Chemistry · Redox Equilibrium · NEET
At the equivalence point the amount of oxidant exactly matches the amount of reductant in electron terms. The safe rule for NEET is: (n-factor × Molarity × Volume) of oxidant = (n-factor × Molarity × Volume) of reductant. The n-factor is the number of electrons one formula unit gains or loses. Never just equate moles or molarity directly unless both n-factors are the same — that is the most common mistake.
In acidic medium KMnO4 changes Mn from +7 to +2, so it gains 5 electrons — n-factor = 5. Oxalic acid (H2C2O4 or the ion C2O4^2-) has two carbons that each go from +3 to +4, losing 2 electrons total — n-factor = 2. So in acid, 2 KMnO4 react with 5 oxalic acid (the ratio is the inverse of the n-factors, 5:2 flipped to 2:5).
Write equivalents equal: n1·M1·V1 = n2·M2·V2. For KMnO4: 5 × M × 10. For oxalic acid: 2 × 0.25 × 10 = 5. So 50 × M = 5, giving M = 0.10 M. The KMnO4 is 0.10 M. Notice the n-factors 5 and 2 do the real work — if you forget them you get the wrong answer.
Yes, and NEET loves this trap. In acidic medium Mn goes +7 to +2, so n-factor = 5. In neutral or faintly alkaline medium Mn goes +7 to +4 (MnO2 forms), so n-factor = 3. In strongly alkaline medium Mn goes +7 to +6, so n-factor = 1. Always check the medium before you pick the n-factor.
KMnO4 is 'self-indicating'. Its own deep purple colour disappears while it is being used up (it becomes almost colourless Mn2+). The moment even one extra drop of KMnO4 is added past the equivalence point, a lasting light-pink colour appears — the NCERT notes it shows up at MnO4^- as low as 10^-6 mol/L. That first permanent pink is your end point.
The equivalence point is the exact theoretical point where oxidant and reductant are equal in stoichiometry (electrons balanced). The end point is what you actually observe — the signal (colour change) that tells you to stop. A good titration is designed so the end point comes as close as possible to the equivalence point, with almost no overshoot.
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:
For the redox reaction MnO4^- + C2O4^2- + H^+ -> Mn^2+ + CO2 + H2O, the correct coefficients of the reactants (MnO4^-, C2O4^2-, H^+) for the balanced equation are:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Electron balance sets the ratio. KMnO4 gains 5 electrons and oxalic acid loses 2. To make electrons lost equal electrons gained, you need 2 KMnO4 (10 e^- gained) for every 5 oxalic acid (10 e^- lost). So the mole ratio is 2 MnO4^- : 5 C2O4^2-.
Warm. The reaction is slow at room temperature, so the flask is heated to about 60-70°C. Above that the oxalic acid can decompose, so you avoid boiling. Mn2+ formed also speeds it up (autocatalysis), which is why it is slow at first and then fast.
Yes. Normality N = n-factor × Molarity, and at the equivalence point N1·V1 = N2·V2 (equal equivalents). Using n-factor × M × V is the same idea written out, which is safer for NEET because you can see exactly which electron count you used.
KMnO4 is added from the burette. While reductant remains, the purple colour vanishes as MnO4^- turns to colourless Mn2+. The end point is the first appearance of a faint pink colour that stays for about 30 seconds — that means one extra drop of KMnO4 had nothing left to react with.
Redox titration questions are almost pure marks if you know the equivalents rule. NEET repeatedly asks for an unknown molarity, a balanced-equation ratio, or the medium-dependent n-factor of KMnO4. All three come from the same 'electrons in = electrons out' idea, so one concept covers several question types.
