JEE Chemistry Redox Reactions MCQs Set 06

Chemistry Objective Questions and Answers: 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.

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Multiple Choice Questions

Question 1: Which of the following is a disproportionation reaction?
(a) \( \ce{BaCl2 + H2SO4 -> BaSO4 + 2HCl} \)
(b) \( \ce{NH4NO3 -> N2O + 2H2O} \)
(c) \( \ce{4H3PO3 -> PH3 + 3H3PO4} \)
(d) \( \ce{AgCl + 2NH3 -> Ag(NH3)2Cl} \)
Show Answer & Explanation

Answer: (c) \( \ce{4H3PO3 -> PH3 + 3H3PO4} \)

Explanation:
1. P(+3) goes to P(-3) in PH3 and to P(+5) in H3PO4.
2. NH4NO3 is comproportionation (N -3 and +5 to +1).

Teacher's Note:
1. Check whether one oxidation state splits into a higher and a lower one.

Question 2: Which of the following reactions involves neither oxidation nor reduction?
(a) \( \ce{CrO4^2- -> Cr2O7^2-} \)
(b) \( \ce{Cr -> CrCl3} \)
(c) \( \ce{Na -> Na^+} \)
(d) \( \ce{2S2O3^2- -> S4O6^2-} \)
Show Answer & Explanation

Answer: (a) \( \ce{CrO4^2- -> Cr2O7^2-} \)

Explanation:
1. Cr stays +6 in both chromate and dichromate.
2. The others show a change in oxidation state.

Teacher's Note:
1. Chromate and dichromate interconvert with pH changes.

Question 3: In \( \ce{Zn + H2SO4 -> ZnSO4 + H2} \), zinc undergoes
(a) reduction
(b) oxidation
(c) both oxidation and reduction
(d) neither
Show Answer & Explanation

Answer: (b) oxidation

Explanation:
1. Zn: 0 to +2.
2. Zn loses electrons, so it is oxidised.

Teacher's Note:
1. Metals that displace hydrogen from acids are oxidised.

Question 4: The set of coefficients that balances \( \ce{x Cr2O7^2- + y H^+ + z e^- -> a Cr^3+ + b H2O} \) is
(a) x=2, y=14, z=6, a=2, b=7
(b) x=1, y=14, z=6, a=2, b=7
(c) x=2, y=7, z=6, a=2, b=7
(d) x=2, y=7, z=6, a=1, b=7
Show Answer & Explanation

Answer: (b) x=1, y=14, z=6, a=2, b=7

Explanation:
1. \( \ce{Cr2O7^2- + 14H^+ + 6e^- -> 2Cr^3+ + 7H2O} \)
2. Check charge: -2 + 14 - 6 = +6 and 2 x 3 = +6.

Teacher's Note:
1. Dichromate takes 6 electrons for two Cr atoms.

Question 5: In \( \ce{MnO4^- + xH^+ + ne^- -> Mn^2+ + yH2O} \), the value of x is
(a) 5
(b) 8
(c) 6
(d) 3
Show Answer & Explanation

Answer: (b) 8

Explanation:
1. Mn: +7 to +2 so n = 5. Oxygen balance: y = 4.
2. Hydrogen balance: x = 8.

Teacher's Note:
1. Reaction: MnO4^- + 8H^+ + 5e^- gives Mn^2+ + 4H2O.

Question 6: In \( \ce{2KMnO4 + 5H2S + 6H^+ -> 2Mn^2+ + 2K^+ + 5S + 8H2O} \), how many electrons are involved in the oxidation of one mole of the reductant?
(a) Two
(b) Five
(c) Ten
(d) One
Show Answer & Explanation

Answer: (a) Two

Explanation:
1. H2S goes to S: S(-2) to S(0), 2 electrons.
2. Reductant is H2S.

Teacher's Note:
1. Count the electrons per mole of the reductant only.

Question 7: The value of n in \( \ce{Cr(OH)4^- + OH^- -> CrO4^2- + H2O + ne^-} \) is
(a) 3
(b) 6
(c) 5
(d) 2
Show Answer & Explanation

Answer: (a) 3

Explanation:
1. Cr: +3 to +6.
2. n = 3.

Teacher's Note:
1. Electrons lost equals increase in oxidation number.

Question 8: For the redox reaction \( \ce{Zn + NO3^- -> Zn^2+ + NH4^+} \) in basic medium, the coefficients of Zn, \( \ce{NO3^-} \) and \( \ce{OH^-} \) in the balanced equation are respectively
(a) 4, 1, 7
(b) 7, 4, 1
(c) 4, 1, 10
(d) 1, 4, 10
Show Answer & Explanation

Answer: (c) 4, 1, 10

Explanation:
1. Zn gives 2e; N: +5 to -3 takes 8e, so 4 Zn per NO3^-.
2. \( \ce{4Zn + NO3^- + 7H2O -> 4Zn^2+ + NH4^+ + 10OH^-} \): check charge -1 on left, 8 + 1 - 10 = -1 on right.

Teacher's Note:
1. Balance electrons first, then O and H using water and OH^- in basic medium.

