What is the moles of copper formed when a current of copper 0.5A is passed through a copper sulphate for 35 minutes

$$Q = 0.5 \text{ A} \times 2100 \text{ s}$$ $$Q = 1050 \text{ C}$$

$$n_e = \frac{Q}{F}$$ $$n_e = \frac{1050 \text{ C}}{96485 \text{ C/mol e}^-}$$ $$n_e \approx 0.010882 \text{ mol e}^-$$

The reduction of copper ions is: $\text{Cu}^{2+} + 2\text{e}^- \to \text{Cu}$
This means 2 moles of electrons are required to deposit 1 mole of copper.

$$\text{Moles of Cu} = \frac{\text{Moles of e}^-}{2}$$ $$\text{Moles of Cu} = \frac{0.010882 \text{ mol e}^-}{2}$$ $$\text{Moles of Cu} \approx 0.005441 \text{ mol}$$

The moles of copper formed is $\text{0.005441 mol}$.