Total charge required for the oxidation of two moles Mn 3 O 4 into MnO 42 in presence of alkaline medium is-

Question

Total charge required for the oxidation of two moles Mn 3 O 4 into MnO 42 in presence of alkaline medium is- (A) 5 F (B) 10 F (C) 20 F (D) None of these

Tardigrade Admin · Accepted Answer

16 OH -+ Mn 3 O 4 arrow 3 MnO 42-+8 H 2 O +10 e - 1 mol electron required is 10 2 mol =2 × 10=20 Charge required =20 × F =20 F

Q. Total charge required for the oxidation of two moles $M n_{3} O_{4}$ into $M n O_{4}^{2}$ in presence of alkaline medium is-

5 F

10 F

20 F

None of these

$16 O H^{-} + M n_{3} O_{4} \to 3 M n O_{4}^{2 -} + 8 H_{2} O + 10 e^{-}$

1 mol electron required is 10

$2 m o l = 2 \times 10 = 20$

Charge required

$= 20 \times F = 20 F$

Q. Total charge required for the oxidation of two moles Mn3​O4​ into MnO42​ in presence of alkaline medium is-

5 F

10 F

20 F

None of these

16OH−+Mn3​O4​→3MnO42−​+8H2​O+10e− 1 mol electron required is 10 2mol=2×10=20 Charge required =20×F=20F

Q. Total charge required for the oxidation of two moles $M n_{3} O_{4}$ into $M n O_{4}^{2}$ in presence of alkaline medium is-

$16 O H^{-} + M n_{3} O_{4} \to 3 M n O_{4}^{2 -} + 8 H_{2} O + 10 e^{-}$

1 mol electron required is 10

$2 m o l = 2 \times 10 = 20$

Charge required

$= 20 \times F = 20 F$