2.

Teacher's Note:
1. Use equivalents: moles x n-factor on both sides.

Question 12: The oxidation product of the substance \( \ce{Na3AsO3} \) is
(a) \( \ce{As2O3} \)
(b) \( \ce{AsO3^3-} \)
(c) \( \ce{AsO4^3-} \)
(d) \( \ce{AsO2^-} \)
Show Answer & Explanation

Answer: (c) \( \ce{AsO4^3-} \)

Explanation:
1. Arsenic is +3 in AsO3^3- and rises to +5 on oxidation.
2. AsO4^3- has As = +5.

Teacher's Note:
1. Arsenite is oxidised to arsenate.

Question 13: The equivalent weight of \( \ce{MnSO4} \) is half of its molecular weight when it is converted to
(a) \( \ce{Mn2O3} \)
(b) \( \ce{MnO2} \)
(c) \( \ce{MnO4^-} \)
(d) \( \ce{MnO4^2-} \)
Show Answer & Explanation

Answer: (b) \( \ce{MnO2} \)

Explanation:
1. E = M/2 means n = 2.
2. Mn: +2 to +4 in MnO2 gives 2 electrons.

Teacher's Note:
1. Mn2O3 gives n = 1, MnO4^2- gives n = 4, MnO4^- gives n = 5.

Question 14: In the change \( \ce{3Fe + 4H2O -> Fe3O4 + 4H2} \), the equivalent weight of iron (atomic weight 56) is
(a) 42
(b) 21
(c) 63
(d) 84
Show Answer & Explanation

Answer: (b) 21

Explanation:
1. Fe3O4 contains Fe(II) + 2 Fe(III): total oxidation = 8 electrons for 3 Fe.
2. n per Fe = 8/3, E = 56 x 3/8 = 21.

Teacher's Note:
1. Fractional n-factors are fine for average changes.

Question 15: In \( \ce{Cr2O7^2- + I^- + H^+ -> Cr^3+ + I2 + H2O} \), the equivalent weight of the reductant (I = 127) is
(a) 26
(b) 127
(c) 63.5
(d) 10.4
Show Answer & Explanation

Answer: (b) 127

Explanation:
1. The reductant is I^- (to I2), losing 1 electron per ion.
2. E = 127/1 = 127.

Teacher's Note:
1. The oxidant is dichromate, with E = M/6.

Question 16: How many moles of \( \ce{KMnO4} \) are reduced by one mole of KI in alkaline medium (KI changes to \( \ce{KIO3} \), KMnO4 to \( \ce{MnO2} \))?
(a) One
(b) Two
(c) Five
(d) One fifth
Show Answer & Explanation

Answer: (b) Two

Explanation:
1. I^- to IO3^-: 6 electrons lost. MnO4^- to MnO2: 3 electrons gained.
2. n x 3 = 1 x 6, so n = 2.

Teacher's Note:
1. In alkaline medium KMnO4 has n = 3.

Question 17: Which one of the following is NOT a redox reaction?
(a) \( \ce{CaCO3 -> CaO + CO2} \)
(b) \( \ce{2H2 + O2 -> 2H2O} \)
(c) \( \ce{Na + H2O -> NaOH + 1/2 H2} \)
(d) \( \ce{MnCl3 -> MnCl2 + 1/2 Cl2} \)
Show Answer & Explanation

Answer: (a) \( \ce{CaCO3 -> CaO + CO2} \)

Explanation:
1. In CaCO3 decomposition no oxidation number changes.
2. The other three involve electron transfer.

Teacher's Note:
1. Thermal decomposition is not necessarily redox; check oxidation numbers.

Question 18: In the reaction \( \ce{4Fe + 3O2 -> 4Fe^3+ + 6O^2-} \), which statement is incorrect?
(a) It is a redox reaction
(b) Metallic iron is a reducing agent
(c) \( \ce{Fe^3+} \) is an oxidising agent
(d) Metallic iron is reduced to \( \ce{Fe^3+} \)
Show Answer & Explanation

Answer: (d) Metallic iron is reduced to \( \ce{Fe^3+} \)

Explanation:
1. Fe is oxidised (0 to +3), not reduced.
2. Fe is the reducing agent.

Teacher's Note:
1. Oxidation means increase in oxidation number.

Question 19: In the reaction \( \ce{Cl2 + OH^- -> Cl^- + ClO3^- + H2O} \), chlorine is
(a) oxidised
(b) reduced
(c) oxidised as well as reduced
(d) neither oxidised nor reduced
Show Answer & Explanation

Answer: (c) oxidised as well as reduced

Explanation:
1. Cl: 0 to -1 (reduction) and 0 to +5 (oxidation).
2. This is a disproportionation.

