Question

How many coulombs of electricity are required for the reduction of $1mole$ of $Mn{O}_4^{-}$ to $M{n}^{2+}$?

• A. $96500C$
• B. $1.93\times {10}^{5}C$
• C. $4.83\times {10}^{5}C$
• D. $9.65\times {10}^{6}C$
• E. $5.62\times {10}^{5}C$

Answer 1

The correct option is C $4.83\times {10}^{5}C$

Half Cell reduction reaction:-

$Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{+2}+4H_{2}O$

1 mole of $M{n}^{2+}$ is formed by $5e^{-}$ mole of electrons.

Moles of $M{n}^{2+}=\frac{Q\left(C\right)}{96500C/mol{e}^{-}}\times moleratio$

mole ratio$=\frac{molesofM{n}^{2+}}{molesofelectron}\times \frac{1}{5}$

$\therefore$ 1 moles of $M{n}^{2+}=\frac{Q\left(C\right)}{96500C/mol{e}^{-}}\times \frac{1}{5}$

$\therefore Q\left(C\right)=1\times 96500\times 5=482500C\approx 4.83\times {10}^{5}C$