CBSE Class 11 Chemistry Redox Reactions MCQs Set 03

Answer: (d) All of the above

Explanation:
1. Electrochemical processes for the extraction of highly reactive metals and non-metals, manufacturing of chemical compounds like caustic soda, operation of dry and wet batteries and corrosion of metals all fall within the range of redox processes.
2. Hence, all the given processes are redox processes.

Teacher's Note:
1. Electrolysis of brine (to make caustic soda) involves oxidation of Cl⁻ at the anode and reduction of water at the cathode.
2. Rusting of iron is a slow redox reaction in which iron is oxidised by oxygen in the presence of moisture.

Answer: (a) Addition of oxygen

Explanation:
1. Addition of oxygen or an electronegative element and removal of hydrogen or an electropositive element take place in oxidation.
2. Addition of hydrogen, removal of oxygen and removal of chlorine (an electronegative element) are reduction processes.

Teacher's Note:
1. Remember the classical pairs: oxygen and hydrogen always behave in opposite ways.
2. Removal of an electronegative element such as chlorine is reduction, not oxidation.

Answer: (d) Both (a) and (b)

Explanation:
1. Reduction is a process which involves addition of hydrogen or an electropositive element to a substance, or removal of oxygen or an electronegative element from a substance.
2. So, both removal of oxygen and addition of hydrogen are reduction.

Teacher's Note:
1. Removal of hydrogen is oxidation, e.g. H₂S → S.
2. In organic chemistry, reduction usually means gain of hydrogen or loss of oxygen.

Answer: (a) reduction process

Explanation:
1. In the given reaction, an electronegative element (chlorine) is removed from ferric chloride: FeCl₃ → FeCl₂.
2. Removal of an electronegative element is reduction. Also, the oxidation number of Fe decreases from +3 to +2.
3. Hence, ferric chloride undergoes reduction.

Teacher's Note:
1. H₂ is the reducing agent here; it is itself oxidised to HCl.
2. Check the change in oxidation number of the metal: a decrease always means reduction.

Answer: (a) H₂S is oxidised and Cl₂ is reduced

Explanation:
1. Hydrogen is removed from H₂S (H₂S → S), so H₂S is oxidised (S: −2 → 0).
2. Hydrogen is added to Cl₂ (Cl₂ → HCl), so Cl₂ is reduced (Cl: 0 → −1).
3. Thus, H₂S is oxidised and Cl₂ is reduced.

Teacher's Note:
1. Removal of hydrogen means oxidation and addition of hydrogen means reduction.
2. H₂S acts as the reducing agent and Cl₂ as the oxidising agent.

Answer: (a) Oxidation due to removal of potassium

Explanation:
1. In K₄[Fe(CN)₆] → K₃[Fe(CN)₆], one potassium atom (an electropositive element) is removed from each formula unit.
2. Removal of an electropositive element is oxidation. (Fe changes from +2 to +3.)
3. So, oxidation takes place due to removal of potassium.

Teacher's Note:
1. Potassium ferrocyanide is oxidised to potassium ferricyanide by H₂O₂.
2. Removal of an electropositive element is oxidation; removal of an electronegative element is reduction.

Answer: (a) oxidation

Explanation:
1. Oxidation is a process which involves addition of oxygen or an electronegative element to a substance, or removal of hydrogen or an electropositive element from a substance.
2. Here, oxygen is added to Mg (Mg → MgO), so Mg undergoes oxidation (0 → +2).

Teacher's Note:
1. The overall reaction is redox, but the question asks what happens to Mg alone.
2. Mg loses two electrons to form Mg²⁺ in MgO.

Answer: (b) reduction of ethylene

Explanation:
1. Hydrogen is added to ethylene to form ethane.
2. Addition of hydrogen is reduction, so reduction of ethylene occurs.

Teacher's Note:
1. The oxidation number of each carbon changes from −2 in ethene to −3 in ethane, a decrease.
2. Catalytic hydrogenation of alkenes is a reduction reaction.

Answer: (b) number of electrons decreases

Explanation:
1. In oxidation, electrons are lost, so the number of electrons decreases.
2. As a result, the oxidation number increases (not decreases).

