JEE Chemistry Redox Reactions MCQs Set 18

Multiple Choice Questions (MCQs) for JEE Chemistry: Redox Reactions

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Multiple Choice Questions

Question 1: The oxidation numbers of C in CH4, CH3Cl, CH2Cl2, CHCl3 and CCl4 are respectively:
(a) +4, +2, 0, −2, −4
(b) +2, +4, 0, −4, −2
(c) −4, −2, 0, +2, +4
(d) −2, −4, 0, +4, +2
Show Answer & Explanation

Answer: (c) −4, −2, 0, +2, +4

Explanation:
1. CH4: x + 4(+1) = 0, so x = −4.
2. CH3Cl: x + 3(+1) + (−1) = 0, so x = −2.
3. CH2Cl2: x + 2 − 2 = 0, so x = 0.
4. CHCl3: x + 1 − 3 = 0, so x = +2.
5. CCl4: x − 4 = 0, so x = +4.
6. Answer: (c)

Teacher's Note:
1. Each Cl replaces an H and raises the oxidation number of C by 2.
2. Cl is more electronegative than C, H is less.

Question 2: The oxidation number of Fe in Fe0.94O is:
(a) 200
(b) 200/94
(c) 94/200
(d) None of these
Show Answer & Explanation

Answer: (b) 200/94

Explanation:
1. Oxygen is −2, so 0.94 × (Fe) = +2.
2. Fe = 2/0.94 = 200/94 (a mixture of Fe2+ and Fe3+ ions).
3. Answer: (b)

Teacher's Note:
1. Non-stoichiometric compounds give fractional averages.
2. Total positive charge must equal total negative charge.

Question 3: Two different oxidation states of chlorine are present in the compound:
(a) CaOCl2
(b) KCl
(c) KClO3
(d) Cl2O7
Show Answer & Explanation

Answer: (a) CaOCl2

Explanation:
1. CaOCl2 (bleaching powder) is Ca(OCl)Cl, containing OCl− (Cl = +1) and Cl− (Cl = −1).
2. KCl (−1), KClO3 (+5) and Cl2O7 (+7) each have a single oxidation state.
3. Answer: (a)

Teacher's Note:
1. Bleaching powder is a mixed salt.
2. Mixed salts can show two states of one element.

Question 4: Match the compounds with the oxidation number of the starred atom: (A) KMn*O4, (B) Ni*(CO)4, (C) [Pt*(NH3)2Cl2]Cl2, (D) Na2O2*; with (1) +4, (2) +7, (3) 0, (4) −1. The correct code for A, B, C, D is:
(a) A−1, B−2, C−3, D−4
(b) A−4, B−3, C−2, D−1
(c) A−2, B−3, C−1, D−4
(d) A−4, B−1, C−2, D−3
Show Answer & Explanation

Answer: (c) A−2, B−3, C−1, D−4

Explanation:
1. KMnO4: Mn = +7, which is (2).
2. Ni(CO)4: CO is neutral, so Ni = 0, which is (3).
3. [Pt(NH3)2Cl2]Cl2: the cation is [Pt(NH3)2Cl2]2+, so Pt + 2(−1) = +2 and Pt = +4, which is (1).
4. Na2O2: peroxide, O = −1, which is (4).
5. Answer: (c)

Teacher's Note:
1. Neutral ligands contribute zero.
2. Peroxides have O in the −1 state.

Question 5: The oxidation state +1 for phosphorus is found in:
(a) Phosphorous acid (H3PO3)
(b) Orthophosphoric acid (H3PO4)
(c) Hypophosphorous acid (H3PO2)
(d) Hypophosphoric acid (H4P2O6)
Show Answer & Explanation

Answer: (c) Hypophosphorous acid (H3PO2)

Explanation:
1. H3PO3: 3 + x − 6 = 0, so x = +3.
2. H3PO4: x = +5.
3. H3PO2: 3 + x − 4 = 0, so x = +1.
4. H4P2O6: 4 + 2x − 12 = 0, so x = +4.
5. Answer: (c)

Teacher's Note:
1. Count H and O atoms in the formula.
2. Hypophosphorous acid is a strong reducing agent.

Question 6: The oxidation state of cobalt in [Co(NH3)4(H2O)Cl]SO4 is:
(a) 0
(b) +4
(c) −3
(d) +3
Show Answer & Explanation

Answer: (d) +3

Explanation:
1. SO42− is outside, so the complex ion has charge +2.
2. NH3 and H2O are neutral and Cl− is −1.
3. x + 0 + 0 − 1 = +2, so x = +3.
4. Answer: (d)

Teacher's Note:
1. First find the charge on the complex ion.
2. Neutral ligands add nothing.

