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NEET Chemistry: Solutions, Redox, Electrochemistry and Kinetics · Chapter 2

Redox Reactions

What to remember

  • Oxidation is loss of electrons (increase in oxidation number); reduction is gain of electrons (decrease in oxidation number). Both always occur together.
  • Oxidising agent gains electrons and is itself reduced; reducing agent loses electrons and is itself oxidised.
  • In a balanced redox equation, total electrons lost equals total electrons gained. Use the oxidation-number method or the half-reaction method.
  • A species with a more negative standard electrode potential is a stronger reducing agent; one with a more positive value is a stronger oxidising agent. A metal above hydrogen in the electrochemical series displaces hydrogen from dilute acids.

Classical and electron-transfer ideas

Classically, oxidation meant addition of oxygen or an electronegative element, or removal of hydrogen or an electropositive element. Reduction meant the reverse. For example, in 2Mg + O₂ → 2MgO, magnesium is oxidised and oxygen is reduced.

In the electron-transfer view, 2Na + Cl₂ → 2NaCl is split into two half-reactions:

  • Na → Na⁺ + e⁻ (oxidation, loss of electrons)
  • Cl₂ + 2e⁻ → 2Cl⁻ (reduction, gain of electrons)

Sodium is the reducing agent. Chlorine is the oxidising agent. A reaction in which both processes occur together is a redox reaction.

Oxidation number and its rules

Oxidation number is the charge an atom would have if all bonding electrons were assigned to the more electronegative atom.

  • 1. An element in its free state (Na, O₂, P₄, S₈) has oxidation number 0.
  • 2. For a monatomic ion, the oxidation number equals its charge.
  • 3. In a neutral compound the sum of oxidation numbers is 0; in an ion it equals the charge on the ion.
  • 4. Fluorine is always −1 in compounds.
  • 5. Oxygen is usually −2. Exceptions: peroxides (H₂O₂, Na₂O₂) −1; superoxides (KO₂) −1/2; OF₂ +2.
  • 6. Hydrogen is usually +1. In metal hydrides (NaH, CaH₂) it is −1.
  • 7. Alkali metals are +1 and alkaline earth metals are +2 in compounds.
SpeciesElementOxidation numberWorking
K₂Cr₂O₇Cr+62(+1) + 2x + 7(−2) = 0
KMnO₄Mn+7+1 + x − 8 = 0
MnO₄²⁻Mn+6x − 8 = −2
H₂O₂O−12(+1) + 2x = 0
KO₂O−1/2+1 + 2x = 0
OF₂O+2x + 2(−1) = 0
Fe₃O₄Fe+8/3 (average)3x + 4(−2) = 0
S₄O₆²⁻S+2.5 (average)4x + 6(−2) = −2
CrO₅Cr+6See below

Fractional oxidation numbers are averages. In Fe₃O₄, iron is present as Fe²⁺ and Fe³⁺ (one Fe²⁺ and two Fe³⁺ per formula unit). In S₄O₆²⁻ (tetrathionate) two sulphur atoms are at 0 and two are at +5, giving an average of +2.5. CrO₅ has a butterfly structure with one oxide oxygen and two peroxide linkages (four oxygen atoms at −1 each). So x + 4(−1) + (−2) = 0, which gives chromium +6. A value of +10 is wrong, because chromium cannot go above +6.

Oxidising and reducing agents

  • A species in its highest oxidation state can only be an oxidising agent (KMnO₄, K₂Cr₂O₇, HNO₃, H₂SO₄ concentrated, F₂).
  • A species in its lowest oxidation state can only be a reducing agent (metals, I⁻, H₂S, NH₃).
  • A species in an intermediate state can act as both. H₂O₂ (oxygen −1, can go to −2 or to 0) can be an oxidising agent and a reducing agent. SO₂ and HNO₂ also act as both.
  • Fluorine is the strongest oxidising agent; lithium is the strongest reducing agent in aqueous solution.