Teacher's Note:
1. OIL RIG: Oxidation Is Loss, Reduction Is Gain (of electrons).
2. Loss of electrons and increase in oxidation number always go together.

Answer: (c) Cd

Explanation:
1. Oxidation numbers: Co in Co(OH)₃ = +3 → Co in Co(OH)₂ = +2 (decrease, reduction).
2. Cd = 0 → Cd in Cd(OH)₂ = +2 (increase in oxidation number).
3. Here, the oxidation number of Cd increases from 0 to +2, hence Cd is oxidised.

Teacher's Note:
1. In a battery, the species oxidised acts as the anode material.
2. Co³⁺ is the oxidising agent and Cd is the reducing agent here.

Answer: (c) H₂ + Br₂ → 2HBr

Explanation:
1. In H₂ + Br₂ → 2HBr: H changes from 0 to +1 (oxidised, H₂ is the reducing agent) and Br changes from 0 to −1 (reduced, Br₂ is the oxidising agent).
2. In the other options, neither oxidation nor reduction takes place because the oxidation numbers of all elements remain the same (they are double decomposition / acid–base reactions).

Teacher's Note:
1. A reaction in which an element changes from its free state (zero) to a compound is always redox.
2. Double displacement and neutralisation reactions are generally not redox reactions.

Answer: (b) oxygen and chlorine

Explanation:
1. Oxidation numbers in KClO₃: K = +1, Cl = +5, O = −2. In the products: KCl (K = +1, Cl = −1) and O₂ (O = 0).
2. Oxygen: −2 → 0 (loss of electrons, oxidation).
3. Chlorine: +5 → −1 (gain of electrons, reduction).
4. So, oxygen is oxidised and chlorine is reduced.

Teacher's Note:
1. This is an intramolecular redox reaction: both elements are in the same compound.
2. Potassium stays +1, so it is neither oxidised nor reduced.

Answer: (b) H₂O₂

Explanation:
1. Oxygen in H₂O₂ has oxidation number −1, which can increase (to 0 in O₂) or decrease (to −2 in H₂O).
2. Hence, H₂O₂ can act both as an oxidising agent and as a reducing agent.
3. In KMnO₄ (Mn +7), Fe₂(SO₄)₃ (Fe +3) and K₂Cr₂O₇ (Cr +6) the metal is in a high oxidation state, so they act only as oxidising agents.

Teacher's Note:
1. A species with an element in an intermediate oxidation state can usually act both ways.
2. Other examples are SO₂, HNO₂ and NO₂⁻.

Answer: (a) Electrons are transferred from H to O-atom

Explanation:
1. H: 0 → +1, so hydrogen is oxidised by loss of 1 electron per atom.
2. O: 0 → −2, so oxygen is reduced by gain of 2 electrons per atom.
3. In this reaction, hydrogen has transferred electrons to oxygen.

Teacher's Note:
1. In covalent molecules the transfer is partial: the shared pair shifts towards the more electronegative O atom.
2. This is why oxidation numbers are assigned by giving shared electrons to the more electronegative atom.

Answer: (d) BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl

Explanation:
1. (a) Cu: +2 → 0 and H: 0 → +1, a redox reaction.
2. (b) Fe: +3 → 0 and C: +2 → +4, a redox reaction.
3. (c) K: 0 → +1 and F: 0 → −1, a redox reaction.
4. (d) Ba (+2), Cl (−1), H (+1), S (+6) and O (−2) keep their oxidation numbers; it is a double displacement (precipitation) reaction, not a redox reaction.

Teacher's Note:
1. Precipitation reactions like BaSO₄ formation simply exchange ions.
2. Quick test: if any element appears in its free state on one side only, the reaction is redox.

Answer: (b) A-3, B-4, C-1, D-2

Explanation:
1. A. 2Mg + O₂ → 2MgO: oxygen is added to Mg (oxidation by addition of oxygen) → 3.
2. B. Mg + Cl₂ → MgCl₂: chlorine, an electronegative element, is added to Mg → 4.
3. C. 2H₂S + O₂ → 2S + 2H₂O: hydrogen is removed from H₂S, which is oxidised to S → 1.
4. D. 2KI + H₂O + O₃ → 2KOH + I₂ + O₂: potassium, an electropositive element, is removed from KI, which is oxidised to I₂ → 2.
5. Correct match: A → 3, B → 4, C → 1, D → 2.

