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Chemistry
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C6H6 reacts with KMnO4 in cold aqueous KOH to yield MnO2 and C6H12O4. I am supposed to balance this with formulas of real compounds you can get from a bottle i think the answer is 3 C6H8 + 4 KMnO4 + 8 H20 --> 4 MnO2+3 C6H12O4 + 4KOH What I don"t understand is the Half Oxidation equation? Is it C6H8---> C6H12O4 + (2/3)e- ?
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