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 = 63.5u, N$$_A$$ = Avogadro's constant):

To determine the number of electrons required to deposit 6.35 g of copper during electrolysis, we start by recalling the reaction at the cathode. Copper ions are reduced to copper metal: $$Cu^{2+ + 2e^- \rightarrow Cu}$$. This indicates that 1 mole of copper atoms (Cu) requires 2 moles of electrons for deposition.

The atomic mass of copper is given as 63.5 u, so the molar mass is 63.5 g/mol. We need to find the number of moles in 6.35 g of copper. The moles of copper are calculated as: $$\text{moles of Cu} = \frac{\text{mass}}{\text{molar mass}} = \frac{6.35}{63.5}$$.

Simplifying the fraction: $$\frac{6.35}{63.5} = \frac{6.35 \div 6.35}{63.5 \div 6.35} = \frac{1}{10} = 0.1$$. Thus, we have 0.1 moles of copper.

Since 1 mole of copper requires 2 moles of electrons, the moles of electrons needed for 0.1 moles of copper are: $$0.1 \times 2 = 0.2$$ moles of electrons.

Avogadro's constant, denoted as $$N_A$$, represents the number of electrons in one mole of electrons. Therefore, the number of electrons required is: $$0.2 \times N_A$$.

Expressing 0.2 as a fraction: $$0.2 = \frac{2}{10} = \frac{1}{5}$$. So, the number of electrons is: $$\frac{1}{5} \times N_A = \frac{N_A}{5}$$.

Comparing with the options:

• A: $$\frac{N_A}{20}$$
• B: $$\frac{N_A}{10}$$
• C: $$\frac{N_A}{5}$$
• D: $$\frac{N_A}{2}$$

The expression $$\frac{N_A}{5}$$ matches option C.

Hence, the correct answer is Option C.