Teacher's Note:
1. All four underlined species are oxidised here; only the classical way of describing it differs.
2. Iodide is oxidised to iodine by ozone – this reaction is used to test for ozone.

Answer: (a) K

Explanation:
1. K has only one electron in its outermost shell and a very low ionisation enthalpy, so it has the highest tendency to donate it, i.e. the highest tendency to get oxidised.
2. It also has the most negative standard electrode potential (E° = −2.93 V) among the given metals.
3. Therefore, K is the strongest reducing agent among the given elements.

Teacher's Note:
1. The more negative the E° value of Mⁿ⁺/M, the stronger the metal is as a reducing agent.
2. Ba (E° = −2.91 V) is very close, but K is still slightly stronger.

Answer: (b) reducing agent

Explanation:
1. In the above reaction, Na changes into the Na⁺ ion (0 → +1), i.e. Na donates its electron to the oxygen atom (loss of electron, oxidation).
2. So, sodium behaves as a reducing agent.

Teacher's Note:
1. The species that is oxidised is always the reducing agent.
2. Alkali metals are strong reducing agents because they lose their single valence electron easily.

Answer: (c) C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O

Explanation:
1. In C₆H₆ + HNO₃ → C₆H₅—NO₂ + H₂O, the oxidation number of N changes from +5 (in HNO₃) to +3 (in C₆H₅NO₂), i.e. HNO₃ is reduced (gain of electrons).
2. So, in this reaction HNO₃ behaves as an oxidising agent.
3. In the other reactions (neutralisation of NaOH and Ca(OH)₂, and the reaction with NaCl) the oxidation number of N remains +5, so HNO₃ behaves neither as an oxidising agent nor as a reducing agent.

Teacher's Note:
1. HNO₃ acts as an acid in neutralisation reactions; its oxidising action needs a change in N's oxidation number.
2. In the nitro group N is bonded to C, which is less electronegative, so N is assigned +3.

Answer: (d) Ni

Explanation:
1. Ni: 0 → +2, an increase in oxidation number (oxidation).
2. Cr: +6 (in Cr₂O₇²⁻) → +3, a decrease in oxidation number (reduction).
3. As Ni is oxidised to Ni²⁺ and reduces Cr₂O₇²⁻ to Cr³⁺, Ni acts as the reducing agent.

Teacher's Note:
1. Cr₂O₇²⁻ is the oxidising agent; H⁺ only provides the acidic medium.
2. Electron check: 3 Ni lose 6 electrons, and 2 Cr each gain 3 electrons.

Answer: (b) Both oxidised and reduced

Explanation:
1. Br in Br₂ = 0; Br in HBr = −1 (gain of electron, reduction); Br in HOBr = +1 (loss of electron, oxidation).
2. Here, the oxidation number of bromine increases as well as decreases, i.e. bromine is oxidised as well as reduced.
3. This is a disproportionation reaction.

Teacher's Note:
1. In disproportionation, the same element in one oxidation state goes to a higher and a lower state.
2. Chlorine reacts with water in the same way: Cl₂ + H₂O → HCl + HOCl.

Answer: (c) N₃H

Explanation:
1. Let the oxidation state of N be x in each compound.
2. (a) N₂H₄: 2x + 4(+1) = 0 ⇒ x = −2
3. (b) NH₃: x + 3(+1) = 0 ⇒ x = −3
4. (c) N₃H: 3x + (+1) = 0 ⇒ x = −1/3
5. (d) NH₂OH: x + 2(+1) + (−2) + (+1) = 0 ⇒ x = −1
6. The oxidation state of nitrogen is highest (−1/3) in N₃H.

Teacher's Note:
1. “Highest” compares signed values: −1/3 is greater than −1, −2 and −3.
2. N₃H is hydrazoic acid; −1/3 is only an average value for the three N atoms.

Answer: (b) XY₂Z₆

Explanation:
1. In a neutral compound, the algebraic sum of the oxidation states is always zero.
2. XY₂Z₆: +2 + (5 × 2) + (−2 × 6) = +2 + 10 − 12 = 0 ✔
3. X₂YZ₆: +4 + 5 − 12 = −3; XY₅: +2 + 25 = +27; X₃YZ₄: +6 + 5 − 8 = +3 (none is zero).
4. So, the formula is XY₂Z₆.

