Redox Reactions — Practice Quiz

2H₂O₂ → 2H₂O + O₂

A disproportionation reaction is one where the SAME element in the SAME substance is simultaneously oxidised and reduced. In 2H₂O₂ → 2H₂O + O₂: Oxygen in H₂O₂ has O.N. = −1. In H₂O, O has O.N. = −2 (reduction). In O₂, O has O.N. = 0 (oxidation). So oxygen goes from −1 to both −2 and 0 simultaneously – this is disproportionation. Option A is a displacement reaction. Option C is a combination reaction. Option D is also a combination reaction. None of these involve one element being both oxidised and reduced simultaneously.

+6

Let the oxidation number of S = x. In H₂SO₄: H has O.N. = +1 (two H atoms: +2 total). O has O.N. = −2 (four O atoms: −8 total). Since H₂SO₄ is a neutral molecule: (+2) + x + (−8) = 0 → x − 6 = 0 → x = +6. So sulphur in H₂SO₄ has oxidation number +6. Option A (+4) is the O.N. of S in SO₂ and H₂SO₃ – a common student confusion. Option B (+2) is incorrect and does not satisfy the equation. Option D (−2) is the O.N. of S in H₂S (sulphide) where S is fully reduced.

Mg is oxidised and acts as a reducing agent

In 2Mg + O₂ → 2MgO: Mg starts with oxidation number 0 (free element). In MgO, Mg exists as Mg²⁺ with oxidation number +2. Since the oxidation number INCREASES from 0 to +2, Mg undergoes OXIDATION. The substance that gets oxidised LOSES electrons and is called the REDUCING AGENT. So Mg is oxidised and acts as a reducing agent. Option A is completely opposite. Option C incorrectly combines reduction with reducing agent. Option D is wrong because Mg clearly changes from O.N. 0 to +2.

S + O₂ → SO₂ (sulphur adds oxygen)

According to the classical definition, oxidation is the process in which a substance adds oxygen or loses hydrogen. In S + O₂ → SO₂, sulphur is adding oxygen to form SO₂. This is oxidation by the classical definition. Option A (CuO + C → Cu + CO): Here C is removing oxygen from CuO. C undergoes oxidation (it adds O to form CO), while Cu is REDUCED (loses oxygen). Option B: Cl₂ adding hydrogen is reduction of Cl₂. Option D: Iron oxide losing oxygen is reduction. The classical and modern concepts both agree that S + O₂ → SO₂ is oxidation of sulphur.