Types of redox reactions

TypeDescriptionExample
CombinationTwo or more substances form one product2Na + Cl₂ → 2NaCl; C + O₂ → CO₂
DecompositionOne compound breaks into two or more, with a change in oxidation number2KClO₃ → 2KCl + 3O₂ (heat)
Displacement: metalA more reactive metal displaces a less reactive metal from its saltZn + CuSO₄ → ZnSO₄ + Cu
Displacement: metal and hydrogenA metal above hydrogen displaces hydrogen from acid or waterZn + 2HCl → ZnCl₂ + H₂
Displacement: non-metalA more reactive halogen displaces a less reactive halideCl₂ + 2Br⁻ → 2Cl⁻ + Br₂
DisproportionationThe same element is oxidised and reduced at once2H₂O₂ → 2H₂O + O₂; P₄ + 3OH⁻ + 3H₂O → PH₃ + 3H₂PO₂⁻; Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O

In Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O (cold, dilute alkali), chlorine changes from 0 to −1 and +1. In 2H₂O₂ → 2H₂O + O₂, oxygen goes from −1 to −2 and to 0.

Decomposition of a compound such as CaCO₃ → CaO + CO₂ is not redox, because no oxidation number changes. Displacement of hydrogen by a metal: the metal is oxidised and hydrogen of the acid is reduced.

Balancing redox equations

Oxidation-number method. (1) Write the skeleton equation. (2) Assign oxidation numbers and find the atoms that change. (3) Calculate the increase and the decrease per formula unit. (4) Multiply so that total increase equals total decrease. (5) Balance the remaining atoms, first other elements, then oxygen with H₂O and hydrogen with H⁺ (acid) or with OH⁻ and H₂O (base).

Worked example 1. Fe₂O₃ + CO → Fe + CO₂. Iron goes from +3 to 0: a decrease of 3 per Fe, so 6 for two Fe. Carbon goes from +2 to +4: an increase of 2. Multiply CO and CO₂ by 3. Result: Fe₂O₃ + 3CO → 2Fe + 3CO₂.

Half-reaction (ion-electron) method. (1) Split into oxidation and reduction half-reactions. (2) Balance atoms other than O and H. (3) Balance O with H₂O. (4) Balance H with H⁺. (5) Balance charge with electrons. (6) Multiply to equalise electrons and add. (7) In basic medium, add OH⁻ equal in number to H⁺ on both sides, combine H⁺ and OH⁻ into H₂O and cancel.

Worked example 2 (acidic medium). MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺.

  • Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
  • Oxidation: Fe²⁺ → Fe³⁺ + e⁻ (multiply by 5)
  • Sum: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

Dichromate with iron(II) in acid: Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O. Here six electrons are transferred.

Worked example 3 (basic medium). MnO₄⁻ + I⁻ → MnO₂ + I₂. Mn goes from +7 to +4 (3 electrons); 2I⁻ → I₂ loses 2 electrons. Taking 2 MnO₄⁻ and 6 I⁻ gives 6 electrons each way. Balanced: 2MnO₄⁻ + 6I⁻ + 4H₂O → 2MnO₂ + 3I₂ + 8OH⁻. Check charge: −8 on both sides. Check oxygen: 12 on both sides.

Redox titrations (in brief)

Redox reactions allow titrations in which the oxidising agent and reducing agent react in a fixed mole ratio.

TitrantMediumReduced productElectrons per formula unitNote
KMnO₄Acid (dilute H₂SO₄)Mn²⁺5Self-indicator: pink colour persists at the end point
KMnO₄Neutral or basicMnO₂3
K₂Cr₂O₇Acid2Cr³⁺6Needs an external or internal indicator
I₂Neutral2I⁻2Titrated against thiosulphate with starch

Iodine with thiosulphate: I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻. One mole of iodine reacts with two moles of thiosulphate.

Hydrochloric acid is not used to acidify permanganate, because KMnO₄ oxidises Cl⁻ to Cl₂ and extra permanganate is used.

