How many electrons would be required to deposit $6.35 \mathrm{~g}$ of copper at the cathode during the…
How many electrons would be required to deposit $6.35 \mathrm{~g}$ of copper at the cathode during the electrolysis of an aqueous solution of copper sulphate? (Atomic mass of copper $=63.5 \mathrm{u}, \mathrm{N}_{\mathrm{A}}=$ Avogadro's constant):
$\frac{\mathrm{N}_{\mathrm{A}}}{20}$
$\frac{\mathrm{N}_{\mathrm{A}}}{10}$
$\frac{\mathrm{N}_{\mathrm{A}}}{5}$
$\frac{\mathrm{N}_{\mathrm{A}}}{2}$
Solution
$\mathrm{Cu} \longrightarrow \mathrm{Cu}^{++}+2 \mathrm{e}^{-}$
i.e, to deposit 1 mole of $\mathrm{Cu}$ at cathode from $\mathrm{Cu}^{2+} \mathrm{SO}_4^{2-}$ solution $=2$ moles of electrons are required
i.e, To deposit
$6.35 \mathrm{~g}=\frac{6.35}{63.5} \times 2=\frac{2}{10}=\frac{1}{5}$ moles
Thus total no. of electrons required $=\frac{\mathrm{N}_{\mathrm{A}}}{5}$