A current of 2 A flows for 10 minutes. Calculate the quantity of electricity in Coulombs and Faradays, and the mass of copper deposited from a CuSO4 solution.

$$t = 10 \text{ min} \times 60 \text{ s/min}$$ $$t = 600 \text{ s}$$

The formula for quantity of electricity is $Q = I \times t$.

$$Q = 2 \text{ A} \times 600 \text{ s}$$ $$Q = \boxed{\text{1200 C}}$$

We know that 1 Faraday ($F$) = 96,500 C.

$$\text{Quantity in Faradays} = \frac{\text{Quantity in Coulombs}}{96,500 \text{ C/F}}$$ $$\text{Quantity in Faradays} = \frac{1200 \text{ C}}{96,500 \text{ C/F}}$$ $$\text{Quantity in Faradays} \approx \boxed{\text{0.0124 F}}$$

The copper ions in $CuSO_4$ are $Cu^{2+}$.

$$Cu^{2+}(aq) + 2e^- \to Cu(s)$$

From the reaction, 2 moles of electrons are required to deposit 1 mole of copper.

2 moles of electrons correspond to $2 \times 96,500 \text{ C}$ or 2 Faradays.
1 mole of copper has a mass of 64 g.
So, 2 Faradays deposit 64 g of copper.

Using the quantity of electricity in Faradays from Step 3:

$$\text{Mass of Cu} = \left( \text{Quantity in Faradays} \right) \times \frac{\text{Molar mass of Cu}}{\text{Moles of electrons}}$$ $$\text{Mass of Cu} = 0.0124 \text{ F} \times \frac{64 \text{ g/mol}}{2 \text{ F/mol}}$$ $$\text{Mass of Cu} = 0.0124 \times 32 \text{ g}$$ $$\text{Mass of Cu} \approx \boxed{\text{0.3968 g}}$$