How many coulombs of electricity are required for the reduction of 1 mole of MnO4 to Mn2+?

Question

How many coulombs of electricity are required for the reduction of 1 mole of MnO4 to Mn2+? (A) 96500 C (B) 1.93 × 105 C (C) 4.83 × 105C (D) 5.62 × 105C

Tardigrade Admin · Accepted Answer

MnO-4 + 5e- ⇒ Mn2+ It is obvious from above equation that reduction of 1 mole of MnO-4 to Mn2+ requires 5 Faraday i.e., 5 × 96500 ≈ 4.83 × 105 C

Q. How many coulombs of electricity are required for the reduction of $1 m o l e$ of $M n O_{4}$ to $M n^{2 +}$ ?

$96500 C$

$1.93 \times 1 0^{5} C$

$4.83 \times 1 0^{5} C$

$5.62 \times 1 0^{5} C$

$M n O_{4}^{-} + 5 e^{-} \Rightarrow M n^{2 +}$
It is obvious from above equation that reduction of $1$ mole of $M n O_{4}^{-}$ to $M n^{2 +}$ requires $5$ Faraday i.e., $5 \times 96500 \approx 4.83 \times 1 0^{5} C$

Q. How many coulombs of electricity are required for the reduction of 1mole of MnO4​ to Mn2+?

96500C

1.93×105C

4.83×105C

5.62×105C