Teacher's Note:
1. Always test each option by adding up the oxidation numbers.
2. A real example of this type is Ca(NO₃)₂ = CaN₂O₆.

Answer: (c) Fe(CO)₅

Explanation:
1. (a) K₄Fe(CN)₆: 4(+1) + x + 6(−1) = 0 ⇒ x = +2
2. (b) Fe₃O₄ = FeO·Fe₂O₃ ⇒ Fe is +2 and +3 (average +8/3)
3. (c) Fe(CO)₅: x + 5(0) = 0 ⇒ x = 0
4. (d) FeSO₄(NH₄)₂SO₄·6H₂O: x + (−2) + 0 + 0 = 0 ⇒ x = +2
5. Thus, in Fe(CO)₅, Fe shows the minimum oxidation number (zero).

Teacher's Note:
1. CO is a neutral ligand, so in metal carbonyls the metal is usually in zero oxidation state.
2. Mohr’s salt, FeSO₄(NH₄)₂SO₄·6H₂O, contains Fe²⁺.

Answer: (d) H₂SO₅ > H₂SO₃ > SCl₂ > H₂S

Explanation:
1. (a) Na₂S₄O₆ (+2.5), H₂S₂O₇ (+6), Na₂S₂O₃ (+2), S₈ (0) – not in decreasing order.
2. (b) H₂SO₄ (+6), SO₂ (+4), H₂S (−2), H₂S₂O₈ (+6) – not in decreasing order.
3. (c) SO₂²⁻ (+2), SO₄²⁻ (+6), SO₃²⁻ (+4), HSO₄⁻ (+6) – not in decreasing order.
4. (d) H₂SO₅ (+6), H₂SO₃ (+4), SCl₂ (+2), H₂S (−2) – decreasing order.
5. Thus, in option (d), the compounds are arranged in the decreasing order of oxidation number of sulphur.

Teacher's Note:
1. In peroxo acids like H₂SO₅ and H₂S₂O₈, S is +6 because the peroxide oxygens are −1.
2. Sulphur shows oxidation states from −2 to +6.

Answer: (b) NH₄NO₃

Explanation:
1. NH₄NO₃ is an ionic compound made of NH₄⁺ and NO₃⁻ ions, and the oxidation number of nitrogen in the two ions is different.
2. In NH₄⁺: x + 4(+1) = +1 ⇒ x = −3
3. In NO₃⁻: x + 3(−2) = −1 ⇒ x = +5
4. In NH₂OH (−1), N₂H₄ (−2) and N₃H (−1/3 average) all N atoms are given a single value.

Teacher's Note:
1. Always split ionic compounds into their ions before finding oxidation numbers.
2. On heating, NH₄NO₃ gives N₂O (N = +1): the −3 and +5 nitrogens meet in the middle.

Answer: (b) 0 and +2

Explanation:
1. Let the oxidation number of the carbonyl carbon in methanal (HCHO) and methanoic acid (HCOOH) be x and y respectively.
2. In HCHO: 2(+1) + x + (−2) = 0 ⇒ x = 0
3. In HCOOH: 2(+1) + y + 2(−2) = 0 ⇒ y = +2

Teacher's Note:
1. Going from methanal (0) to methanoic acid (+2) is an oxidation, as expected for aldehyde → acid.
2. Carbon can show any oxidation state from −4 (CH₄) to +4 (CO₂).

Answer: (b) Fe (26)

Explanation:
1. Electronic configurations in the +3 state:
2. The +3 oxidation state of Fe has extra stability due to half-filled d-orbitals (3d⁵).
3. Hence, +3 oxidation state is most common in Fe.

Teacher's Note:
1. Half-filled (d⁵) and completely filled (d¹⁰) configurations are extra stable.
2. Zn shows only +2 (3d¹⁰), and Cu commonly shows +1 and +2.

Answer: (b) reducing agent

Explanation:
1. Na: 0 → +1 in Na₂O (loss of electron, oxidation).
2. In this reaction, Na converts into the Na⁺ ion, i.e. Na donates its electron to the oxygen atom. So, it behaves as a reducing agent.

