Question

The charge required for reducing $1\,mole$ of $MnO^-_4$ to $Mn^{2+}$ is

• A. $1.93 \times 10^5\, C$
• B. $2.895 \times 10^5\, C$
• C. $4.28 \times 10^5\, C$
• D. $4.825 \times 10^5\, C$

Answer 1

Answer: (D)

$\underset{\text{1 mole}}{MnO^{-}_4}+\underset{\text{5 mole}}{5e-}\to \underset{\text{1 mole}}{Mn^{2+}}$
$5\, moles$ of electrons are needed for reduction of $1 \,mole$ of $MnO^-_4$ to $Mn^{2+}$
$5 \,moles$ of electrons $= 5$ Faradays
Quantity of charge required $= 5 \times 96500$
$= 4.825 \times 10^5$ Coulombs