JEE Chemistry Redox Reactions MCQs Set 14

Multiple Choice Questions (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.

Practice Redox Reactions MCQs for JEE Chemistry

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Multiple Choice Questions

Question 1: The oxidation number of oxygen in Na2O2 is
(a) +2
(b) -1
(c) -2
(d) +1
Show Answer & Explanation

Answer: (b) -1

Explanation:
1. Na2O2 contains the peroxide ion O22-.
2. Each O carries -1, so the oxidation number is -1.

Teacher's Note:
1. Peroxides have O = -1.
2. Check: 2(+1) + 2(-1) = 0.

Question 2: The oxidation states of sulphur in SO32-, S2O42- and S2O62- follow the order
(a) SO32- < S2O42- < S2O62-
(b) S2O42- < SO32- < S2O62-
(c) S2O42- < S2O62- < SO32-
(d) S2O62- < S2O42- < SO32-
Show Answer & Explanation

Answer: (b) S2O42- < SO32- < S2O62-

Explanation:
1. SO32-: x - 6 = -2, so S = +4.
2. S2O42-: 2x - 8 = -2, so S = +3.
3. S2O62-: 2x - 12 = -2, so S = +5.
4. Order: S2O42- (+3) < SO32- (+4) < S2O62- (+5).

Teacher's Note:
1. Write the charge balance for each ion.

Question 3: The average oxidation state of Fe in Fe3O4 is
(a) 8/3
(b) 3
(c) -8/3
(d) 2
Show Answer & Explanation

Answer: (a) 8/3

Explanation:
1. 3x + 4(-2) = 0, so x = 8/3.

Teacher's Note:
1. It is a mixed oxide of FeO and Fe2O3.

Question 4: When xHI + yHNO3 → NO + I2 + H2O is balanced with whole-number coefficients, the values of x and y are
(a) x = 6, y = 1
(b) x = 6, y = 2
(c) x = 3, y = 2
(d) x = 2, y = 3
Show Answer & Explanation

Answer: (b) x = 6, y = 2

Explanation:
1. I goes from -1 to 0 (1 e- per HI); N goes from +5 to +2 (3 e-).
2. 6HI + 2HNO3 → 2NO + 3I2 + 4H2O, so x = 6 and y = 2.

Teacher's Note:
1. Balance electrons first, then H and O.

Question 5: When xP4 + yHNO3 → H3PO4 + NO2 + H2O is balanced with whole numbers, x and y are
(a) x = 1, y = 15
(b) x = 1, y = 20
(c) x = 2, y = 10
(d) x = 1, y = 5
Show Answer & Explanation

Answer: (b) x = 1, y = 20

Explanation:
1. Each P goes from 0 to +5, so P4 loses 20 e-; each N gains 1 e-.
2. P4 + 20HNO3 → 4H3PO4 + 20NO2 + 4H2O.

Teacher's Note:
1. Electrons lost by P4 = 20.

Question 6: CN- is oxidised by NO3- in acid: aCN- + bNO3- + cH+ → (a + b)NO + aCO2 + (c/2)H2O. The whole-number values of a, b, c in that order are
(a) 3, 7, 7
(b) 3, 10, 7
(c) 3, 10, 10
(d) 3, 7, 10
Show Answer & Explanation

Answer: (d) 3, 7, 10

Explanation:
1. In CN-, C is +2 and N is -3. They become C = +4 in CO2 and N = +2 in NO, so each CN- loses 2 + 5 = 7 e-; each NO3- (N = +5 to +2) gains 3 e-.
2. Balancing electrons gives a : b = 3 : 7, and charge balance gives c = 10.
3. Check O: 21 = 10 + 6 + 5.

Teacher's Note:
1. Balance electrons first, then charge.
2. The net equation is 3CN- + 7NO3- + 10H+ → 10NO + 3CO2 + 5H2O.