Question 7: The oxidation number of carbon in graphite is:
(a) Zero
(b) +1
(c) +4
(d) +2
Show Answer & Explanation

Answer: (a) Zero

Explanation:
1. Graphite is an allotrope of carbon in its free elemental state.
2. Every atom of a free element has oxidation number zero.
3. Answer: (a)

Teacher's Note:
1. All allotropes of an element have zero oxidation number.
2. Free elements are always zero.

Question 8: The oxidation number of arsenic in H3AsO4 is:
(a) −1
(b) −3
(c) +3
(d) +5
Show Answer & Explanation

Answer: (d) +5

Explanation:
1. 3(+1) + x + 4(−2) = 0
2. x = +5
3. Answer: (d)

Teacher's Note:
1. Same method as H3PO4.
2. Group 15 shows +5 maximum.

Question 9: The oxidation number of iron in potassium ferricyanide, K3[Fe(CN)6], is:
(a) Two
(b) Six
(c) Three
(d) Four
Show Answer & Explanation

Answer: (c) Three

Explanation:
1. 3(+1) + x + 6(−1) = 0
2. x = +3, i.e. three.
3. Answer: (c)

Teacher's Note:
1. CN− carries charge −1.
2. Ferricyanide means Fe(III).

Question 10: The oxidation numbers of phosphorus in PH4+, PO23−, PO43− and PO33− are respectively:
(a) −3, +1, +3, +5
(b) −3, +3, +5, +1
(c) +3, −3, +5, +1
(d) −3, +1, +5, +3
Show Answer & Explanation

Answer: (d) −3, +1, +5, +3

Explanation:
1. PH4+: x + 4 = +1, so x = −3.
2. PO23−: x − 4 = −3, so x = +1.
3. PO43−: x − 8 = −3, so x = +5.
4. PO33−: x − 6 = −3, so x = +3.
5. Answer: (d)

Teacher's Note:
1. Sum of oxidation numbers equals the ion charge.
2. Keep the order of ions as given.

Question 11: The oxidation number of sodium in sodium amalgam is:
(a) +1
(b) 0
(c) −1
(d) +2
Show Answer & Explanation

Answer: (b) 0

Explanation:
1. An amalgam is an alloy of the metal with mercury.
2. In alloys all components have oxidation number zero.
3. Answer: (b)

Teacher's Note:
1. Alloys are metallic mixtures, not compounds.
2. Hence zero.

Question 12: If an element is in its lowest oxidation state, under proper conditions it can act as:
(a) A reducing agent
(b) An oxidising agent
(c) Both oxidising and reducing agent
(d) Neither oxidising nor reducing agent
Show Answer & Explanation

Answer: (a) A reducing agent

Explanation:
1. In the lowest state the element cannot gain more electrons.
2. It can only lose electrons, so it is a reducing agent.
3. Answer: (a)

Teacher's Note:
1. Lowest state: only oxidation possible.
2. Highest state: only reduction possible.

Question 13: A compound contains atoms A, B and C. The oxidation number of A is +2, of B is +5 and of C is −2. The possible formula of the compound is:
(a) ABC2
(b) B2(AC3)2
(c) A3(BC4)2
(d) A3(B4C)2
Show Answer & Explanation

Answer: (c) A3(BC4)2

Explanation:
1. ABC2: 2 + 5 − 4 = +3, not zero.
2. B2(AC3)2: 10 + 2(2 − 6) = +2, not zero.
3. A3(BC4)2: 6 + 2(5 − 8) = 0, valid.
4. A3(B4C)2: 6 + 2(20 − 2) = 42, not zero.
5. Answer: (c)

Teacher's Note:
1. Total oxidation number must be zero.
2. Test each option.

Question 14: In the reaction 2S2O32− + I2 → S4O62− + 2I−, the equivalent weight of Na2S2O3 is equal to its:
(a) Molecular weight
(b) Mol. wt./2
(c) 2 × Mol. wt.
(d) Mol. wt./6
Show Answer & Explanation

Answer: (a) Molecular weight

Explanation:
1. 2 thiosulphate ions lose 2 electrons in total (S: +2 to +2.5 in each of 4 S atoms, total change 4 × 0.5 = 2).
2. n-factor per Na2S2O3 = 1, so equivalent weight = M.
3. Answer: (a)

Teacher's Note:
1. Calculate the total change for the balanced equation, then divide by moles.
2. This is the basis of iodometry.

