JEE Chemistry Redox Reactions MCQs Set 09

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.

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Multiple Choice Questions

Question 1: The oxidation number of cobalt in \( \ce{K[Co(C2O4)2(NH3)2} \)] is
(a) +3
(b) +2
(c) +4
(d) +1
Show Answer & Explanation

Answer: (a) +3

Explanation:
1. K is +1. Each oxalate is 2-, so two give -4. \( \ce{NH3} \) is neutral.
2. 1 + x - 4 = 0, so x = +3.

Teacher's Note:
1. The sum of oxidation numbers in a neutral compound is zero.

Question 2: The oxidation number of phosphorus in \( \ce{K4P2O7} \) is
(a) +3
(b) +5
(c) +4
(d) +7
Show Answer & Explanation

Answer: (b) +5

Explanation:
1. 4(+1) + 2x + 7(-2) = 0.
2. 2x = 10, so x = +5.

Teacher's Note:
1. Set the sum of oxidation numbers to zero and solve for the unknown.

Question 3: The oxidation number of iron in \( \ce{Na2[Fe(CN)5(NO+)} \)] is
(a) +2
(b) +3
(c) +1
(d) 0
Show Answer & Explanation

Answer: (a) +2

Explanation:
1. 2(+1) + x + 5(-1) + 1 = 0 (NO is present as the nitrosonium ion, NO+).
2. x = +2.

Teacher's Note:
1. NO+ carries +1 charge, unlike NO or NO-. Account for ligand charges separately.

Question 4: The oxidation number of iron in \( \ce{[Fe(NO+)(H2O)5]SO4} \) is
(a) +2
(b) +3
(c) 0
(d) +1
Show Answer & Explanation

Answer: (d) +1

Explanation:
1. Sulphate is 2-, so the complex cation is 2+.
2. x + 1 (NO+) + 0 (water) = +2, so x = +1.

Teacher's Note:
1. Find the charge of the complex ion from the counter ion and then split it between the metal and the ligands.

Question 5: The average oxidation number of iron in \( \ce{Fe0.93O} \) is
(a) +2
(b) +2.5
(c) +3
(d) +2.15
Show Answer & Explanation

Answer: (d) +2.15

Explanation:
1. O is -2, so 0.93 x (Fe) = +2.
2. Fe = 2/0.93 = +2.15 (a mixture of Fe(II) and Fe(III)).

Teacher's Note:
1. A non-stoichiometric compound gives a fractional average oxidation number.

Question 6: In the reaction \( \ce{KMnO4 + KCl + H2SO4 -> MnSO4 + K2SO4 + H2O + Cl2} \), the oxidant and the reductant are respectively
(a) KCl and \( \ce{KMnO4} \)
(b) \( \ce{H2SO4} \) and KCl
(c) \( \ce{KMnO4} \) and \( \ce{H2SO4} \)
(d) \( \ce{KMnO4} \) and KCl
Show Answer & Explanation

Answer: (d) \( \ce{KMnO4} \) and KCl

Explanation:
1. Mn goes from +7 to +2 (reduced), so \( \ce{KMnO4} \) is the oxidant.
2. Cl goes from -1 to 0 (oxidised), so KCl is the reductant.

Teacher's Note:
1. The oxidant is the species that is reduced, and it gains electrons.

Question 7: In the reaction \( \ce{Na2HAsO3 + KBrO3 + HCl -> NaCl + KBr + H3AsO4} \), the oxidant and the reductant are respectively
(a) \( \ce{KBrO3} \) and \( \ce{Na2HAsO3} \)
(b) \( \ce{Na2HAsO3} \) and \( \ce{KBrO3} \)
(c) HCl and \( \ce{KBrO3} \)
(d) \( \ce{KBrO3} \) and HCl
Show Answer & Explanation

Answer: (a) \( \ce{KBrO3} \) and \( \ce{Na2HAsO3} \)

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

Teacher's Note:
1. Track the oxidation numbers of the two elements that change.

