Chemistry · Redox Equilibrium · NEET
In acidic medium (dilute H2SO4) KMnO4 is reduced fully. Mn changes from +7 to +2, forming the almost colourless Mn2+ ion. The half reaction is MnO4^- + 8H+ + 5e- -> Mn2+ + 4H2O, so it gains 5 electrons and the n-factor is 5. Plenty of H+ ions are available, which lets the reduction go all the way to the most reduced common state, Mn2+. This is why the deep purple colour disappears.
In neutral or faintly alkaline medium there are not enough H+ ions to reduce Mn all the way to +2. Reduction stops at Mn(+4), which appears as a brown solid MnO2. The half reaction is MnO4^- + 2H2O + 3e- -> MnO2 + 4OH-, so it gains 3 electrons and the n-factor is 3. A classic NEET example: KMnO4 oxidises iodide (I^-) to iodate (IO3^-) in this medium while Mn falls from +7 to +4.
In strongly alkaline medium Mn is reduced only by one step, from +7 to +6, giving the green manganate ion MnO4^2-. The half reaction is MnO4^- + e- -> MnO4^2-, so it gains just 1 electron and the n-factor is 1. Because it accepts only one electron, KMnO4 is a much weaker oxidising agent in strong base than in acid.
n-factor equals the number of electrons gained per MnO4^- ion. Acidic medium: n = 5 (Mn +7 to +2). Neutral or faintly alkaline: n = 3 (Mn +7 to +4). Strongly basic: n = 1 (Mn +7 to +6). Equivalent weight = molar mass / n-factor, so the equivalent weight is smallest in acid (M/5 = 158/5 = 31.6 g) and largest in strong base (M/1 = 158 g). Getting the medium right is essential for titration calculations.
In acidic medium KMnO4 gains 5 electrons per ion, giving the highest n-factor and the sharpest, most complete oxidation. The product Mn2+ is colourless, so the first excess drop of purple KMnO4 gives a clear pink end point (self-indicator). In neutral or basic medium a brown MnO2 precipitate forms, which masks the colour change and makes the end point unclear. Dilute H2SO4 is used, not HCl (HCl would be oxidised to Cl2, wasting KMnO4) and not HNO3 (it is itself an oxidiser).
Name the gas that can readily decolourise acidified KMnO4 solution.
When neutral or faintly alkaline KMnO4 is treated with potassium iodide, iodide ion is converted into 'X'. 'X' is:
In the neutral or faintly alkaline medium, KMnO4 oxidises iodide into iodate. The change in oxidation state of manganese in this reaction is from:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is a much stronger oxidiser in acidic medium, where it gains 5 electrons (Mn +7 to +2). In strongly basic medium it gains only 1 electron (Mn +7 to +6), so it is a weaker oxidiser there.
Mn2+ (acidic, +2) is almost colourless. MnO2 (neutral/faintly alkaline, +4) is a brown solid. MnO4^2- manganate (strongly basic, +6) is green. The starting MnO4^- is deep purple.
HCl would be oxidised to Cl2 by KMnO4, using up some of it and giving a wrong reading. HNO3 is itself an oxidising agent and interferes. Dilute H2SO4 provides H+ without side reactions, so it is the correct acid.
Its own deep purple colour disappears as it is reduced to colourless Mn2+ during the titration. The very first drop of excess KMnO4 after the end point turns the solution light pink, signalling completion. No separate indicator is needed.
Equivalent weight = molar mass (158) / n-factor. Acidic: 158/5 = 31.6 g. Neutral/faintly alkaline: 158/3 = 52.7 g. Strongly basic: 158/1 = 158 g. It rises as you move from acidic to basic medium.