Mole ratios from the equations above: 1 mol MnO₄⁻ oxidises 5 mol Fe²⁺; 1 mol Cr₂O₇²⁻ oxidises 6 mol Fe²⁺.

Redox reactions and electrode processes

In Zn + Cu²⁺ → Zn²⁺ + Cu, electrons move directly from zinc to copper ions. If the two half-reactions occur in separate vessels connected by a wire and a salt bridge, the chemical energy appears as electrical energy. This arrangement is the Daniell cell.

  • Anode (oxidation): Zn → Zn²⁺ + 2e⁻
  • Cathode (reduction): Cu²⁺ + 2e⁻ → Cu
  • Electrons flow through the wire from the zinc electrode to the copper electrode. The salt bridge completes the circuit and keeps solutions neutral.

Standard electrode potential (E°) is the potential of a half-cell measured against the standard hydrogen electrode (SHE) under standard conditions: 1 M concentration of ions, gases at 1 bar pressure and temperature 298 K. The SHE is a platinum electrode coated with platinum black, dipping in 1 M H⁺ with hydrogen gas at 1 bar. Its potential is taken as 0.00 V.

E°(cell) = E°(cathode) − E°(anode), using reduction potentials. For the Daniell cell: 0.34 − (−0.76) = 1.10 V. A positive E°(cell) means that the reaction is feasible as written.

Electrochemical series

Half-reaction (reduction)E° (V)
F₂ + 2e⁻ → 2F⁻+2.87
Cl₂ + 2e⁻ → 2Cl⁻+1.36
Br₂ + 2e⁻ → 2Br⁻+1.09
Ag⁺ + e⁻ → Ag+0.80
I₂ + 2e⁻ → 2I⁻+0.54
Cu²⁺ + 2e⁻ → Cu+0.34
2H⁺ + 2e⁻ → H₂0.00
Fe²⁺ + 2e⁻ → Fe−0.44
Zn²⁺ + 2e⁻ → Zn−0.76
Mg²⁺ + 2e⁻ → Mg−2.36
Na⁺ + e⁻ → Na−2.71
Ca²⁺ + 2e⁻ → Ca−2.87
K⁺ + e⁻ → K−2.93
Li⁺ + e⁻ → Li−3.05

Uses of the series:

  • The higher in the series (more positive E°), the stronger the oxidising agent of the species on the left; the lower (more negative), the stronger the reducing agent of the metal on the right.
  • A metal with lower E° displaces a metal with higher E° from its salt solution. Zn displaces Cu²⁺, and Fe displaces Cu²⁺. Copper displaces Ag⁺ but cannot displace Zn²⁺ or Fe²⁺.
  • Metals with negative E° (above hydrogen) liberate hydrogen from dilute acids; Cu and Ag (E° positive) do not.
  • A halogen with a higher E° displaces a halogen with a lower E° from its halide: Cl₂ displaces Br⁻ and I⁻.

NEET traps

  • 1. Oxidation is loss of electrons (OIL RIG); do not reverse it.
  • 2. The oxidising agent is reduced; the reducing agent is oxidised.
  • 3. Oxygen is −1 in H₂O₂, −1/2 in KO₂ and +2 in OF₂, not always −2.
  • 4. Hydrogen is −1 in NaH and CaH₂.
  • 5. Fractional oxidation numbers (Fe₃O₄, S₄O₆²⁻) are averages.
  • 6. CrO₅ has chromium +6, not +10, because of two peroxide linkages.
  • 7. H₂O₂ acts as both an oxidising and a reducing agent because oxygen is at the intermediate state −1.
  • 8. In Zn + CuSO₄ → ZnSO₄ + Cu, the reaction is displacement, not disproportionation.
  • 9. A more negative E° means a stronger reducing agent, not a stronger oxidising agent.
  • 10. In acid, MnO₄⁻ goes to Mn²⁺ (5 electrons); in neutral or basic medium it goes to MnO₂ (3 electrons).