Question 7: In xNO3- + yI- + zH+ → 2NO + 3I2 + 4H2O, x, y and z respectively are
(a) 1, 6, 4
(b) 2, 6, 8
(c) 0, 6, 8
(d) 2, 3, 4
Show Answer & Explanation

Answer: (b) 2, 6, 8

Explanation:
1. N atoms: x = 2. I atoms: y = 6. H atoms: z = 8.
2. Check charge: -2 - 6 + 8 = 0 on the left and 0 on the right.

Teacher's Note:
1. Count atoms and check the charge.

Question 8: The sum of the coefficients of all reactants and products in KMnO4 + H2O2 + H2SO4 → MnSO4 + O2 + H2O + K2SO4, when balanced with the smallest whole numbers, is
(a) 26
(b) 23
(c) 28
(d) 22
Show Answer & Explanation

Answer: (a) 26

Explanation:
1. Mn gains 5 e-; each H2O2 loses 2 e-, so KMnO4 : H2O2 = 2 : 5.
2. 2KMnO4 + 5H2O2 + 3H2SO4 → 2MnSO4 + 5O2 + 8H2O + K2SO4.
3. Sum = 2 + 5 + 3 + 2 + 5 + 8 + 1 = 26.

Teacher's Note:
1. Balance electrons first, then K, S, H and O.

Question 9: Which of the following reactions is NOT a redox reaction? (A) H2O2 + KOH → KHO2 + H2O (B) Cr2O72- + 2OH- → 2CrO42- + H2O (C) Ca(HCO3)2 → CaCO3 + CO2 + H2O (D) H2O2 → H2O + 1/2 O2. Select the option listing all the non-redox reactions.
(a) A, B, C and D
(b) B and D only
(c) A and C only
(d) A, B and C
Show Answer & Explanation

Answer: (d) A, B and C

Explanation:
1. (A) is an acid-base reaction with no change in oxidation numbers.
2. (B) Cr stays +6 and (C) has no change in oxidation numbers (only decomposition).
3. (D) O goes from -1 to -2 and 0, so it is a redox (disproportionation) reaction.

Teacher's Note:
1. Look for any element whose oxidation number changes.

Question 10: Which of the following are examples of disproportionation? (A) HgO → Hg + O2 (B) KClO3 → KCl + O2 (C) KClO3 → KClO4 + KCl (D) Cl2 + OH- → ClO- + Cl- + H2O
(a) B and C
(b) C and D
(c) A and D
(d) A and B
Show Answer & Explanation

Answer: (b) C and D

Explanation:
1. Disproportionation: one element is oxidised and reduced at the same time.
2. (C) Cl in KClO3 (+5) goes to +7 and -1. (D) Cl2 (0) goes to +1 and -1.
3. (A) and (B) involve different elements being oxidised and reduced.

Teacher's Note:
1. The same element must go both up and down.

Question 11: Consider xCu3P + yCr2O72- + zH+ → Cu2+ + H3PO4 + Cr3+ + H2O. The values of x and y, and the electrons lost per formula unit of Cu3P, are respectively
(a) x = 6, y = 11, 11 e-
(b) x = 6, y = 11, 9 e-
(c) x = 11, y = 6, 11 e-
(d) x = 3, y = 6, 8 e-
Show Answer & Explanation

Answer: (a) x = 6, y = 11, 11 e-

Explanation:
1. In Cu3P: Cu = +1 and P = -3. Cu goes to +2 (3 e- for three Cu) and P to +5 (8 e-), so 11 e- are lost per formula unit.
2. Dichromate gains 6 e-, so 6 Cu3P (66 e-) reacts with 11 Cr2O72- (66 e-).
3. So x = 6 and y = 11.

Teacher's Note:
1. Calculate electrons lost by each element in Cu3P.

Question 12: A compound contains elements P (oxidation number +2), Q (+5) and R (-2). The possible formula of the compound is
(a) P3(Q4R)2
(b) P3[QR4]2
(c) PQR2
(d) Q2(PR3)2
Show Answer & Explanation

Answer: (b) P3[QR4]2

Explanation:
1. Check that the total oxidation number is zero.
2. P3[QR4]2: 3(+2) + 2(+5) + 8(-2) = 6 + 10 - 16 = 0.