Teacher's Note:
1. Oxygen gains electrons (0 → −2) and is the oxidising agent.
2. A reducing agent is itself oxidised.

Answer: (d) 0, +1, +5

Explanation:
1. The oxidation number of Cl in the molecular state (Cl₂) is zero.
2. In NaOCl: +1 + (−2) + x = 0 ⇒ x = +1
3. In ClO₃⁻: x + 3(−2) = −1 ⇒ x = +5

Teacher's Note:
1. NaOCl (sodium hypochlorite) is the active component of household bleach.
2. Chlorine shows oxidation states −1, 0, +1, +3, +5 and +7.

Answer: (a) +\( \frac{4}{3} \)

Explanation:
1. Structure of carbon suboxide (C₃O₂) with oxidation states:
2. In C₃O₂, the two C-atoms linked with oxygen atoms are in +2 oxidation state and the central carbon is in zero oxidation state.
3. So, the average oxidation state of carbon = \( \frac{2 + 0 + 2}{3} = +\frac{4}{3} \). (Also from the formula: 3x + 2(−2) = 0 ⇒ x = +4/3.)

Teacher's Note:
1. A fractional oxidation number is only an average; no single atom has it.
2. Similar cases are Br₃O₈ (16/3) and S₄O₆²⁻ (2.5).

Answer: (a) +6, +4, +6

Explanation:
1. H₂SO₄: 2(+1) + x + 4(−2) = 0 ⇒ x = +6
2. HSO₃⁻: (+1) + x + 3(−2) = −1 ⇒ x = +4
3. SO₂Cl₂: x + 2(−2) + 2(−1) = 0 ⇒ x = +6

Teacher's Note:
1. In SO₂Cl₂ (sulphuryl chloride), both O and Cl are more electronegative than S, so both count as negative.
2. In an ion, equate the sum to the charge, not to zero.

Answer: (d) 3d⁵ 4s²

Explanation:
1. Highest oxidation number of a transition element = number of (n − 1)d electrons + number of ns electrons.
2. (a) 3d¹ 4s² = 1 + 2 = 3
3. (b) 3d³ 4s² = 3 + 2 = 5
4. (c) 3d⁵ 4s¹ = 5 + 1 = 6
5. (d) 3d⁵ 4s² = 5 + 2 = 7
6. Thus, option (d) is correct.

Teacher's Note:
1. 3d⁵ 4s² is manganese, which shows +7 in KMnO₄.
2. This rule works up to Mn in the 3d series; after Mn the maximum oxidation state falls.

Answer: (d) Mg₃X₂

Explanation:
1. X (1s² 2s² 2p³) is nitrogen; it needs 3 electrons to complete its octet, so its valency is 3 (X³⁻).
2. The valency of Mg is 2 (Mg²⁺).
3. Balancing the charges, 3(+2) + 2(−3) = 0, so the ionic compound has the formula Mg₃X₂.

Teacher's Note:
1. The real compound is magnesium nitride, Mg₃N₂, formed when Mg burns in air.
2. Cross the valencies: Mg²⁺ and X³⁻ give Mg₃X₂.

Answer: (a) 3

Explanation:
1. Let the oxidation state of Cr in [Cr(H₂O)₄Cl₂]⁺ be x.
2. x + 4(0) + 2(−1) = +1 ⇒ x = +3
3. So, the valency (oxidation state) of Cr in the ion is 3.

Teacher's Note:
1. H₂O is a neutral ligand and Cl⁻ carries −1 charge.
2. Here “valency” means the primary valency (oxidation state); the coordination number of Cr is 6.

Answer: (c) A-2, B-4, C-1, D-3

Explanation:
1. NH₂OH: x + 2(+1) + (−2) + (+1) = 0 ⇒ x = −1 → 2
2. Mg₃N₂: 3(+2) + 2x = 0 ⇒ x = −3 → 4
3. N₂O: 2x + (−2) = 0 ⇒ x = +1 → 1
4. N₂O₅: 2x + 5(−2) = 0 ⇒ x = +5 → 3
5. Correct match: A → 2, B → 4, C → 1, D → 3.

