The charge required for reducing 1 mole of MnO-4 to Mn2+ is

Question

The charge required for reducing 1 mole of MnO-4 to Mn2+ is (A) 1.93 × 105 C (B) 2.895 × 105 C (C) 4.28 × 105 C (D) 4.825 × 105 C

Tardigrade Admin · Accepted Answer

underset text1 moleMnO-4+ underset text5 mole5e-→ underset text1 moleMn2+ 5 moles of electrons are needed for reduction of 1 mole of MnO-4 to Mn2+ 5 moles of electrons = 5 Faradays Quantity of charge required = 5 × 96500 = 4.825 × 105 Coulombs

Q. The charge required for reducing $1 m o l e$ of $M n O_{4}^{-}$ to $M n^{2 +}$ is

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

$2.895 \times 1 0^{5} C$

$4.28 \times 1 0^{5} C$

$4.825 \times 1 0^{5} C$

$1 mole M n O_{4}^{-} + 5 mole 5 e - \to 1 mole M n^{2 +}$
$5 m o l es$ of electrons are needed for reduction of $1 m o l e$ of $M n O_{4}^{-}$ to $M n^{2 +}$
$5 m o l es$ of electrons $= 5$ Faradays
Quantity of charge required $= 5 \times 96500$
$= 4.825 \times 1 0^{5}$ Coulombs

Q. The charge required for reducing 1mole of MnO4−​ to Mn2+ is

1.93×105C

2.895×105C

4.28×105C

4.825×105C