One-liners

  • 1. Oxidation number of an element in its free state is zero.
  • 2. Oxidation number of fluorine in compounds is always −1.
  • 3. Oxidation number of oxygen in peroxides is −1.
  • 4. Oxidation number of oxygen in superoxides is −1/2.
  • 5. Oxidation number of hydrogen in metal hydrides is −1.
  • 6. Mn is +7 in KMnO₄; Cr is +6 in K₂Cr₂O₇.
  • 7. KMnO₄ in acid becomes Mn²⁺ and gains 5 electrons.
  • 8. K₂Cr₂O₇ in acid becomes Cr³⁺ and gains 6 electrons.
  • 9. Disproportionation: the same element is oxidised and reduced.
  • 10. SHE has E° = 0.00 V.
  • 11. E°(cell) = E°(cathode) − E°(anode).
  • 12. Daniell cell E°(cell) = 1.10 V.
  • 13. In the Daniell cell, zinc is the anode and copper is the cathode.
  • 14. Fluorine is the strongest oxidising agent (+2.87 V).
  • 15. Lithium has the lowest E° (−3.05 V), so it is the strongest reducing agent.

Practice questions

  1. What is the oxidation number of chromium in K₂Cr₂O₇?

    1. +6
    2. +3
    3. +5
    4. +7
    Answer

    A. +6

    2(+1) + 2x + 7(−2) = 0, so x = +6.

  2. What is the average oxidation number of sulphur in the tetrathionate ion, S₄O₆²⁻?

    1. +2
    2. +3
    3. +2.5
    4. +5
    Answer

    C. +2.5

    4x + 6(−2) = −2, so x = +10/4 = +2.5.

  3. What is the oxidation number of oxygen in potassium superoxide, KO₂?

    1. −1
    2. −1/2
    3. −2
    4. +1/2
    Answer

    B. −1/2

    K is +1, so the two oxygen atoms together are −1, giving −1/2 each.

  4. What is the average oxidation number of iron in Fe₃O₄?

    1. +2
    2. +3
    3. +4
    4. +8/3
    Answer

    D. +8/3

    3x + 4(−2) = 0, so x = +8/3. It is a mixture of one Fe²⁺ and two Fe³⁺.

  5. CrO₅ contains two peroxide linkages. What is the oxidation number of chromium in it?

    1. +5
    2. +4
    3. +6
    4. +10
    Answer

    C. +6

    Four oxygen atoms are at −1 (peroxide) and one at −2: x − 4 − 2 = 0, so x = +6.

  6. What is the oxidation number of manganese in the manganate ion, MnO₄²⁻?

    1. +4
    2. +6
    3. +7
    4. +2
    Answer

    B. +6

    x + 4(−2) = −2, so x = +6.

  7. In which of the following compounds does oxygen have a positive oxidation number?

    1. H₂O₂
    2. Na₂O₂
    3. KO₂
    4. OF₂
    Answer

    D. OF₂

    Fluorine is more electronegative than oxygen, so oxygen is +2 in OF₂. It is −1 in peroxides and −1/2 in KO₂.

  8. Which of the following is a disproportionation reaction?

    1. Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O
    2. 2Na + Cl₂ → 2NaCl
    3. CaCO₃ → CaO + CO₂
    4. Zn + CuSO₄ → ZnSO₄ + Cu
    Answer

    A. Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O

    Chlorine (0) is converted to Cl⁻ (−1) and ClO⁻ (+1): the same element is oxidised and reduced.

  9. Given E° values F₂/F⁻ = +2.87 V, Cl₂/Cl⁻ = +1.36 V, Br₂/Br⁻ = +1.09 V and I₂/I⁻ = +0.54 V, which is the strongest oxidising agent?

    1. Cl₂
    2. Br₂
    3. F₂
    4. I₂
    Answer

    C. F₂

    The species with the most positive E° is the strongest oxidising agent.

