To calculate the quantity of electricity required, we will use Faraday's laws of electrolysis. We need the molar mass of copper and Faraday's constant. We will assume copper is in the +2 oxidation state (Cu²⁺), which is common in copper compounds. Step 1: Determine the molar mass of copper and Faraday's constant. • Molar mass of copper ($M_{Cu}$) = $63.5 \text{ g/mol}$ • Faraday's constant ($F$) = $96485 \text{ C/mol e⁻}$ Step 2: Write the half-reaction for the liberation of copper. Assuming copper is in the +2 oxidation state, the reduction half-reaction is: $$\text{Cu}^{2+}(aq) + 2\text{e}^- \to \text{Cu}(s)$$ This reaction shows that 2 moles of electrons are required to liberate 1 mole of copper. Step 3: Calculate the number of moles of copper to be liberated. Given mass of copper ($m$) = $10 \text{ g}$. Number of moles ($n_{Cu}$) = $\frac{\text{mass}}{\text{molar mass}}$ $$n_{Cu} = \frac{10 \text{ g}}{63.5 \text{ g/mol}}$$ $$n_{Cu} \approx 0.15748 \text{ mol}$$ Step 4: Calculate the total moles of electrons required. From the half-reaction, 2 moles of electrons are needed per mole of copper. Moles of electrons ($n_e$) = $n_{Cu} \times 2$ $$n_e = 0.15748 \text{ mol} \times 2$$ $$n_e \approx 0.31496 \text{ mol e⁻}$$ Step 5: Calculate the quantity of electricity (charge) in coulombs. Quantity of electricity ($Q$) = $n_e \times F$ $$Q = 0.31496 \text{ mol e⁻} \times 96485 \text{ C/mol e⁻}$$ $$Q \approx 30398.9 \text{ C}$$ Rounding to three significant figures: $$Q \approx 30400 \text{ C}$$ The quantity of electricity required is: $$\boxed{\text{30400 C}}$$ What's next?