Answer: D (purple acidified potassium manganate(VII) turning colourless). Reduction means a gain of electrons, shown by a decrease in oxidation state of the named reagent. In option D, the manganese in the manganate(VII) ion, MnO₄⁻, changes from +7 to +2 in the colourless Mn²⁺ ion, so the acidified KMnO₄ changes from purple to colourless as it is reduced. That is the reduction of the named reagent.
In option B, colourless aqueous potassium iodide turns brown as iodide ions, I⁻ (−1), are oxidised to iodine, I₂ (0) — oxidation, not reduction. In option C, green iron(II) chloride turns yellow-brown iron(III) chloride as Fe²⁺ is oxidised to Fe³⁺. In option A, copper stays at +2 throughout (oxide to sulfate), so that is not a redox change at all. The answer is D.
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