Question 8: In the reaction \( \ce{I2 + Na2S2O3 -> Na2S4O6 + NaI} \), the oxidant and the reductant are respectively
(a) \( \ce{Na2S2O3} \) and \( I_2 \)
(b) \( I_2 \) and \( \ce{Na2S2O3} \)
(c) \( I_2 \) and NaI
(d) \( \ce{Na2S4O6} \) and \( I_2 \)
Show Answer & Explanation

Answer: (b) \( I_2 \) and \( \ce{Na2S2O3} \)

Explanation:
1. I goes from 0 to -1 (reduced): \( I_2 \) is the oxidant.
2. S goes from +2 to +2.5 (oxidised): \( \ce{Na2S2O3} \) is the reductant.

Teacher's Note:
1. Thiosulphate is a classic reductant for iodine in titrations.

Question 9: Match List-I (compounds) with List-II (oxidation states of nitrogen): (a) \( NaN_3 \) (b) \( N_2H_2 \) (c) NO (d) \( N_2O_5 \) with (1) +5 (2) +2 (3) -1/3 (4) -1. The correct codes for (a), (b), (c), (d) are
(a) 3, 4, 2, 1
(b) 4, 3, 2, 1
(c) 3, 4, 1, 2
(d) 4, 3, 1, 2
Show Answer & Explanation

Answer: (a) 3, 4, 2, 1

Explanation:
1. \( NaN_3 \): 3x + 1 = 0, so N = -1/3 (3).
2. \( N_2H_2 \): 2x + 2 = 0, so N = -1 (4).
3. NO: N = +2 (2). \( N_2O_5 \): N = +5 (1).

Teacher's Note:
1. Solve each compound for nitrogen with H = +1, O = -2, Na = +1.

Question 10: 1 mol of \( \ce{N2H4} \) loses ten moles of electrons to form a new compound Y. Assuming all nitrogen appears in Y, and there is no change in the oxidation state of hydrogen, the oxidation state of nitrogen in Y is
(a) -1
(b) -3
(c) +3
(d) +5
Show Answer & Explanation

Answer: (c) +3

Explanation:
1. In \( \ce{N2H4} \), N is -2. Two N atoms lose 10 electrons, so each N loses 5.
2. New oxidation state = -2 + 5 = +3.

Teacher's Note:
1. Divide the electrons lost by the number of atoms losing them.

Question 11: In the reaction xHI + y\( \ce{HNO3} \) -> NO + \( I_2 \) + \( H_2O \), upon balancing with whole number coefficients
(a) x = 3, y = 2
(b) x = 2, y = 3
(c) x = 6, y = 2
(d) x = 6, y = 1
Show Answer & Explanation

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

Explanation:
1. I changes from -1 to 0 (1 e per I), N from +5 to +2 (3 e).
2. Balanced: \( \ce{6HI + 2HNO3 -> 2NO + 3I2 + 4H2O} \).
3. So x = 6 and y = 2.

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

Question 12: For the redox reaction \( MnO_4^- \) + \( C_2O_4^{2-} \) + \( H^+ \) -> \( Mn^{2+} \) + \( CO_2 \) + \( H_2O \), the correct whole number coefficients of \( MnO_4^- \), \( C_2O_4^{2-} \) and \( H^+ \) are respectively
(a) 2, 5, 16
(b) 16, 5, 2
(c) 5, 16, 2
(d) 2, 16, 5
Show Answer & Explanation

Answer: (a) 2, 5, 16

Explanation:
1. Mn gains 5 e; each oxalate loses 2 e. So ratio 2 : 5.
2. \( \ce{2MnO4- + 5C2O4^2- + 16H+ -> 2Mn^2+ + 10CO2 + 8H2O} \).
3. Check charge: -2 - 10 + 16 = +4 on both sides.

Teacher's Note:
1. Equalise electrons gained and lost, then balance charge with H+ and atoms with water.

