Question

The charge required for the oxidation of 1 mole Mn3O4 into MnO4 2- in presence of alkaline...

Answer 1

Oxidation of Mn₃O₄ into MnO₄²⁻ takes place as follows:
2Mn₃O₄ + 2OH⁻ → 3Mn₂O₃ + 2e + H₂O
3Mn₂O₃ + 6OH⁻ → 6MnO₂ + 6e + H₂O
6MnO₂ + 24OH⁻ → 6MnO₄²⁻ + 12e + 12H₂O

Total moles of electrons 20
We have to calculate the charge required for the oxidation of 1 mole.
Therefore,
n = 10
Charge required = n x F
Charge required = 10 x 96500 C

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