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)

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