Number of Faradays (F) required to reduce 3 mol of $M n O_{4}^{⊖}$ to $M n^{2 +}$ is:

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Solution

$8 H^{\oplus} + 5 e^{-} + M n O_{4}^{⊖} \to M n^{2 +} + 4 H_{2} O$
1 mole of $M n O_{4}^{⊖} \equiv 5 e^{-} = 5 F$
3 moles of $M n O_{4}^{⊖} = 5 \times 3 = 15 F$

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Number of Faradays (F) required to reduce 3 mol of MnO⊖4 to Mn2+ is:

8H⊕+5e−+MnO⊖4→Mn2++4H2O1 mole of MnO⊖4≡5e−=5F3 moles of MnO⊖4=5×3=15F

Number of Faradays (F) required to reduce 3 mol of $M n O_{4}^{⊖}$ to $M n^{2 +}$ is:

$8 H^{\oplus} + 5 e^{-} + M n O_{4}^{⊖} \to M n^{2 +} + 4 H_{2} O$
1 mole of $M n O_{4}^{⊖} \equiv 5 e^{-} = 5 F$
3 moles of $M n O_{4}^{⊖} = 5 \times 3 = 15 F$