Question 9: In \( \ce{H2O + SO3^2- -> SO4^2- + 2H^+ + X} \), the value of X is
(a) 4e
(b) 3e
(c) 2e
(d) 1e
Show Answer & Explanation

Answer: (c) 2e

Explanation:
1. S: +4 to +6, so 2 electrons are released.
2. Charge check: -2 = -2 + 2 - 2.

Teacher's Note:
1. Electrons appear on the product side in an oxidation half-reaction.

Question 10: The number of electrons required to balance \( \ce{NO3^- + 4H^+ -> 2H2O + NO} \) is
(a) 2 on the right side
(b) 3 on the left side
(c) 3 on the right side
(d) 5 on the left side
Show Answer & Explanation

Answer: (b) 3 on the left side

Explanation:
1. N: +5 to +2, so 3 electrons are gained and appear on the reactant (left) side.
2. Charge: -1 + 4 - 3 = 0.

Teacher's Note:
1. A reduction half-reaction has electrons on the left.

Question 11: The oxidation numbers of P in \( \ce{PO4^3-} \), S in \( \ce{SO4^2-} \) and Cr in \( \ce{Cr2O7^2-} \) are respectively
(a) -3, +6, +6
(b) +5, +6, +6
(c) +3, +6, +5
(d) +5, +3, +6
Show Answer & Explanation

Answer: (b) +5, +6, +6

Explanation:
1. P: x - 8 = -3, x = +5.
2. S: x - 8 = -2, x = +6.
3. Cr: 2x - 14 = -2, x = +6.

Teacher's Note:
1. Each of these ions shows its element at its highest state.

Question 12: The correct order of oxidising strength is
(a) \( \ce{MnO4^-} \) > \( \ce{VO2^+} \) > \( \ce{Cr2O7^2-} \)
(b) \( \ce{Cr2O7^2-} \) > \( \ce{MnO4^-} \) > \( \ce{VO2^+} \)
(c) \( \ce{Cr2O7^2-} \) > \( \ce{VO2^+} \) > \( \ce{MnO4^-} \)
(d) \( \ce{MnO4^-} \) > \( \ce{Cr2O7^2-} \) > \( \ce{VO2^+} \)
Show Answer & Explanation

Answer: (d) \( \ce{MnO4^-} \) > \( \ce{Cr2O7^2-} \) > \( \ce{VO2^+} \)

Explanation:
1. Standard reduction potentials: MnO4^-/Mn^2+ = +1.51 V, Cr2O7^2-/Cr^3+ = +1.33 V, VO2^+/VO^2+ = +1.00 V.
2. Higher potential means stronger oxidant.

Teacher's Note:
1. Permanganate is stronger than dichromate in acid.

Question 13: The amount of \( \ce{CuSO4.5H2O} \) required to liberate 2.54 g of iodine when titrated with KI is (Cu = 63.5)
(a) 2.5 g
(b) 4.99 g
(c) 2.4 g
(d) 1.2 g
Show Answer & Explanation

Answer: (b) 4.99 g

Explanation:
1. \( \ce{2Cu^2+ + 4I^- -> 2CuI + I2} \): 2 mol Cu^2+ per mole I2.
2. n(I2) = 2.54/254 = 0.01 mol, so n(Cu) = 0.02 mol. Mass = 0.02 x 249.5 = 4.99 g.

Teacher's Note:
1. Cu(II) is reduced by iodide to Cu(I).

Question 14: In which of the following compounds does nitrogen exhibit the highest oxidation state?
(a) \( \ce{N3H} \)
(b) \( \ce{NH2OH} \)
(c) \( \ce{N2H4} \)
(d) \( \ce{NH3} \)
Show Answer & Explanation

Answer: (a) \( \ce{N3H} \)

Explanation:
1. N3H: -1/3, NH2OH: -1, N2H4: -2, NH3: -3.
2. The highest is -1/3.

Teacher's Note:
1. All are negative; choose the one nearest to zero.

Question 15: \( \ce{K2Cr2O7} \) reacts with hydrazine to form a product. The oxidation state of chromium in the product is
(a) +4
(b) +3
(c) +5
(d) +2
Show Answer & Explanation

Answer: (b) +3

Explanation:
1. Dichromate is reduced to Cr^3+ by hydrazine (which is oxidised to N2).

Teacher's Note:
1. Hydrazine is a strong reductant.

Question 16: The roles of \( \ce{H2O2} \) in (a) \( \ce{H2O2 + O3 -> H2O + 2O2} \) and (b) \( \ce{H2O2 + Ag2O -> 2Ag + H2O + O2} \) are respectively
(a) oxidising in (a) and reducing in (b)
(b) reducing in (a) and oxidising in (b)
(c) reducing in (a) and (b)
(d) oxidising in (a) and (b)
Show Answer & Explanation

Answer: (c) reducing in (a) and (b)

Explanation:
1. In both reactions H2O2 is converted to O2 (oxygen -1 to 0).
2. So it acts as a reductant in both.

Teacher's Note:
1. O3 and Ag2O are the oxidants.