  10. E° for Zn²⁺/Zn is −0.76 V and for Cu²⁺/Cu is +0.34 V. What is E°(cell) of the Daniell cell?

    1. +0.76 V
    2. +1.10 V
    3. +0.42 V
    4. −1.10 V
    Answer

    B. +1.10 V

    E°(cell) = E°(cathode) − E°(anode) = 0.34 − (−0.76) = 1.10 V.

  11. In acidic medium, how many moles of KMnO₄ are needed to oxidise 1 mole of FeSO₄?

    1. 1.0 mol
    2. 0.5 mol
    3. 0.167 mol
    4. 0.2 mol
    Answer

    D. 0.2 mol

    MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O: 1 mol MnO₄⁻ per 5 mol Fe²⁺, so 1/5 = 0.2 mol.

  12. How many electrons does one MnO₄⁻ ion gain when it is reduced to MnO₂ in basic medium?

    1. 1
    2. 5
    3. 3
    4. 7
    Answer

    C. 3

    Mn changes from +7 to +4, a gain of 3 electrons.

  13. Iodine reacts with thiosulphate as I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻. How many moles of thiosulphate react with 0.01 mol of iodine?

    1. 0.02 mol
    2. 0.04 mol
    3. 0.01 mol
    4. 0.005 mol
    Answer

    A. 0.02 mol

    The mole ratio I₂ : S₂O₃²⁻ is 1 : 2, so 0.01 × 2 = 0.02 mol.

  14. Which of the following metals does not liberate hydrogen from dilute hydrochloric acid?

    1. Magnesium
    2. Iron
    3. Zinc
    4. Copper
    Answer

    D. Copper

    Copper has a positive standard electrode potential (+0.34 V); the other metals have negative values and lie above hydrogen in the series.

  15. Which of the following is NOT a feature of the standard hydrogen electrode?

    1. Hydrogen gas at 2 bar pressure
    2. 1 M concentration of H⁺ ions
    3. Platinum electrode coated with platinum black
    4. Hydrogen gas at 1 bar pressure
    Answer

    A. Hydrogen gas at 2 bar pressure

    The standard hydrogen electrode uses hydrogen at 1 bar and 1 M H⁺ with a platinum black electrode.

  16. In the reaction Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O, how many moles of water form per mole of dichromate used?

    1. 6
    2. 7
    3. 14
    4. 2
    Answer

    B. 7

    The balanced equation gives 7 mol of H₂O for each mol of Cr₂O₇²⁻.

  17. Given below are two statements: Statement I: The oxidation number of an element in its free state is zero. Statement II: Hydrogen has oxidation number −1 in sodium hydride (NaH).

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    Both are correct: H is −1 when bonded to a more electropositive metal.

  18. Given below are two statements: Statement I: Fluorine shows only the oxidation number −1 in its compounds. Statement II: Oxygen has oxidation number −2 in all of its compounds.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    Oxygen is −1 in peroxides, −1/2 in superoxides and +2 in OF₂, so Statement II is incorrect.

  19. Given below are two statements: Statement I: Oxidation is the gain of electrons. Statement II: Reduction is a decrease in oxidation number.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    Oxidation is loss of electrons, so Statement I is incorrect. Statement II is correct.

  20. Given below are two statements: Statement I: The reaction Zn + CuSO₄ → ZnSO₄ + Cu is a disproportionation reaction. Statement II: In this reaction zinc is reduced.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    B. Both Statement I and Statement II are incorrect

    It is a displacement reaction, and zinc is oxidised (0 to +2). Both statements are incorrect.

  21. Given below are two statements: Statement I: In a disproportionation reaction the same element is oxidised and reduced at the same time. Statement II: Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O is an example of a disproportionation reaction.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    Chlorine goes from 0 to −1 and +1. Both are correct.

