KMnO4 + K2Cr2O7 Reactions
d-block & f-block Elements
38
JEE Qs
10%
Hard
90
min
Master the n-factors of KMnO4 and K2Cr2O7 in various media and practice balancing their redox reactions with common reducing agents thoroughly.
🧮 Key Formulas
✅ Key Points for JEE
- 1The n-factor (number of electrons gained/lost) for KMnO4 depends critically on the pH of the medium, leading to different reduction products (Mn^2+, MnO2, MnO4^2-).
- 2K2Cr2O7 is a strong oxidizing agent predominantly used in acidic medium, where Cr(VI) (orange) is reduced to Cr(III) (green).
- 3Memorize the common reducing agents that react with KMnO4 and K2Cr2O7, along with their oxidation products (e.g., Fe^2+ to Fe^3+, C2O4^2- to CO2, I^- to I2, SO3^2- to SO4^2-).
- 4Balancing redox reactions for these compounds in different media (acidic, basic) is crucial, often using the ion-electron method.
- 5Understand the characteristic color changes: MnO4^- (purple) to Mn^2+ (colorless), MnO2 (brown precipitate), MnO4^2- (green); Cr2O7^2- (orange) to Cr^3+ (green).
⚠️ Common Mistakes
- ✕Using an incorrect n-factor for MnO4^- by not considering the pH of the reaction medium.
- ✕Incorrectly balancing the half-reactions, especially regarding H+, OH-, and H2O, in complex redox equations.
- ✕Confusing the oxidizing power or suitable reaction conditions for KMnO4 versus K2Cr2O7.
- ✕Failing to remember the color changes associated with the reactants and products, which is vital for identification and titrations.
📝 Practice Questions
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Q52.Consider the following reactions K2Cr2O7 KOH→ [A] H2SO4→ [ B] + K2SO4 The products [A] and [B], −H2O −H2O respectively are : (1) K2CrO4 and CrO (2) K2CrO4 and Cr2O3 (3) K2CrO4 and K2Cr2O7 (4) K2Cr(OH)6 and Cr2O3
Q57.Preparation of potassium permanganate from MnO2 involves two step process in which the 1st step is a reaction with KOH and KNO3 to produce (1) K3MnO4 (2) K4 [Mn(OH)6] (3) KMnO4 (4) K2MnO4
NCERT Chapters
- Class 11 Chemistry Part 1 Ch 8: Redox Reactions
- Class 12 Chemistry Part 1 Ch 8: The d- and f-Block Elements