Question

How many electrons would be required to deposit $6.35 \,g$ of copper at the cathode during the electrolysis of an aqueous solution of copper sulphate ? (Atomic mass of copper $= 63.5\, u, N_A =$ Avogadro's constant) :

• A. $\frac{N_{A}}{20}$
• B. $\frac{N_{A}}{10}$
• C. $\frac{N_{A}}{5}$
• D. $\frac{N_{A}}{2}$

Answer 1

Answer: (C)

$\frac{N_{A}}{5}$

Answer 2

$W =\frac{ E }{96500} \times Q$
$\Rightarrow 6.35 =\frac{63.5}{2 \times 96500} \times Q$
$\Rightarrow Q =2 \times 9650$ coulomb

$\Rightarrow 1 F =$ charge of $1$ mol of $e ^{-}=96500$
$\therefore$ No. of $e ^{-} =\frac{ N _{ A }}{10} \times 2$
$=\frac{ N _{ A }}{5}$