Question

Total charge required for the oxidation of two moles mn3o4 into mno4-2 in presence of alkaline medium is

Answer 1

Answer:

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