Which of the following reactions is an oxidation process?

Analyze the question: We need to find which half-reaction is an oxidation process.

Concept: Oxidation is the loss of electrons (remember OIL RIG: Oxidation Is Loss, Reduction Is Gain). In a half-equation, electrons appear on the right-hand side for oxidation and on the left-hand side for reduction.

Step-by-step check:

• A. 2H⁺ + 2e⁻ → H₂: electrons are gained, so this is reduction.
• B. 2O²⁻ → O₂ + 4e⁻ is the correct way to write oxygen ions losing electrons (oxidation number of O goes from -2 to 0). The option as printed, 2O²⁻ + 4e⁻ → O₂, has the electrons on the wrong side. Charge is not balanced as written: the left side is 2(-2) + (-4) = -8, and the right side is 0. Even so, the change O²⁻ → O₂ is clearly an oxidation, since oxygen's oxidation number rises from -2 to 0. This is almost certainly a typing slip in the question, and it is the only option where the species changes by oxidation.
• C. OCl⁻ + H₂O + 2e⁻ → 2OH⁻ + Cl⁻: chlorine goes from +1 to -1, so electrons are gained. This is reduction.
• D. Cl + e⁻ → Cl⁻: electrons are gained, so this is reduction.

Answer: Options A, C and D are all reductions. Only B involves oxidation (O²⁻ → O₂, loss of electrons). I should be honest that, as printed, the electrons in B are written on the reactant side, which is a flaw in the question.

Tip: Look at the oxidation number change, not just the equation's layout. If the oxidation number increases (-2 → 0), it is oxidation. Also, in any oxidation half-equation, the electrons should be on the product side. If you see electrons on the left, it is reduction, so in an exam, choose the option where the species is actually being oxidised.