Question 15: In the reaction VO + Fe2O3 → FeO + V2O5, the equivalent weight of V2O5 is equal to its:
(a) Molecular weight
(b) Mol. wt./8
(c) Mol. wt./6
(d) None of these
Show Answer & Explanation

Answer: (c) Mol. wt./6

Explanation:
1. V goes from +2 (in VO) to +5 (in V2O5): 3 electrons per V.
2. V2O5 contains 2 V atoms, so 6 electrons are involved per formula unit.
3. Equivalent weight = M/6.
4. Answer: (c)

Teacher's Note:
1. Use the species as it appears in the formula.
2. Two V atoms means a factor of 2.

Question 16: In the reaction I2 + 2S2O32− → 2I− + S4O62−, the equivalent weight of iodine is:
(a) Its molecular weight
(b) Mol. wt./2
(c) Mol. wt./4
(d) None of these
Show Answer & Explanation

Answer: (b) Mol. wt./2

Explanation:
1. I2 + 2e− → 2I−, so n = 2.
2. Equivalent weight = M/2.
3. Answer: (b)

Teacher's Note:
1. Two iodine atoms each gain one electron.
2. Always check the half reaction.

Question 17: The molecular weight of KBrO3 is M. Its equivalent weight in the change BrO3− → Br− (acidic medium) is:
(a) M
(b) M/4
(c) M/6
(d) 6M
Show Answer & Explanation

Answer: (c) M/6

Explanation:
1. Br goes from +5 to −1, so n = 6.
2. Equivalent weight = M/6.
3. Answer: (c)

Teacher's Note:
1. Oxidation number change gives n.
2. Equivalent weight is M divided by n.

Question 18: The number of moles of oxalate ions oxidised by one mole of MnO4− in acidic medium is:
(a) 1/5
(b) 2/5
(c) 5/2
(d) 5
Show Answer & Explanation

Answer: (c) 5/2

Explanation:
1. 2MnO4− + 5C2O42− + 16H+ → 2Mn2+ + 10CO2 + 8H2O
2. 1 mol MnO4− oxidises 5/2 mol oxalate.
3. Answer: (c)

Teacher's Note:
1. Equate electrons: 5 per MnO4− and 2 per oxalate.
2. Ratio is 5 : 2.

Question 19: In a reaction, 4 mol of electrons are transferred to one mole of HNO3 when it acts as an oxidant. The possible reduction product is:
(a) 1/2 mole of N2
(b) 1/2 mole of N2O
(c) 1 mole of NO2
(d) 1 mole of NH3
Show Answer & Explanation

Answer: (b) 1/2 mole of N2O

Explanation:
1. N in HNO3 is +5; gaining 4 electrons gives +1.
2. N in +1 state is N2O, and one N gives 1/2 mole of N2O.
3. Answer: (b)

Teacher's Note:
1. Final state = initial state minus electrons gained.
2. +1 is found in N2O.

Question 20: How many moles of KMnO4 are reduced by 1 mole of ferrous oxalate in acidic medium?
(a) 1/5
(b) 5/3
(c) 1/3
(d) 3/5
Show Answer & Explanation

Answer: (d) 3/5

Explanation:
1. FeC2O4 loses 1 electron (Fe2+ → Fe3+) and 2 electrons (oxalate → 2CO2): n = 3.
2. Each MnO4− accepts 5 electrons.
3. Moles of KMnO4 = 3/5.
4. Answer: (d)

Teacher's Note:
1. Equivalents of oxidant equal equivalents of reductant.
2. Both species of ferrous oxalate are oxidised.

Question 21: Which of the following is not a redox change?
(a) 2H2S + SO2 → 2H2O + 3S
(b) 2BaO + O2 → 2BaO2
(c) BaO2 + H2SO4 → BaSO4 + H2O2
(d) 2KClO3 → 2KCl + 3O2
Show Answer & Explanation

Answer: (c) BaO2 + H2SO4 → BaSO4 + H2O2

Explanation:
1. In (c), Ba stays +2, S stays +6, O stays −1 (peroxide) and H stays +1, so no oxidation number changes.
2. The other three involve changes in S, O or Cl.
3. Answer: (c)

Teacher's Note:
1. Look for any change in oxidation number.
2. Double decomposition reactions are not redox.

