Consider the following reaction : CH₄ + 2O₂ ⟶ CO₂ + 2H₂O Which of the following about the reaction given above is/are correct ? 1. Carbon is oxidized. 2. Hydrogen is oxidized. 3. Hydrogen is reduced. 4. Carbon is reduced. Select the correct answer using the code given below :
- (a)1 only
- (b)1 and 2 only
- (c)2 and 3 only
- (d)2 and 4 only
Correct — B, statements 1 and 2 only. This is the complete combustion of methane, and the classical definition of oxidation that school chemistry uses is combination with oxygen. Follow the two elements of methane through the equation: the carbon leaves as carbon dioxide, joined to oxygen, and the hydrogen leaves as water, also joined to oxygen. On that definition both the carbon and the hydrogen of methane have been oxidised, so statements 1 and 2 hold. Nothing in methane has been reduced — reduction here happens to the oxygen of O₂, which gains electrons and ends up in the two products. That disposes of statements 3 and 4, and the code that carries only 1 and 2 is option (b).
- (a)1 only — This keeps the carbon and drops the hydrogen. It would be the answer if one insisted on strict oxidation-number bookkeeping, but the reaction is set on the addition-of-oxygen definition, on which the hydrogen of methane is oxidised too when it becomes part of water.
- (c)2 and 3 only — Self-contradictory. Hydrogen cannot be both oxidised and reduced in the same step, so any code pairing statements 2 and 3 can be struck out on sight.
- (d)2 and 4 only — It gets the hydrogen right but reverses the carbon. Carbon moves from methane, where it is bonded only to hydrogen, to carbon dioxide, where it is bonded to two oxygens — that is a gain of oxygen and so an oxidation, never a reduction.
Oxidation and reduction can be defined at three levels. The oldest is in terms of oxygen and hydrogen — oxidation is gain of oxygen or loss of hydrogen, reduction the reverse. The electronic definition is loss and gain of electrons. The most general is in terms of oxidation number, which rises on oxidation and falls on reduction. In any redox reaction the two happen together: whatever is oxidised has an oxidising agent opposite it, and here that agent is molecular oxygen.
It is worth being clear about which definition the question is using, because the two do not agree for the hydrogen. On the oxygen-and-hydrogen definition the hydrogen of methane gains oxygen when it becomes water, so it is oxidised, and that is the reading the key uses. If instead you assign oxidation numbers, carbon goes from -4 in methane to +4 in carbon dioxide, an unambiguous oxidation, while hydrogen is +1 on both sides and does not formally change, so on that stricter accounting only the carbon is oxidised and the oxygen reduced. Statements 3 and 4 fail on either reading, and a candidate who notices that options (c) and (d) each contain a reduction of something inside methane can eliminate them at once.
- In the classical definition, oxidation is gain of oxygen or loss of hydrogen and reduction is the reverse.
- In the complete combustion of methane the carbon leaves as carbon dioxide and the hydrogen as water — both combined with oxygen.
- The oxidation number of carbon rises from -4 in CH₄ to +4 in CO₂.
- Hydrogen is +1 in both CH₄ and H₂O, so its oxidation number does not change.
- Molecular oxygen is the oxidising agent and is itself reduced from 0 to -2.
- Oxidation and reduction always occur together in a redox reaction.
Statements 1 and 2 hold and nothing in methane is reduced, so the code is option (b).
- Choosing a code in which the same element is both oxidised and reduced.
- Assuming that because oxygen appears on the product side it must have been oxidised; it is the species reduced.
- Mixing the oxygen-based and oxidation-number definitions without noticing that they can give different verdicts for hydrogen.
Redox appears as a statement-code item on a given equation, as identification of the oxidising agent, or as an oxidation-number calculation.
Match List I (Oxidation number) with List II (The element) and select the correct answer using the codes given below the lists: List I (Oxidation number) A. 2 B. 3 C. 4 D. 6 — List II (The element) 1. Oxidation number of Mn in MnO₂ 2. Oxidation number of S in H₂S₂O₇ 3. Oxidation number of Ca in CaO₂ 4. Oxidation number of Al in NaAlH₄
- (a) A-3, B-4, C-1, D-2
- (b) A-4, B-3, C-1, D-2
- (c) A-3, B-4, C-2, D-1
- (d) A-4, B-3, C-2, D-1
Answer(a) A-3, B-4, C-1, D-2
Drills the oxidation-number machinery that gives the stricter reading of what happens to carbon and hydrogen here.
Which one of the following is an oxidation-reduction reaction?
- (a) NaOH + HCl → NaCl + H₂O
- (b) CaO + H₂O → Ca(OH)₂
- (c) 2Mg + O₂ → 2MgO
- (d) Na₂SO₄ + BaCl₂ → BaSO₄ + 2NaCl
Answer(c) 2Mg + O₂ → 2MgO
The same test applied to four equations at once — spot the one in which something gains oxygen.
- practice — not a real PYQ
In the reaction 2Mg + O₂ → 2MgO, the substance oxidised is
- (a)oxygen
- (b)magnesium
- (c)magnesium oxide
- (d)neither of the reactants
Answer(b) magnesium — it combines with oxygen and its oxidation number rises from 0 to +2.
- practice — not a real PYQ
The oxidation number of carbon in carbon dioxide is
- (a)-4
- (b)0
- (c)+2
- (d)+4
Answer(d) +4 — each oxygen is -2, so carbon must be +4 for a neutral molecule.