Redox reaction between acidified permanganate(VII) and excess iodide ions Steps:

• Acidified KMnO4 (purple) acts as a strong oxidising agent, with MnO4⁻ in +7 oxidation state.
• Excess I⁻ (colourless) is oxidised to I2 (yellow-brown) via the half-equation 2I⁻ → I2 + 2e⁻.
• MnO4⁻ is reduced to Mn²⁺ (colourless) via MnO4⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H2O.
• Overall, purple solution decolourises as MnO4⁻ is consumed, though I3⁻ forms a brown tint with excess I⁻.

Not enough information: The "row" likely refers to a table describing changes for either manganate(VII) or iodide, but choices are ambiguous without the full table.

Final answer: B