Write the cathode reaction

In aqueous $CuSO_4$, copper ions are reduced at the cathode: $Cu^{2+} + 2e^- \rightarrow Cu$

Use Faraday's law of electrolysis

• $1$ mole of $Cu$ requires $2$ moles of electrons.
• $1$ Faraday $= 1$ mole of electrons.

Therefore:

• $2$ Faraday $= 2$ moles of electrons
• This will deposit: $\frac{2\text{ mol } e^-}{2\text{ mol } e^- / 1\text{ mol Cu}} = 1\text{ mol Cu}$

Convert moles of copper to mass

Atomic mass of copper $= 63.5\,g/mol$

So deposited mass: $1 \times 63.5 = 63.5\,g$

Match with the options

The correct option is: $\boxed{A:\ 63.5\,g}$

Compare with stored correct answer

Stored correct answer: $A$

My derived answer is also $A$, so they agree.