Chemistry · Redox Equilibrium · NEET
Each Fe2+ loses 1 electron to become Fe3+ (n-factor = 1). Each MnO4- gains 5 electrons to become Mn2+ (Mn goes from +7 to +2, n-factor = 5). To balance electrons lost and gained, you need 5 Fe2+ for every 1 MnO4-. Balanced: MnO4- + 5Fe2+ + 8H+ -> Mn2+ + 5Fe3+ + 4H2O.
KMnO4 needs H+ ions to give its full 5-electron oxidation to Mn2+, so an acid is required. HCl is avoided because KMnO4 would oxidise the Cl- to Cl2, using up extra KMnO4 and giving a wrong (higher) reading. HNO3 is avoided because it is itself an oxidiser and would oxidise Fe2+ on its own. Dilute H2SO4 is neutral to this reaction, so it is the safe choice.
Mohr salt is ferrous ammonium sulphate, FeSO4.(NH4)2SO4.6H2O. Plain FeSO4 in air slowly gets oxidised by oxygen from Fe2+ to Fe3+, so its concentration is not reliable. Mohr salt is a stable double salt that keeps Fe2+ from oxidising, so it gives a fixed, accurate concentration for titration.
KMnO4 is a self-indicator. As long as Fe2+ is present, the purple MnO4- is decolourised to almost colourless Mn2+. At the endpoint all Fe2+ is used up, so one extra drop of KMnO4 stays and turns the solution light pink. That first permanent pink is the endpoint, so no external indicator is needed.
Equivalent weight = molar mass / n-factor. Molar mass of KMnO4 is 158 g/mol and n-factor in acidic medium is 5, so equivalent weight = 158/5 = 31.6 g/equiv. For Fe2+ (or Mohr salt) the n-factor is 1, so its equivalent weight equals its molar mass.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
MnO4- + 5Fe2+ + 8H+ -> Mn2+ + 5Fe3+ + 4H2O. In molecular form: 2KMnO4 + 10FeSO4 + 8H2SO4 -> 2MnSO4 + 5Fe2(SO4)3 + K2SO4 + 8H2O.
The solution changes from colourless (or light green from Fe2+) to a permanent light pink when the last drop of excess KMnO4 remains unreacted.
Its own intense purple colour marks the endpoint. Below the endpoint the purple is decolourised by Fe2+; at the endpoint the first excess drop gives a lasting pink, so no separate indicator dye is needed.
5 moles of Fe2+ react with 1 mole of KMnO4, because Fe2+ has n-factor 1 and KMnO4 has n-factor 5 in acidic medium.
KMnO4 would oxidise the chloride ions of HCl to chlorine gas, consuming extra KMnO4 and giving a falsely high titre value. Dilute H2SO4 is used instead.