Teacher's Note:
1. The sum of oxidation numbers of all atoms in a neutral formula is zero.
2. The other options do not sum to zero.

Question 13: Which of the following is a redox reaction?
(a) NaCl + KNO3 → NaNO3 + KCl
(b) CaC2O4 + 2HCl → CaCl2 + H2C2O4
(c) Zn + 2AgCN → 2Ag + Zn(CN)2
(d) Mg(OH)2 + 2NH4Cl → MgCl2 + 2NH4OH
Show Answer & Explanation

Answer: (c) Zn + 2AgCN → 2Ag + Zn(CN)2

Explanation:
1. Only in the last reaction do oxidation numbers change: Zn 0 to +2 and Ag +1 to 0.

Teacher's Note:
1. Look for a change in oxidation number.

Question 14: In the reaction xMnO4- + yC2O42- + zH+ → xMn2+ + 2yCO2 + (z/2)H2O, the values of x, y and z respectively are
(a) 2, 5 and 8
(b) 5, 2 and 16
(c) 2, 5 and 16
(d) 5, 2 and 8
Show Answer & Explanation

Answer: (c) 2, 5 and 16

Explanation:
1. Electron balance: Mn gains 5, oxalate loses 2, so x : y = 2 : 5.
2. Charge balance: -x - 2y + z = 2x gives z = 16 for x = 2, y = 5.

Teacher's Note:
1. Balance electrons then charge.

Question 15: Which of the following is a redox reaction?
(a) XeF6 + H2O → XeOF4 + 2HF
(b) XeF2 + PF5 → [XeF]+[PF6]-
(c) XeF6 + 2H2O → XeO2F2 + 4HF
(d) XeF4 + O2F2 → XeF6 + O2
Show Answer & Explanation

Answer: (d) XeF4 + O2F2 → XeF6 + O2

Explanation:
1. In the last reaction Xe goes from +4 to +6 and O from +1 (in O2F2) to 0, so electrons are transferred.
2. The others are acid-base or hydrolysis reactions with no change in oxidation numbers.

Teacher's Note:
1. Check oxidation numbers on both sides.

Question 16: Which of the following equations represents an oxidation-reduction reaction?
(a) 2H2SO4 + Cu → CuSO4 + 2H2O + SO2
(b) H2SO4 + Na2CO3 → Na2SO4 + H2O + CO2
(c) H2SO4 + 2NH3 → (NH4)2SO4
(d) 2K2CrO4 + H2SO4 → K2Cr2O7 + K2SO4 + H2O
Show Answer & Explanation

Answer: (a) 2H2SO4 + Cu → CuSO4 + 2H2O + SO2

Explanation:
1. In the last equation Cu goes from 0 to +2 and S from +6 to +4.
2. The others show no change in oxidation numbers.

Teacher's Note:
1. Look for changes in oxidation numbers.

Question 17: In aF2 + bOH- → cF- + dOF2 + eH2O, the values of a, b and c are respectively
(a) 3, 4, 2
(b) 3, 2, 4
(c) 2, 2, 2
(d) 2, 2, 4
Show Answer & Explanation

Answer: (c) 2, 2, 2

Explanation:
1. 2F2 + 2OH- → 2F- + OF2 + H2O: F: 4 on both sides (2 + 2); O: 2 on the left and 1 + 1 on the right; H: 2 on both sides; charge: -2 on both sides.
2. So a = 2, b = 2 and c = 2.

Teacher's Note:
1. Balance the charge and each atom.

Question 18: In the oxidation-state analysis of K[Co(C2O4)2(NH3)2], the oxidation number of cobalt is (oxalate ion = -2, ammonia is neutral)
(a) +1
(b) +2
(c) +3
(d) +4
Show Answer & Explanation

Answer: (c) +3

Explanation:
1. +1 + x + 2(-2) + 0 = 0, so x = +3.

