NEET Redox Reactions Previous Year Questions with Solutions

1. Potassium permanganate (KMnO₄): Manganese is reduced from +7 in MnO₄⁻ to +2 in Mn²⁺. The change in oxidation state per mole is |+7 - (+2)| = 5. 2. Oxalic acid (H₂C₂O₄): Carbon is oxidized from +3 in C₂O₄²⁻ to +4 in CO₂. For two carbon atoms per molecule, total electron loss = 2 × 1 = 2.

• • n-factor of KMnO₄ in acidic medium = 5.
• • n-factor of oxalic acid = 2.

At the equivalence (end) point of a titration, the number of gram equivalents of oxidizing agent equals the number of gram equivalents of reducing agent: Number of equivalents = Molarity × n-factor × Volume

• • V₁ = 10 mL.
• • V₂ = 10 mL.
• • M₂ = 0.25 M.

Substitute the given values into the equivalence equation.

• • Strength = 0.10 M.
• • Matches Option (1).

1. In N₂O₅: 2x + 5(-2) = 0 ⇒ 2x = +10 ⇒ x = +5 for Nitrogen. 2. In Al₂O₃: 2x + 3(-2) = 0 ⇒ 2x = +6 ⇒ x = +3 for Aluminium. 3. In H₂S: 2(+1) + x = 0 ⇒ x = -2 for Sulfur.

• • The values +5, +3, -2 follow a strictly decreasing sequence.

In P₄O₁₀, 4x + 10(-2) = 0 ⇒ x = +5. In SO₃, x + 3(-2) = 0 ⇒ x = +6. In H₂O, 2(+1) + x = 0 ⇒ x = -2. The sequence +5, +6, -2 is NOT decreasing because +5 < +6.

• • Violates decreasing order.

In Option (3): PbO₂ (+4), N₂O₃ (+3), SO₃ (+6). Here +3 < +6 (invalid). In Option (4): P₄O₆ (+3), Cl₂O₇ (+7), AlH₃ (+3). Here +3 < +7 (invalid).

• • Matches Option (2).

Potassium (K) is an alkali metal (Group 1). By the canonical rules for oxidation numbers, alkali metals strictly exhibit an oxidation state of +1 in all their compounds:

• • The compound is potassium superoxide, consisting of K⁺ and O₂⁻ ions.
• • The average oxidation state of each oxygen is -1/2.

In hydrogen peroxide, hydrogen has an oxidation number of +1. Using the charge neutrality rule:

• • In peroxides containing the (-O-O-) single bond, oxygen has an oxidation number of -1.

In sulfuric acid, hydrogen is +1 and oxygen is -2:

• • Sulfur achieves its maximum group oxidation state of +6.
• • Consolidated result: +1, -1, and +6, matching Option (1).

A redox (reduction-oxidation) reaction is a chemical process in which one or more elements undergo a change in oxidation number resulting from an electron transfer between reacting species:

Assigning oxidation numbers to each atom in BaCl2 + Na2SO4 -> BaSO4 + 2 NaCl: - Barium: +2 in BaCl2 and +2 in BaSO4 (Δ = 0) - Chlorine: -1 in BaCl2 and -1 in NaCl (Δ = 0) - Sodium: +1 in Na2SO4 and +1 in NaCl (Δ = 0) - Sulfur: +6 in Na2SO4 and +6 in BaSO4 (Δ = 0) - Oxygen: -2 in Na2SO4 and -2 in BaSO4 (Δ = 0) Because no element undergoes oxidation or reduction, it is purely a precipitation (ionic partner exchange) reaction, NOT a redox reaction.

Checking the other options to confirm they are indeed redox reactions: - Reaction (1): H increases from 0 to +1 (oxidation); Cl decreases from 0 to -1 (reduction). - Reaction (3): Zn increases from 0 to +2 (oxidation); Cu decreases from +2 to 0 (reduction). - Reaction (4): I increases from 0 to +5 in KIO3 (oxidation); Cl decreases from +5 in KClO3 to 0 in Cl2 (reduction).

Dichromate ion Cr₂O₇²⁻ is reduced to Cr³⁺ in acidic medium. Balance Cr atoms, then oxygen with H₂O, and hydrogen with H⁺:

Sulphite ion SO₃²⁻ is oxidized to sulphate ion SO₄²⁻. Add H₂O to the left and H⁺ to the right:

Dichromate accepts 6 electrons while sulphite releases 2 electrons. Multiply the oxidation half-reaction by 3:

Add both half-reactions and cancel net H₂O (7 - 3 = 4) and net H⁺ (14 - 6 = 8):

In NCERT Redox Reactions, a metal displacement reaction is defined as a redox process in which a metal in a compound is displaced by another metal in the uncombined elementary state. The displacing metal is more reducing (has a more negative standard reduction potential) than the displaced metal.

In the aluminothermic (Goldschmidt) reduction process: - Aluminum is oxidized from oxidation state $0$ to $+3$ in $Al_2O_3$. - Chromium is reduced from oxidation state $+3$ in $Cr_2O_3$ to $0$ in elemental $Cr$ metal. Because elemental aluminum (a metal) displaces chromium (another metal) from its compound:

- Reaction 1: Decomposition reaction (decomposition of $KClO_3$). - Reaction 3: Hydrogen displacement (non-metal displacement) reaction where iron displaces hydrogen gas from an acid. - Reaction 4: Thermal decomposition of lead nitrate.

When assigning oxidation states, bonding electrons are completely assigned to the more electronegative atom: - Electronegativity values on the Pauling scale: $$\text{Cl (3.16)} > \text{C (2.55)} > \text{H (2.20)}$$.

Because carbon is more electronegative than hydrogen, hydrogen has an oxidation number of $$+1$$: $$x + 4(+1) = 0$$ $$x = -4$$ Thus, carbon is in the $$-4$$ oxidation state in methane ($$\text{CH}_4$$).

Because chlorine is more electronegative than carbon, each chlorine atom has an oxidation number of $$-1$$: $$x + 4(-1) = 0$$ $$x = +4$$ Thus, carbon is in the $$+4$$ oxidation state in carbon tetrachloride ($$\text{CCl}_4$$).

The oxidation state of carbon increases from $$-4$$ in $$\text{CH}_4$$ to $$+4$$ in $$\text{CCl}_4$$, representing an 8-electron oxidation. (Chlorine is concurrently reduced from $$0$$ in $$\text{Cl}_2$$ to $$-1$$ in $$\text{CCl}_4$$ and $$\text{HCl}$$).