Question 22: In the reaction MnO4− + SO32− + H+ → SO42− + Mn2+ + H2O:
(a) MnO4− and H+ both are reduced
(b) MnO4− is reduced and H+ is oxidised
(c) MnO4− is reduced and SO32− is oxidised
(d) MnO4− is oxidised and SO32− is reduced
Show Answer & Explanation

Answer: (c) MnO4− is reduced and SO32− is oxidised

Explanation:
1. Mn goes from +7 to +2: reduction.
2. S goes from +4 to +6: oxidation.
3. H+ only supplies the acidic medium and does not change.
4. Answer: (c)

Teacher's Note:
1. Compare oxidation numbers before and after.
2. H+ is a spectator for electrons.

Question 23: The balanced half reaction for CrO42− → CrO2− + OH− (basic medium) is:
(a) CrO42− + 2H2O + 3e− → CrO2− + 4OH−
(b) 2CrO42− + 8H2O → CrO2− + 4H2O + 8OH−
(c) CrO42− + H2O → CrO2− + H2O + OH−
(d) 3CrO42− + 4H2O + 6e− → 2CrO2− + 8OH−
Show Answer & Explanation

Answer: (a) CrO42− + 2H2O + 3e− → CrO2− + 4OH−

Explanation:
1. Cr goes from +6 to +3, so 3 electrons are gained.
2. Option (a): charge left = −2 − 3 = −5, right = −1 − 4 = −5; O: 4 + 2 = 6 on both sides; H: 4 on both sides.
3. The other options are unbalanced in charge or atoms.
4. Answer: (a)

Teacher's Note:
1. Balance atoms, then charge, using H2O and OH−.
2. Check O, H and charge in turn.

Question 24: The coefficients x, y, z, a, b that balance x Cr2O72− + y H+ + z e− → a Cr3+ + b H2O are:
(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. Cr2O72− + 14H+ + 6e− → 2Cr3+ + 7H2O
2. Charge: −2 + 14 − 6 = +6 on the left and 2(+3) = +6 on the right; O: 7 on both sides; H: 14 on both sides.
3. Answer: (b)

Teacher's Note:
1. Balance O with water, H with H+, then charge with electrons.
2. This is the standard dichromate half reaction.

Question 25: The number of electrons n required to balance the equation NO3− + 4H+ + ne− → 2H2O + NO is:
(a) 5
(b) 4
(c) 3
(d) 2
Show Answer & Explanation

Answer: (c) 3

Explanation:
1. Charge on the left: −1 + 4 − n; charge on the right: 0.
2. −1 + 4 − n = 0, so n = 3.
3. (Check: N goes from +5 to +2, gain of 3 electrons.)
4. Answer: (c)

Teacher's Note:
1. Balance charge using electrons.
2. Cross-check with the change in oxidation number.

Question 26: The value of n in MnO4− + 8H+ + ne− → Mn2+ + 4H2O is:
(a) 5
(b) 4
(c) 3
(d) 2
Show Answer & Explanation

Answer: (a) 5

Explanation:
1. Left charge: −1 + 8 − n = 7 − n; right charge: +2.
2. 7 − n = 2, so n = 5.
3. Answer: (a)

Teacher's Note:
1. Charge balance fixes n.
2. Mn goes from +7 to +2, which is also 5 electrons.

Question 27: In the balanced equation Zn + H+ + NO3− → NH4+ + Zn2+ + H2O, the coefficient of NH4+ is:
(a) 4
(b) 3
(c) 2
(d) 1
Show Answer & Explanation

Answer: (d) 1

Explanation:
1. 4Zn + 10H+ + NO3− → NH4+ + 4Zn2+ + 3H2O
2. Charge: +10 − 1 = +9 on the left; +1 + 8 = +9 on the right.
3. The coefficient of NH4+ is 1.
4. Answer: (d)

Teacher's Note:
1. One N atom takes 8 electrons.
2. 4 Zn atoms supply those 8 electrons.

Question 28: In the reaction H2O + SO32− → SO42− + 2H+ + X, the value of X is:
(a) 4e−
(b) 3e−
(c) 2e−
(d) 1e−
Show Answer & Explanation

Answer: (c) 2e−

Explanation:
1. Charge on the left: −2.
2. Charge on the right: −2 + 2 = 0, plus X.
3. So X must carry −2, i.e. 2e− (S: +4 to +6).
4. Answer: (c)

Teacher's Note:
1. Balance charge on both sides.
2. Oxidation produces electrons on the right.

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