Here's the solution to questions 2 and 3: 2. During purification of copper by electrolysis, $1.48 \text{ g}$ of copper were deposited when a current was passed through aqueous copper (II) sulphate for $2 \frac{1}{2}$ hours. Calculate the amount of current passed. ($\text{Cu} = 63.5$, $1 \text{ Faraday} = 96500 \text{ C}$) Step 1: Write the half-reaction for the deposition of copper. $$\text{Cu}^{2+}(aq) + 2\text{e}^- \to \text{Cu}(s)$$ This shows that 2 moles of electrons are required to deposit 1 mole of copper. Step 2: Calculate the moles of copper deposited. $$\text{Moles of Cu} = \frac{\text{Mass of Cu}}{\text{Molar mass of Cu}}$$ $$\text{Moles of Cu} = \frac{1.48 \text{ g}}{63.5 \text{ g/mol}} \approx 0.023307 \text{ mol}$$ Step 3: Calculate the moles of electrons required. $$\text{Moles of electrons} = \text{Moles of Cu} \times 2$$ $$\text{Moles of electrons} = 0.023307 \text{ mol} \times 2 = 0.046614 \text{ mol}$$ Step 4: Calculate the total charge (Q) passed. $$\text{Q} = \text{Moles of electrons} \times \text{Faraday constant}$$ $$\text{Q} = 0.046614 \text{ mol} \times 96500 \text{ C/mol} \approx 4500.85 \text{ C}$$ Step 5: Convert the time to seconds. $$\text{Time (t)} = 2 \frac{1}{2} \text{ hours} = 2.5 \text{ hours}$$ $$\text{t} = 2.5 \text{ hours} \times 3600 \text{ s/hour} = 9000 \text{ s}$$ Step 6: Calculate the current (I). $$\text{Q} = \text{I} \times \text{t}$$ $$\text{I} = \frac{\text{Q}}{\text{t}}$$ $$\text{I} = \frac{4500.85 \text{ C}}{9000 \text{ s}} \approx 0.500 \text{ A}$$ The amount of current passed is $\boxed{0.500 \text{ A}}$ 3. The diagram below represents a set-up that can be used for the electrolysis of dilute sulphuric acid a) Name the electrodes A and B Electrode A is connected to the positive terminal of the battery. Electrode B is connected to the negative terminal of the battery. Anode: Electrode A Cathode: Electrode B b) Write an equation for the reaction taking place at electrode B Electrode B is the cathode (negative electrode). In dilute sulphuric acid, the species present are $\text{H}^+$, $\text{SO}_4^{2-}$, and $\text{H}_2\text{O}$. At the cathode, reduction occurs. Hydrogen ions are preferentially reduced over water. $$\boxed{2\text{H}^+(aq) + 2\text{e}^- \to \text{H}_2(g)}$$ c) What happens to the concentration dilute sulphuric acid as the reaction continues? At the cathode (B), $\text{H}^+$ ions are consumed to produce hydrogen gas. At the anode (A), water is oxidized to produce oxygen gas and $\text{H}^+$ ions: $2\text{H}_2\text{O}(l) \to \text{O}_2(g) + 4\text{H}^+(aq) + 4\text{e}^-$. The overall reaction is the decomposition of water: $2\text{H}_2\text{O}(l) \to 2\text{H}_2(g) + \text{O}_2(g)$. Since water is consumed and sulphuric acid acts as a catalyst (its ions are regenerated or not consumed), the amount of solvent (water) decreases. Therefore, the concentration of the dilute sulphuric acid increases. That's 2 down. 3 left today — send the next one.