Teacher's Note:
1. Same element oxidised and reduced equals disproportionation.

Question 20: Which of the following does NOT represent disproportionation?
(a) \( \ce{MnO2 + 4HCl -> MnCl2 + Cl2 + 2H2O} \)
(b) \( \ce{2H2O2 -> 2H2O + O2} \)
(c) \( \ce{4KClO3 -> 3KClO4 + KCl} \)
(d) \( \ce{3Cl2 + 6NaOH -> 5NaCl + NaClO3 + 3H2O} \)
Show Answer & Explanation

Answer: (a) \( \ce{MnO2 + 4HCl -> MnCl2 + Cl2 + 2H2O} \)

Explanation:
1. In MnO2 + HCl, Mn is reduced and Cl is oxidised, but they are in different species.
2. The other three have a single species both oxidised and reduced.

Teacher's Note:
1. Disproportionation involves one element in one species going both ways.

Question 21: Which of the following changes represents disproportionation?
(a) \( \ce{Cl2 + 2OH^- -> ClO^- + Cl^- + H2O} \)
(b) \( \ce{Cu2O + 2H^+ -> Cu + Cu^2+ + H2O} \)
(c) \( \ce{2HCuCl2 -> Cu + Cu^2+ + 4Cl^- + 2H^+} \) (on dilution with water)
(d) All of the above
Show Answer & Explanation

Answer: (d) All of the above

Explanation:
1. (a) Cl2: 0 to +1 and -1. (b) Cu(+1) to Cu(0) and Cu(+2). (c) Cu(+1) again to Cu(0) and Cu(+2).
2. All three are disproportionation reactions.

Teacher's Note:
1. Look for the same element in the middle state going to both a higher and a lower one.

Question 22: One mole of iron (55.8 g) when oxidised to the +2 state gives up
(a) \( 1N_A \) electron
(b) \( 2N_A \) electrons
(c) \( 3N_A \) electrons
(d) 0.5 mole of electrons
Show Answer & Explanation

Answer: (b) \( 2N_A \) electrons

Explanation:
1. One mole of Fe loses 2 moles of electrons.
2. 2 moles of electrons = 2 N_A electrons.

Teacher's Note:
1. Moles of electrons times Avogadro number gives number of electrons.

Question 23: In \( \ce{MnO4^- + SO3^2- + H^+ -> SO4^2- + Mn^2+ + H2O} \), which statement is correct?
(a) MnO4^- and H^+ are both reduced
(b) MnO4^- is reduced and H^+ is oxidised
(c) MnO4^- is reduced and SO3^2- is oxidised
(d) MnO4^- is oxidised and SO3^2- is reduced
Show Answer & Explanation

Answer: (c) MnO4^- is reduced and SO3^2- is oxidised

Explanation:
1. Mn: +7 to +2 (reduction). S: +4 to +6 (oxidation).
2. H^+ only supplies protons.

Teacher's Note:
1. Always check which atoms change oxidation number.

Question 24: In the reaction \( \ce{I2 + KI -> KI3} \),
(a) only oxidation takes place
(b) only reduction takes place
(c) both oxidation and reduction take place
(d) neither oxidation nor reduction takes place
Show Answer & Explanation

Answer: (d) neither oxidation nor reduction takes place

Explanation:
1. I3^- is formed by the donor-acceptor combination I2 + I^-.
2. The average state of iodine is -1/3 but no electron transfer occurs.

Teacher's Note:
1. This is complex formation, not redox.

Question 25: Which of the following reactions shows the oxidising behaviour of \( \ce{H2SO4} \)?
(a) \( \ce{2PCl5 + H2SO4 -> 2POCl3 + 2HCl + SO2Cl2} \)
(b) \( \ce{2NaOH + H2SO4 -> Na2SO4 + 2H2O} \)
(c) \( \ce{NaCl + H2SO4 -> NaHSO4 + HCl} \)
(d) \( \ce{2HI + H2SO4 -> I2 + SO2 + 2H2O} \)
Show Answer & Explanation

Answer: (d) \( \ce{2HI + H2SO4 -> I2 + SO2 + 2H2O} \)

Explanation:
1. Only in the last reaction S is reduced (+6 to +4) while I^- is oxidised.
2. The others show acidic or dehydrating behaviour.

Teacher's Note:
1. SO2, S or H2S as products indicate that H2SO4 acted as an oxidant.

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Multiple Choice Questions (MCQs) for JEE Chemistry Redox Reactions

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