Teacher's Note:
1. Oxalate carries a charge of -2.

Question 19: The oxidation number of iron in sodium nitroprusside, Na2[Fe(CN)5(NO+)], is
(a) +3
(b) +2
(c) +4
(d) +1
Show Answer & Explanation

Answer: (b) +2

Explanation:
1. Charges: 2(+1) + x + 5(-1) + 1 = 0, so x = +2.

Teacher's Note:
1. The NO ligand is present as NO+ here.

Question 20: In Fe0.93O, the average oxidation number of iron is about
(a) +2.5
(b) +2
(c) +2.15
(d) +3
Show Answer & Explanation

Answer: (c) +2.15

Explanation:
1. 0.93x - 2 = 0, so x = 2/0.93 = +2.15.

Teacher's Note:
1. This non-stoichiometric oxide has some Fe in the +3 state.

Question 21: In FeCl2 + H2O2 + HCl → FeCl3 + H2O, the oxidant and reductant are respectively
(a) H2O2 and FeCl2
(b) FeCl2 and H2O2
(c) HCl and FeCl2
(d) H2O2 and HCl
Show Answer & Explanation

Answer: (a) H2O2 and FeCl2

Explanation:
1. Fe goes from +2 to +3 (oxidised), so FeCl2 is the reductant.
2. O goes from -1 to -2 (reduced), so H2O2 is the oxidant.

Teacher's Note:
1. Identify which species gains electrons.

Question 22: In Cu + HNO3(dil) → Cu(NO3)2 + H2O + NO, the oxidant and reductant are respectively
(a) Cu(NO3)2 and Cu
(b) HNO3 and Cu
(c) HNO3 and NO
(d) Cu and HNO3
Show Answer & Explanation

Answer: (b) HNO3 and Cu

Explanation:
1. Cu goes from 0 to +2 (oxidised), so Cu is the reductant.
2. N goes from +5 to +2 (reduced), so HNO3 is the oxidant.

Teacher's Note:
1. Reductant is oxidised.

Question 23: In Na2HAsO3 + KBrO3 + HCl → NaCl + KBr + H3AsO4, the oxidant and reductant are respectively
(a) HCl and KBrO3
(b) Na2HAsO3 and KBrO3
(c) KBrO3 and HCl
(d) KBrO3 and Na2HAsO3
Show Answer & Explanation

Answer: (d) KBrO3 and Na2HAsO3

Explanation:
1. As goes from +3 to +5 (oxidised), so Na2HAsO3 is the reductant.
2. Br goes from +5 to -1 (reduced), so KBrO3 is the oxidant.

Teacher's Note:
1. Compare oxidation numbers of As and Br.

Question 24: In I2 + Na2S2O3 → Na2S4O6 + NaI, the oxidant and reductant are respectively
(a) Na2S2O3 and I2
(b) Na2S4O6 and I2
(c) I2 and NaI
(d) I2 and Na2S2O3
Show Answer & Explanation

Answer: (d) I2 and Na2S2O3

Explanation:
1. I goes from 0 to -1 (reduced), so I2 is the oxidant.
2. S goes from +2 to +2.5 (oxidised), so thiosulphate is the reductant.

Teacher's Note:
1. This is the iodometric titration reaction.

Question 25: When IO3- + Re → ReO4- + I- is balanced in acidic solution, the coefficient of H2O is
(a) 3
(b) 6
(c) 1
(d) 2
Show Answer & Explanation

Answer: (a) 3

Explanation:
1. Re goes 0 to +7 (7 e-); I goes +5 to -1 (6 e-), so Re : IO3- = 6 : 7.
2. 6Re + 7IO3- + 3H2O → 6ReO4- + 7I- + 6H+, so the coefficient of water is 3.

Teacher's Note:
1. Balance electrons, then O with water, then H with H+.

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Multiple Choice Questions (MCQs) for JEE Chemistry Redox Reactions

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