Question 13: For the redox reaction x\( P_4 \) + y\( \ce{HNO3} \) -> \( H_3PO_4 \) + \( NO_2 \) + \( H_2O \), upon balancing with whole number coefficients
(a) x = 1, y = 5
(b) x = 2, y = 10
(c) x = 1, y = 20
(d) x = 1, y = 15
Show Answer & Explanation

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

Explanation:
1. P4 loses 20 e (4 P go from 0 to +5). Each N gains 1 e.
2. \( \ce{P4 + 20HNO3 -> 4H3PO4 + 20NO2 + 4H2O} \).
3. So x = 1 and y = 20.

Teacher's Note:
1. Count total electrons for the whole \( \ce{P4} \) molecule.

Question 14: \( CN^- \) is oxidised by \( NO_3^- \) in presence of acid: a\( CN^- \) + b\( NO_3^- \) + c\( H^+ \) -> (a + b)NO + a\( CO_2 \) + (c/2)\( H_2O \). The whole number values of a, b, c in that order are
(a) 3, 7, 7
(b) 3, 7, 10
(c) 3, 10, 10
(d) 3, 10, 7
Show Answer & Explanation

Answer: (b) 3, 7, 10

Explanation:
1. CN-: C goes +2 to +4 (2 e) and N goes -3 to +2 (5 e), so 7 e per CN-. \( NO_3^- \) to NO gains 3 e.
2. 7a = 3b gives a = 3, b = 7.
3. Charge balance: -3 - 7 + c = 0, so c = 10.

Teacher's Note:
1. Count electrons lost by both C and N in cyanide.

Question 15: The sum of the average oxidation numbers of sulphur in \( \ce{H2SO5} \) (peroxymonosulphuric acid) and in \( \ce{Na2S2O3} \) (sodium thiosulphate) is
(a) 6
(b) 10
(c) 8
(d) 4
Show Answer & Explanation

Answer: (c) 8

Explanation:
1. \( \ce{H2SO5} \) has one peroxide link (two O at -1): 2 + x + 3(-2) + 2(-1) = 0, so S = +6.
2. \( \ce{Na2S2O3} \): 2 + 2x - 6 = 0, so average S = +2.
3. Sum = 6 + 2 = 8.

Teacher's Note:
1. Remember that two of the five oxygen atoms in \( \ce{H2SO5} \) are peroxide oxygens at -1.

Question 16: The reaction \( \ce{Cl2 + S2O3^2- -> SO4^2- + Cl-} \) is carried out in basic medium. Starting with 0.15 mol of \( \ce{Cl2} \), 0.01 mol of \( S_2O_3^{2-} \) and 0.3 mol of \( OH^- \), the moles of \( OH^- \) left after the reaction is complete are (assume no other reaction occurs)
(a) 0.2
(b) 0.1
(c) 0.3
(d) 0.05
Show Answer & Explanation

Answer: (a) 0.2

Explanation:
1. Balanced: \( \ce{4Cl2 + S2O3^2- + 10OH- -> 2SO4^2- + 8Cl- + 5H2O} \).
2. Thiosulphate (0.01 mol) is the limiting reagent since \( \ce{Cl2} \) is in excess. OH- used = 10 x 0.01 = 0.1 mol.
3. OH- left = 0.3 - 0.1 = 0.2 mol.

Teacher's Note:
1. Balance in basic medium and find the limiting species before counting the hydroxide used.

Question 17: The composition of a sample is \( Fe_{0.93}O_{1.00} \). If Fe is present in +2 and +3 oxidation states, the percentage of Fe present in the +3 state is
(a) 85%
(b) 30%
(c) 15%
(d) 60%
Show Answer & Explanation

Answer: (c) 15%

Explanation:
1. Let Fe(II) = x and Fe(III) = y: x + y = 0.93; 2x + 3y = 2.
2. y = 2 - 1.86 = 0.14.
3. Percentage = 0.14/0.93 x 100 = 15%.

Teacher's Note:
1. Use the atom count and the charge balance together.

Question 18: When ZnS is boiled with strong nitric acid, the products are zinc nitrate, sulphuric acid and nitrogen dioxide. The changes in the oxidation numbers of Zn, S and N are
(a) +2, +4, -1
(b) +2, +6, -2
(c) 0, +4, -2
(d) 0, +8, -1
Show Answer & Explanation

Answer: (d) 0, +8, -1

Explanation:
1. Zn stays at +2, so the change is 0.
2. S goes from -2 to +6, a change of +8.
3. N goes from +5 to +4, a change of -1.

