In the electrosynthesis, potassium manganate (VII) is converted to manganese (IV) dioxide. By passage of 1 F of electrolysis, one mole of potassium manganate (VII) will form manganese dioxide.

Question

In the electrosynthesis, potassium manganate (VII) is converted to manganese (IV) dioxide. By passage of 1 F of electrolysis, one mole of potassium manganate (VII) will form manganese dioxide. (A) 1.00 mol (B) 0.33 mol (C) 0.66 mol (D) 0.50 mol

Tardigrade Admin · Accepted Answer

Manganate (VII) is MnO 4- with oxidation number of Mn =+7 underset1 mol MnO 4-+3 e - longrightarrow underset+4 MnO 2 1 mole required 3 F of electricity thus, 1 F will reduce only (1 / 3) mole MnO 4- to MnO 2 Thus, MnO 2 formed =0.33 mol.

Q. In the electrosynthesis, potassium manganate (VII) is converted to manganese (IV) dioxide. By passage of $1 F$ of electrolysis, one mole of potassium manganate (VII) will form manganese dioxide.

1.00 mol

0.33 mol

0.66 mol

0.50 mol

Manganate (VII) is $M n O_{4}^{-}$ with oxidation number of $M n = + 7$

$1 m o l M n O_{4}^{-} + 3 e^{-} ⟶ + 4 M n O_{2}$

1 mole required $3 F$ of electricity thus, $1 F$ will reduce only $(1/3)$ mole $M n O_{4}^{-}$ to $M n O_{2}$

Thus, $M n O_{2}$ formed $= 0.33 m o l$ .

Q. In the electrosynthesis, potassium manganate (VII) is converted to manganese (IV) dioxide. By passage of 1F of electrolysis, one mole of potassium manganate (VII) will form manganese dioxide.