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Question

Total charge required for oxidation of 2 moles Mn3O4 into MnO42- in the presence of alkaline medium is:

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Solution

Oxidation of Mn3O4 to MnO42- takes place in following steps:

2Mn3O4+ 2OH-3Mn2O3+ 2e +H2O3Mn2O3+ 6OH-6MnO2+ 6e +H2O6MnO2+ 24OH-6MnO42-+ 12H2O +12e

Total moles of electrons = 20
So, total charge required = 20 x 96500 F

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