Teacher's Note:
1. Look at the change in oxidation number, not the final values.

Question 19: In the redox reaction \( \ce{2S2O3^2- + I2 -> S4O6^2- + 2I-} \), which statements are correct? (1) \( S_2O_3^{2-} \) is reduced to \( S_4O_6^{2-} \) (2) \( S_2O_3^{2-} \) is oxidised to \( S_4O_6^{2-} \) (3) \( I_2 \) is reduced to \( I^- \) (4) \( I_2 \) is oxidised to \( I^- \)
(a) 2 and 3 only
(b) 1 and 4 only
(c) 1 and 3 only
(d) 2 and 4 only
Show Answer & Explanation

Answer: (a) 2 and 3 only

Explanation:
1. S goes from +2 to +2.5 (oxidation); I goes from 0 to -1 (reduction).

Teacher's Note:
1. Check the oxidation number change of S and I.

Question 20: Among the following, the pair having both metals in their highest oxidation state is
(a) \( Fe(CN)_6^{3-} \) and \( Co(CN)_6^{3-} \)
(b) \( CrO_2Cl_2 \) and \( MnO_4^- \)
(c) \( TiO_2 \) and \( MnO_2 \)
(d) \( MnCl_4^{2-} \) and \( NiF_6^{2-} \)
Show Answer & Explanation

Answer: (b) \( CrO_2Cl_2 \) and \( MnO_4^- \)

Explanation:
1. Cr in \( \ce{CrO2Cl2} \) is +6 (its maximum) and Mn in \( \ce{MnO4-} \) is +7 (its maximum).
2. Fe(+3), Co(+3), Mn(+4 or +2) and Ni(+4) are not at their highest states. Ti is +4 (its maximum) but Mn in \( \ce{MnO2} \) is +4, not +7.

Teacher's Note:
1. The highest oxidation state of a transition metal equals its number of valence electrons (d + s).

Question 21: When KMnO4 acts as an oxidising agent and ultimately forms \( MnO_4^{2-} \), \( MnO_2 \), \( Mn_2O_3 \) and \( Mn^{2+} \), the number of electrons transferred in each case is respectively
(a) 4, 3, 1, 5
(b) 1, 5, 3, 7
(c) 1, 3, 4, 5
(d) 3, 5, 7, 1
Show Answer & Explanation

Answer: (c) 1, 3, 4, 5

Explanation:
1. Mn(VII) to Mn(VI): 1 e. To Mn(IV): 3 e. To Mn(III): 4 e. To Mn(II): 5 e.

Teacher's Note:
1. Electrons gained = initial oxidation number minus final.

Question 22: Which of the following is a redox reaction?
(a) \( \ce{NaCl + KNO3 -> NaNO3 + KCl} \)
(b) \( \ce{CaC2O4 + 2HCl -> CaCl2 + H2C2O4} \)
(c) \( \ce{Mg(OH)2 + 2NH4Cl -> MgCl2 + 2NH4OH} \)
(d) \( \ce{Zn + 2AgCN -> 2Ag + Zn(CN)2} \)
Show Answer & Explanation

Answer: (d) \( \ce{Zn + 2AgCN -> 2Ag + Zn(CN)2} \)

Explanation:
1. In the last reaction Zn goes from 0 to +2 and Ag from +1 to 0.
2. The others only exchange ions without any change in oxidation number.

Teacher's Note:
1. A redox reaction shows a change in oxidation numbers.

Question 23: Which of the following chemical reactions depicts the oxidising behaviour of \( H_2SO_4 \)?
(a) \( \ce{2HI + H2SO4 -> I2 + SO2 + 2H2O} \)
(b) \( \ce{Ca(OH)2 + H2SO4 -> CaSO4 + 2H2O} \)
(c) \( \ce{NaCl + H2SO4 -> NaHSO4 + HCl} \)
(d) \( \ce{2PCl5 + H2SO4 -> 2POCl3 + 2HCl + SO2Cl2} \)
Show Answer & Explanation

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

Explanation:
1. In the first reaction S goes from +6 in \( \ce{H2SO4} \) to +4 in \( \ce{SO2} \), so it is reduced and \( \ce{H2SO4} \) acts as an oxidant.
2. The other reactions involve no change in oxidation number.