  22. Which of the following statements are correct? 1. A metal with a more negative E° is a stronger reducing agent. 2. A metal with a lower E° can displace a metal of higher E° from its salt solution. 3. Copper can displace zinc from zinc sulphate solution.

    1. 1 and 3 only
    2. 2 and 3 only
    3. 1 and 2 only
    4. 1, 2 and 3
    Answer

    C. 1 and 2 only

    Copper (+0.34 V) has a higher E° than zinc (−0.76 V), so it cannot displace zinc. Statement 3 is wrong.

  23. Given below are two statements: Statement I: The potential of the standard hydrogen electrode is taken as zero. Statement II: A metal with a negative E° is a stronger oxidising agent than H⁺.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    Metals with negative E° are stronger reducing agents than H₂; Statement II is incorrect.

  24. Given below are two statements: Statement I: Hydrogen peroxide acts only as an oxidising agent. Statement II: Oxygen in hydrogen peroxide is in an intermediate oxidation state, −1.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    Oxygen at −1 can be lowered to −2 or raised to 0, so H₂O₂ is both an oxidising and a reducing agent.

  25. Which of the following statements are correct? 1. Oxygen has oxidation number −1/2 in KO₂. 2. Oxygen has oxidation number −2 in Na₂O₂. 3. Oxygen has oxidation number +2 in OF₂.

    1. 1 and 2 only
    2. 2 and 3 only
    3. 1, 2 and 3
    4. 1 and 3 only
    Answer

    D. 1 and 3 only

    Na₂O₂ is a peroxide, so oxygen is −1. Statements 1 and 3 are correct.

  26. Given below are two statements: Statement I: In acidic medium, the permanganate ion is reduced to MnO₂. Statement II: In acidic medium, the dichromate ion is reduced to Cr²⁺.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    B. Both Statement I and Statement II are incorrect

    In acid, MnO₄⁻ forms Mn²⁺ and Cr₂O₇²⁻ forms Cr³⁺. Both statements are incorrect.

  27. Assertion (A): Fluorine shows only the oxidation number −1 in its compounds. Reason (R): Fluorine is the most electronegative element.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    Fluorine always attracts the bonding electrons, so it is −1; R explains A.

  28. Assertion (A): Chlorine displaces bromide ions from potassium bromide solution. Reason (R): Bromine is a liquid at room temperature.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    B. Both A and R are true but R is not the correct explanation of A

    A is true because Cl₂ has a higher E° than Br₂. R is true but does not explain A.

  29. Assertion (A): The oxidation number of chromium in CrO₅ is +6. Reason (R): CrO₅ contains one peroxide linkage.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    C. A is true but R is false

    CrO₅ has two peroxide linkages, so R is false; A is true.

  30. Assertion (A): Silver reacts with dilute hydrochloric acid to liberate hydrogen. Reason (R): Silver has a positive standard electrode potential.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    D. A is false but R is true

    Positive E° means silver cannot displace hydrogen, so A is false; R is true.

  31. Assertion (A): Potassium permanganate titrations are not carried out in the presence of hydrochloric acid. Reason (R): Potassium permanganate oxidises chloride ions to chlorine.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    Some permanganate is used up on chloride, which gives a wrong end point; R explains A.

  32. Assertion (A): In the Daniell cell, the zinc electrode is the anode. Reason (R): Zinc undergoes oxidation at that electrode.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    Anode is where oxidation occurs: Zn → Zn²⁺ + 2e⁻.

  33. Match List I with List II. List I: a. Sulphur in S₄O₆²⁻ b. Iron in Fe₃O₄ c. Oxygen in KO₂ d. Chromium in CrO₅ List II: i. +6 ii. −1/2 iii. +2.5 iv. +8/3

    1. a-ii, b-iv, c-iii, d-i
    2. a-iii, b-iv, c-ii, d-i
    3. a-iii, b-iv, c-i, d-ii
    4. a-iv, b-iii, c-ii, d-i
    Answer

    B. a-iii, b-iv, c-ii, d-i

    S: +2.5; Fe: +8/3; O in superoxide: −1/2; Cr in CrO₅: +6.

