Aqueous CuSO₄ is subjected to electrolysis using copper electrodes. If 3.175g of copper is deposited at cathode then at anode

According to faraday's law:
W=ZQ=(E/96500)∗Q.
Here, we are flowing same charge through cathode and anode, so at cathode and anode same mass reacts.
At cathode if 3.175 gm is reducing then at anode 3.175gm of Cu will be oxidized into solution so there is a decrease of 3.175 g of mass takes place.

$\mathrm{moles} \mathrm{of} \mathrm{Cu} =\frac{\mathrm{mass} \mathrm{of} \mathrm{Cu} \mathrm{deposited}}{\mathrm{Atomic} \mathrm{mass} \mathrm{of} \mathrm{Cu}}= \frac{3.175}{63.5}=0.05 \mathrm{mol}$

Alternatively

Since copper electrode is used it is example of attackable electrode

During attackable electrode process at anode anode it self undergoes oxidation

For same quantity of charge passed same equivalent under go change

No.of equivalents of copper deposited at cathode = number of equivalents of copper oxidized at anode

= 0.1 eq = 0.05 mol