Teacher's Note:
1. Nitrogen shows every oxidation state from −3 (NH₃, nitrides) to +5 (N₂O₅, HNO₃).
2. N₂O is nitrous oxide (laughing gas), with N in +1 state.

Answer: (d) +2 and +1 respectively

Explanation:
1. Fluorine is more electronegative than oxygen, so F is always assigned −1.
2. OF₂: x + 2(−1) = 0 ⇒ oxidation number of O = +2
3. O₂F₂: 2x + 2(−1) = 0 ⇒ oxidation number of O = +1

Teacher's Note:
1. These are the only common compounds in which oxygen has a positive oxidation number.
2. Fluorine is −1 in all its compounds because it is the most electronegative element.

Answer: (a) CrO₂⁻, ClO₃⁻, CrO₄²⁻, MnO₄⁻

Explanation:
1. Oxidation numbers of the central atoms: Cr in CrO₂⁻ = +3; Cl in ClO₃⁻ = +5; Cr in CrO₄²⁻ = +6; Mn in MnO₄⁻ = +7.
2. (a) +3, +5, +6, +7 – increasing
3. (b) +5, +6, +7, +3; (c) +3, +5, +7, +6; (d) +6, +7, +3, +5 – not increasing
4. Only in arrangement (a) does the oxidation number of the central atom increase from left to right.

Teacher's Note:
1. For an oxoanion XO_n^m−: ON of X = 2n − m.
2. Work out each value once and then compare the sequences.

Answer: (c) MnO₄⁻(aq) + I⁻(aq) → MnO₂(s) + I₂(s)

Explanation:
1. In MnO₄⁻(aq) + I⁻(aq) → MnO₂(s) + I₂(s), Mn changes from +7 to +4 and I from −1 to 0.
2. This reaction is represented in basic medium because in basic (or neutral) medium MnO₄⁻ is reduced to MnO₂ (Mn⁷⁺ to Mn⁴⁺), while in acidic medium MnO₄⁻ is reduced to Mn²⁺.
3. Formation of Mn²⁺, Fe³⁺ (from H₂O₂) and Cr³⁺ in the other options shows acidic medium.
4. Balanced form: 2MnO₄⁻ + 6I⁻ + 4H₂O → 2MnO₂ + 3I₂ + 8OH⁻

Teacher's Note:
1. Product of MnO₄⁻: Mn²⁺ in acid, MnO₂ in neutral/weakly basic, MnO₄²⁻ in strongly basic medium.
2. Cr³⁺ and Mn²⁺ ions are the typical products in acidic solution.

Answer: (a) 3, 8, 2

Explanation:
1. U: +4 (in UO²⁺) → +6 (in UO₂²⁺), increase of 2 per U.
2. Cr: +6 → +3, decrease of 3 per Cr, i.e. 6 for Cr₂O₇²⁻. So, 3 UO²⁺ are needed per Cr₂O₇²⁻.
3. The balanced equation is 3UO²⁺ + Cr₂O₇²⁻ + 8H⁺ → 3UO₂²⁺ + 2Cr³⁺ + 4H₂O
4. Check: O = 3 + 7 = 10 = 6 + 4; charge = 6 − 2 + 8 = +12 = 6 + 6.
5. Hence, x, y and z are 3, 8 and 2 respectively.

Teacher's Note:
1. Balance the electron change first, then O with H₂O and H with H⁺.
2. Always check both atoms and total charge after balancing.

Answer: (c) Cl

Explanation:
1. When the mixture is heated, the chlorate is reduced and oxalic acid is oxidised:
2. KClO₃ + H₂C₂O₄ + H₂SO₄ → K₂SO₄ + KCl + CO₂ + H₂O (unbalanced)
3. Oxidation numbers: K +1, Cl +5 (in KClO₃) → −1 (in KCl); C +3 (in H₂C₂O₄) → +4 (in CO₂); H stays +1; S stays +6; O stays −2.
4. Thus, Cl undergoes the maximum change in oxidation number (from +5 to −1, a change of 6).

Teacher's Note:
1. S in H₂SO₄ stays +6 – sulphuric acid only provides the acidic medium here.
2. Even if chlorate is reduced only to ClO₂ (+4), C changes by only 1 per atom.