  34. Match List I with List II. List I: a. Combination b. Decomposition c. Displacement d. Disproportionation List II: i. 2KClO₃ → 2KCl + 3O₂ ii. Zn + CuSO₄ → ZnSO₄ + Cu iii. 2Na + Cl₂ → 2NaCl iv. P₄ + 3OH⁻ + 3H₂O → PH₃ + 3H₂PO₂⁻

    1. a-iii, b-i, c-iv, d-ii
    2. a-iii, b-ii, c-i, d-iv
    3. a-iii, b-i, c-ii, d-iv
    4. a-i, b-iii, c-ii, d-iv
    Answer

    C. a-iii, b-i, c-ii, d-iv

    Two elements form one product (iii); one compound gives two (i); a metal replaces a metal (ii); phosphorus is both oxidised and reduced (iv).

  35. Match List I with List II. List I: a. E° of F₂/F⁻ b. E° of the standard hydrogen electrode c. E°(cell) of the Daniell cell d. E° of Cu²⁺/Cu List II: i. +1.10 V ii. 0.00 V iii. +2.87 V iv. +0.34 V

    1. a-iii, b-ii, c-iv, d-i
    2. a-ii, b-iii, c-i, d-iv
    3. a-i, b-ii, c-iii, d-iv
    4. a-iii, b-ii, c-i, d-iv
    Answer

    D. a-iii, b-ii, c-i, d-iv

    F₂/F⁻ = +2.87 V; SHE = 0.00 V; Daniell cell = 1.10 V; Cu²⁺/Cu = +0.34 V.

  36. Match List I with List II. List I: a. Reduced product of KMnO₄ in acid b. Reduced product of K₂Cr₂O₇ in acid c. Oxidation product of thiosulphate with iodine d. Reduced product of KMnO₄ in neutral or basic medium List II: i. Cr³⁺ ii. Mn²⁺ iii. MnO₂ iv. S₄O₆²⁻

    1. a-i, b-ii, c-iv, d-iii
    2. a-ii, b-i, c-iv, d-iii
    3. a-ii, b-i, c-iii, d-iv
    4. a-iii, b-i, c-iv, d-ii
    Answer

    B. a-ii, b-i, c-iv, d-iii

    MnO₄⁻ gives Mn²⁺ in acid and MnO₂ in neutral/basic medium; dichromate gives Cr³⁺; thiosulphate gives S₄O₆²⁻.

  37. Match List I with List II. List I: a. MnO₄⁻ → Mn²⁺ (number of electrons gained) b. Cr₂O₇²⁻ → 2Cr³⁺ (electrons gained) c. Fe²⁺ → Fe³⁺ (electrons lost) d. I₂ → 2I⁻ (electrons gained) List II: i. 6 ii. 1 iii. 2 iv. 5

    1. a-i, b-iv, c-ii, d-iii
    2. a-iv, b-ii, c-i, d-iii
    3. a-iv, b-i, c-ii, d-iii
    4. a-iv, b-i, c-iii, d-ii
    Answer

    C. a-iv, b-i, c-ii, d-iii

    Mn +7 to +2 gives 5; two Cr +6 to +3 gives 6; Fe +2 to +3 gives 1; I₂ to 2I⁻ gives 2.

  38. Match List I with List II. List I: a. Metal with the lowest E° (−3.05 V) b. Inert electrode material in the standard hydrogen electrode c. Does not liberate hydrogen from dilute HCl d. Anode of the Daniell cell List II: i. Cu ii. Zn iii. Li iv. Pt

    1. a-iii, b-i, c-iv, d-ii
    2. a-iii, b-iv, c-i, d-ii
    3. a-iii, b-iv, c-ii, d-i
    4. a-iv, b-iii, c-i, d-ii
    Answer

    B. a-iii, b-iv, c-i, d-ii

    Lithium has the lowest E°; platinum is the inert electrode; copper has a positive E°; zinc is oxidised at the anode.

