See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

  OH 2 2 2 4 2 4 Cr OH H O CrO H O           Oxidation Half Reduction Half   2 4 2 4 Cr OH CrO 4H O 4 OH 3e             2 2 2 H O 2H O 2 OH    2  + 2H2O 6e– +2e– 3  6e–