The charge required for reducing 1 mole of (MnO\_{4}^{-}) to... | CHEMISTRY - ELECTROLYTIC CELLS AND ELECTROLYSIS | SolveeIt

The charge required for reducing 1 mole of $MnO_{4}^{-}$ to $Mn^{2 +}$ is

$1.93 \times 10^{5}C$

$2.895 \times 10^{5}C$

$4.28 \times 10^{5}C$

$4.825 \times 10^{5}C$

Answer

$4.825 \times 10^{5}C$

Explanation

$\underset{1mole}{MnO_{4}^{-}} + \underset{5moles}{5e^{-}} \rightarrow \underset{1mole}{Mn^{2 +}}$

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 $= 5 \times 96500$

$= 4.825 \times 10^{5}Coulombs$

Question: The charge required for reducing 1 mole of \(MnO_{4}^{-}\) to \(Mn^{2 +}\) is...