Chemistry · Redox Equilibrium · NEET
In KMnO4 the Mn is in its highest oxidation state, +7 (K is +1, each of the 4 oxygens is -2, so Mn = +7). This is why it can only be reduced, making it an oxidising agent. The product changes with the medium: in ACIDIC medium it is reduced to Mn2+ (+2); in NEUTRAL or faintly ALKALINE medium to MnO2 (+4); in strongly ALKALINE medium to MnO4^2- (+6, the green manganate ion). For NEET, memorise these three end-products first — most questions are just asking which one forms.
The n-factor equals the change in oxidation number of Mn. Acidic: +7 to +2 = 5 electrons, so n-factor = 5 (equivalent weight = 158/5 = 31.6 g). Neutral/faintly alkaline: +7 to +4 = 3 electrons, n-factor = 3 (equivalent weight = 158/3 = 52.67 g). Strongly alkaline: +7 to +6 = 1 electron, n-factor = 1 (equivalent weight = 158). This matters for NEET titration sums, where equivalents of KMnO4 = equivalents of the reducing agent.
In acidic medium there are plenty of H+ ions available. These H+ ions help pull the oxygen atoms off the MnO4^- and reduce Mn all the way to the soluble Mn2+ ion (colourless). In neutral or faintly alkaline medium there are not enough H+ ions to do this, so the reduction stops earlier at MnO2, a brown insoluble solid where Mn is +4. So the amount of acid available decides how far Mn is reduced.
KMnO4 solution is deep purple (violet) because of the MnO4^- ion. In acidic medium the product Mn2+ is almost colourless, so the purple colour disappears — the solution becomes decolourised. In neutral/faintly alkaline medium the product MnO2 is a brown precipitate. In strongly alkaline medium the product is the green manganate ion MnO4^2-. NEET loves asking 'which reagent decolourises acidified KMnO4' — the answer is a reducing agent such as SO2, oxalate, Fe2+ or H2O2.
Any good reducing agent will decolourise acidified KMnO4 by reducing purple Mn(+7) to colourless Mn2+. Common NEET examples: SO2 gas (oxidised to sulphate), oxalic acid / oxalate C2O4^2- (oxidised to CO2), ferrous ion Fe2+ (oxidised to Fe3+), and H2O2 (oxidised to O2). CO2 and P2O5 do NOT decolourise it because they are not reducing agents.
In neutral or faintly alkaline medium, KMnO4 oxidises iodide (I-, oxidation state -1) all the way to iodate ion IO3^- (I is +5). At the same time Mn goes from +7 to +4 (MnO2). The balanced reaction is 2MnO4^- + I^- + H2O -> 2MnO2 + IO3^- + 2OH^-. Note: in acidic medium iodide would only be oxidised to I2; the deeper oxidation to iodate is a neutral/alkaline-medium result.
In the neutral or faintly alkaline medium, KMnO4 oxidises iodide into iodate. The change in oxidation state of manganese in this reaction is from:
Name the gas that can readily decolourise acidified KMnO4 solution:
In acidic medium, 10 mL of 0.25 M oxalic acid is titrated with KMnO4 solution. If the volume of KMnO4 required to reach the end point is 10 mL, the strength of the KMnO4 solution is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Because Mn in KMnO4 is already in its highest possible oxidation state, +7. It cannot lose more electrons (be oxidised further), it can only gain electrons (be reduced). Gaining electrons and reducing itself means it oxidises other species, so it acts only as an oxidising agent.
5. In acidic medium Mn is reduced from +7 to +2, a change of 5, so 5 electrons are gained per MnO4^-. The equivalent weight = molar mass 158 / 5 = 31.6 g. This value is used in titration calculations.
In strongly alkaline medium Mn goes from +7 to +6, forming the green manganate ion MnO4^2- (a 1-electron change). Example: 2MnO4^- + 2OH^- -> 2MnO4^2- + H2O + [O]. This is different from neutral/faintly alkaline medium, which gives MnO2 (+4).
Yes. In acidic medium KMnO4 accepts 5 electrons and its oxidising power (and standard electrode potential) is highest, so it is the strongest. In neutral/alkaline medium it accepts fewer electrons (3 or 1) and is a weaker oxidant. For NEET, acidified KMnO4 is the standard strong oxidiser used in titrations.
Dilute sulphuric acid (H2SO4) is used. HCl is avoided because chloride ions (Cl-) are themselves oxidised by KMnO4 to Cl2, which would consume extra KMnO4 and give a wrong titre. HNO3 is avoided because it is itself an oxidising agent.
