Question

How many coulombs of electricity are required for reduction of $1$ mole of $MnO_{4}^{-}$ to $Mn^{2+}$ ?

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

Answer 1

Answer: (C)

$4.83\times 10^{5}C$

Answer 2

$\underset{1 \text { mol }}{ MnO _{4}^{-1}}+8 H ^{+}+\underset{5 mole }{5 e^{-}} \longrightarrow Mn ^{2+}+4 H _{2} O$ ie, $5\, F$ Hence, electricity required $=5 \times 96500\, C$ $=482500\, C$ $=4.83 \times 10^{5} C$