Question

which half-reaction correctly describes an oxidation?
○ ca(s) + 2e⁻ ⟶ ca²⁺(aq)
○ br₂ + 2e⁻ ⟶ 2br⁻
○ na⁺(aq) ⟶ na(s) + e⁻
○ cr(s) ⟶ cr³⁺(aq) + 3e⁻

Explanation:

Step1: Recall oxidation definition

Oxidation is loss of electrons (LEO: Loss of Electrons is Oxidation).

Step2: Analyze each option

• Option 1: $\ce{Ca(s) + 2e^- -> Ca^{2+}(aq)}$: Gaining electrons (reduction), not oxidation.
• Option 2: $\ce{Br_2 + 2e^- -> 2Br^-}$: Gaining electrons (reduction), not oxidation.
• Option 3: $\ce{Na^+(aq) -> Na(s) + e^-}$: Sodium ion is gaining an electron? No, charge here: $\ce{Na^+}$ (charge +1) to $\ce{Na}$ (charge 0) – this is gain of electron (reduction, since oxidation state decreases from +1 to 0). Wait, actually, the equation is written as losing an electron, but charge-wise: left side $\ce{Na^+}$ (oxidation state +1), right side $\ce{Na}$ (0) – so to go from +1 to 0, it gains an electron (reduction). So this is incorrect for oxidation.
• Option 4: $\ce{Cr(s) -> Cr^{3+}(aq) + 3e^-}$: Chromium (oxidation state 0) becomes $\ce{Cr^{3+}}$ (oxidation state +3) – it loses 3 electrons. Loss of electrons is oxidation, so this is correct.

Answer:

D. $\ce{Cr(s) -> Cr^{3+}(aq) + 3e^-}$ (assuming the last option is D, if options are labeled A - D with the last one as D)