Question

How many electrons would be required to deposit 6.35 gm of copper at the cathode during the electrolysis of an aqueous solution of copper sulphate?

[Atomic mass of copper $=63.5u,N_A=$ 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

The correct option is C $\frac{N_A}{5}$

According to the reaction:
$C{u}^{2+}(aq.)+2e^{-}\to Cu\left(s\right)$

We require 2 moles of electrons or $2N_A$ electrons to deposit 1 mol or 63.5 g of $Cu$.

For 6.35 g, we require $\frac{2N_A}{10}$ electrons.

Hence, the number of electrons required to deposit 6.35 gm of copper = $\frac{N_A}{5}$ electrons.