Question 17: The reaction of aqueous \( \ce{KMnO4} \) with \( \ce{H2O2} \) in acidic conditions gives
(a) \( \ce{Mn^4+} \) and \( \ce{O2} \)
(b) \( \ce{Mn^2+} \) and \( \ce{O2} \)
(c) \( \ce{Mn^2+} \) and \( \ce{O3} \)
(d) \( \ce{Mn^4+} \) and \( \ce{MnO2} \)
Show Answer & Explanation

Answer: (b) \( \ce{Mn^2+} \) and \( \ce{O2} \)

Explanation:
1. \( \ce{2MnO4^- + 5H2O2 + 6H^+ -> 2Mn^2+ + 5O2 + 8H2O} \)

Teacher's Note:
1. Acidic permanganate is always reduced to Mn^2+.

Question 18: In weak alkaline medium \( \ce{KMnO4} \) is converted to
(a) \( \ce{MnO2} \)
(b) \( \ce{K2MnO4} \)
(c) \( \ce{Mn(OH)3} \)
(d) \( \ce{MnO} \)
Show Answer & Explanation

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

Explanation:
1. Neutral or weakly alkaline medium: MnO4^- + 2H2O + 3e^- gives MnO2 + 4OH^-.

Teacher's Note:
1. Strongly alkaline medium gives manganate (K2MnO4) instead.

Question 19: When \( \ce{KMnO4} \) reacts with \( \ce{H2O2} \) in acidic medium, which of the following is formed?
(a) \( \ce{Mn^2+} \)
(b) \( \ce{Mn^7+} \)
(c) \( \ce{MnO2} \)
(d) \( \ce{Mn2O3} \)
Show Answer & Explanation

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

Explanation:
1. In acid, MnO4^- is reduced to colourless Mn^2+.

Teacher's Note:
1. The solution is decolourised.

Question 20: The lowest oxidation state of phosphorus is in
(a) \( \ce{H3PO2} \)
(b) \( \ce{H3PO4} \)
(c) \( \ce{H4P2O7} \)
(d) \( \ce{H3PO3} \)
Show Answer & Explanation

Answer: (a) \( \ce{H3PO2} \)

Explanation:
1. H3PO2: +1. H3PO3: +3. H3PO4 and H4P2O7: +5.

Teacher's Note:
1. Count the P-H bonds: more P-H means a lower oxidation state.

Question 21: For the redox reaction \( \ce{MnO4^- + C2O4^2- + H^+ -> Mn^2+ + CO2 + H2O} \), the correct coefficients of the reactants (\( \ce{MnO4^-} \), \( \ce{C2O4^2-} \), \( \ce{H^+} \)) are
(a) 16, 5, 2
(b) 2, 5, 16
(c) 2, 16, 5
(d) 5, 16, 2
Show Answer & Explanation

Answer: (b) 2, 5, 16

Explanation:
1. \( \ce{2MnO4^- + 5C2O4^2- + 16H^+ -> 2Mn^2+ + 10CO2 + 8H2O} \)

Teacher's Note:
1. Electrons: 2 x 5 = 10 gained, 5 x 2 = 10 lost.

Question 22: 2 g of \( \ce{FeSO4} \) is completely oxidised by acidic 0.05 M \( \ce{KMnO4} \). The volume of \( \ce{KMnO4} \) required is (Fe = 56, S = 32, O = 16)
(a) 0.10 L
(b) 0.05 L
(c) 0.30 L
(d) 0.40 L
Show Answer & Explanation

Answer: (b) 0.05 L

Explanation:
1. n(FeSO4) = 2/152 = 0.01316 mol.
2. n(KMnO4) = 0.01316/5 = 0.002632 mol, V = 0.002632/0.05 = 0.0526 L, nearly 0.05 L.

Teacher's Note:
1. 5 moles of Fe^2+ react with 1 mole of MnO4^-.

Question 23: In the reaction \( \ce{CH3OH -> HCOOH} \), the number of electrons that must be added to the right is
(a) 4
(b) 3
(c) 2
(d) 1
Show Answer & Explanation

Answer: (a) 4

Explanation:
1. C: -2 in CH3OH and +2 in HCOOH.
2. 4 electrons are lost, so 4 electrons appear on the right.

Teacher's Note:
1. In an oxidation half-reaction electrons appear on the product side.

Question 24: The oxidation number of Xe in \( \ce{XeF5^-} \) is
(a) +1
(b) +2
(c) +3
(d) +4
Show Answer & Explanation

Answer: (d) +4

Explanation:
1. x - 5 = -1.
2. x = +4.

Teacher's Note:
1. The ion has a pentagonal planar shape.

Question 25: Bromine in \( \ce{H2O + Br2 -> HOBr + HBr} \) behaves as
(a) both oxidised and reduced
(b) oxidised only
(c) reduced only
(d) proton acceptor only
Show Answer & Explanation

Answer: (a) both oxidised and reduced

Explanation:
1. Br: 0 to +1 in HOBr and 0 to -1 in HBr.
2. Disproportionation of bromine.

Teacher's Note:
1. Same kind as chlorine in water.

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Multiple Choice Questions (MCQs) for JEE Chemistry Redox Reactions

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