Teacher's Note:
1. The oxidant is itself reduced.

Question 24: The balanced net ionic equation (redox reaction in acidic solution) for \( IO_3^- \) + Re -> \( ReO_4^- \) + \( I^- \) is
(a) \( \ce{7IO3- + 6Re + 3H2O -> 6ReO4- + 7I- + 6H+} \)
(b) \( \ce{IO3- + Re + H2O -> ReO4- + I- + 2H+} \)
(c) \( \ce{7IO3- + 6Re + 6H+ -> 6ReO4- + 7I- + 3H2O} \)
(d) \( \ce{6IO3- + 7Re + 3H2O -> 7ReO4- + 6I- + 6H+} \)
Show Answer & Explanation

Answer: (a) \( \ce{7IO3- + 6Re + 3H2O -> 6ReO4- + 7I- + 6H+} \)

Explanation:
1. I goes from +5 to -1 (gains 6 e). Re goes from 0 to +7 (loses 7 e).
2. Cross-multiply: 7 IO3- and 6 Re. Oxygen: 21 on the left, 24 on the right, so add 3 H2O to the left.
3. Hydrogen then gives 6 H+ on the right. Charge: -7 on both sides.

Teacher's Note:
1. Equalise electrons by cross-multiplication, then balance O and H.

Question 25: The balanced equation (redox reaction in acidic solution) for the disproportionation \( HNO_2 \) -> \( NO_3^- \) + NO is
(a) \( \ce{3HNO2 -> NO3- + 2NO + H+ + H2O} \)
(b) \( \ce{2HNO2 -> NO3- + NO + H+ + H2O} \)
(c) \( \ce{3HNO2 -> NO3- + 2NO + 2H+} \)
(d) \( \ce{HNO2 -> NO3- + NO + H2O} \)
Show Answer & Explanation

Answer: (a) \( \ce{3HNO2 -> NO3- + 2NO + H+ + H2O} \)

Explanation:
1. N goes from +3 to +5 (loses 2 e) in nitrate, and from +3 to +2 (gains 1 e) in NO.
2. So 1 nitrate forms for every 2 NO: 3 HNO2 in total.
3. Balance O and H with H2O and H+: 3 HNO2 gives NO3- + 2NO + H+ + H2O.

Teacher's Note:
1. Disproportionation: the same species is both oxidised and reduced. Use the electrons lost and gained to fix the ratio.

Question 26: The balanced equation for the reaction of \( H_2O_2 \) with \( Cl_2O_7 \) giving \( ClO_2^- \) and \( O_2 \) in basic solution (redox reaction in basic medium) is
(a) \( \ce{Cl2O7 + H2O2 + 2OH- -> 2ClO2- + O2 + 3H2O} \)
(b) \( \ce{Cl2O7 + 4H2O2 -> 2ClO2- + 4O2 + 4H2O} \)
(c) \( \ce{Cl2O7 + 4H2O2 + 2OH- -> 2ClO2- + 4O2 + 5H2O} \)
(d) \( \ce{Cl2O7 + 4H2O2 + 2OH- -> 2ClO2- + 2O2 + 5H2O} \)
Show Answer & Explanation

Answer: (c) \( \ce{Cl2O7 + 4H2O2 + 2OH- -> 2ClO2- + 4O2 + 5H2O} \)

Explanation:
1. Each Cl goes from +7 to +3 (gains 4 e), so 2 Cl gain 8 e. Each H2O2 loses 2 e, so 4 H2O2 are needed.
2. Balance charge with 2 OH- on the left (charge -2 on both sides).
3. Check O: left 7 + 8 + 2 = 17; right 2(2) + 4(2) + 5 = 17.

Teacher's Note:
1. In basic solution, balance charge with OH- and then balance H and O with water.

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