  39. 20 mL of 0.1 M KMnO₄ solution oxidises a Fe²⁺ solution in acidic medium. How many moles of Fe²⁺ are oxidised?

    1. 0.010 mol
    2. 0.100 mol
    3. 0.020 mol
    4. 0.002 mol
    Answer

    A. 0.010 mol

    Moles of MnO₄⁻ = 0.1 × 0.020 = 0.002; each oxidises 5 Fe²⁺, so 0.002 × 5 = 0.010 mol.

  40. How many moles of Fe²⁺ can be oxidised in acidic medium by 0.02 mol of Cr₂O₇²⁻?

    1. 0.24 mol
    2. 0.02 mol
    3. 0.06 mol
    4. 0.12 mol
    Answer

    D. 0.12 mol

    1 mol Cr₂O₇²⁻ oxidises 6 mol Fe²⁺, so 0.02 × 6 = 0.12 mol.

  41. A cell is made from Zn²⁺/Zn (E° = −0.76 V) and Ag⁺/Ag (E° = +0.80 V). What is its E°(cell)?

    1. +0.80 V
    2. +0.04 V
    3. +2.36 V
    4. +1.56 V
    Answer

    D. +1.56 V

    E°(cell) = 0.80 − (−0.76) = 1.56 V.

  42. Given E° values: Zn²⁺/Zn = −0.76 V, Fe²⁺/Fe = −0.44 V, Cu²⁺/Cu = +0.34 V, Ag⁺/Ag = +0.80 V. Which reaction is feasible?

    1. Fe + CuSO₄ → FeSO₄ + Cu
    2. Ag + CuSO₄ → Ag₂SO₄ + Cu
    3. Cu + FeSO₄ → CuSO₄ + Fe
    4. Cu + ZnSO₄ → CuSO₄ + Zn
    Answer

    A. Fe + CuSO₄ → FeSO₄ + Cu

    Fe (E° = −0.44 V) is a stronger reducing agent than Cu (E° = +0.34 V), so it displaces copper. In the other cases the metal is weaker than the one it must displace.

  43. 0.2 mol of Cr₂O₇²⁻ is completely reduced to Cr³⁺ in acidic medium. How many moles of electrons are gained?

    1. 2.4 mol
    2. 1.2 mol
    3. 1.0 mol
    4. 0.6 mol
    Answer

    B. 1.2 mol

    Each Cr₂O₇²⁻ gains 6 electrons (2 Cr from +6 to +3), so 0.2 × 6 = 1.2 mol.

  44. 25 mL of an iodine solution requires 20 mL of 0.1 M sodium thiosulphate for complete reaction (I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻). What is the molarity of the iodine solution?

    1. 0.08 M
    2. 0.02 M
    3. 0.04 M
    4. 0.05 M
    Answer

    C. 0.04 M

    Thiosulphate = 0.1 × 0.020 = 0.002 mol; I₂ = 0.001 mol; M = 0.001/0.025 = 0.04 M.

  45. A clean copper wire is dipped in silver nitrate solution. Which observation and reason are correct?

    1. A silvery deposit forms and the solution turns blue, because Ag⁺ is a stronger oxidising agent than Cu²⁺
    2. No reaction occurs, because copper has a positive standard electrode potential
    3. A reddish-brown deposit forms, because silver is a stronger reducing agent than copper
    4. Hydrogen gas is evolved, because copper lies above hydrogen in the series
    Answer

    A. A silvery deposit forms and the solution turns blue, because Ag⁺ is a stronger oxidising agent than Cu²⁺

    E°(Ag⁺/Ag) = +0.80 V is higher than E°(Cu²⁺/Cu) = +0.34 V, so Cu is oxidised to blue Cu²⁺ and Ag⁺ is reduced to silver.

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