Here's the solution to question 4: 4. In an electrolysis, a current of $200 \text{ A}$ was passed through molten oxide of metal $\text{Q}$ for $58$ minutes and $64.8 \text{ g}$ of the metal deposited. Determine: $\text{Q} = 27$, $1\text{F} = 96500\text{ C}$, molar gas volume stp $= 22.4\text{ dm}^3$ i) Charge on metal Q Step 1: Calculate the total quantity of electricity passed ($Q_{total}$). First, convert time from minutes to seconds: $$t = 58 \text{ min} \times 60 \text{ s/min} = 3480 \text{ s}$$ Now, calculate the charge: $$Q_{total} = I \times t$$ $$Q_{total} = 200 \text{ A} \times 3480 \text{ s} = 696000 \text{ C}$$ Step 2: Calculate the moles of metal Q deposited. $$\text{Moles of Q} = \frac{\text{Mass of Q}}{\text{Molar mass of Q}}$$ $$\text{Moles of Q} = \frac{64.8 \text{ g}}{27 \text{ g/mol}} = 2.4 \text{ mol}$$ Step 3: Calculate the moles of electrons passed. $$\text{Moles of electrons} = \frac{Q_{total}}{\text{Faraday constant}}$$ $$\text{Moles of electrons} = \frac{696000 \text{ C}}{96500 \text{ C/mol}} \approx 7.2124 \text{ mol}$$ Step 4: Determine the charge on metal Q ($n$). The deposition of metal Q from its oxide involves the reduction of $\text{Q}^{n+}$ ions: $$\text{Q}^{n+} + n\text{e}^- \to \text{Q}(s)$$ From this equation, $n$ moles of electrons are required to deposit 1 mole of Q. Therefore, the ratio of moles of electrons to moles of Q is $n$: $$n = \frac{\text{Moles of electrons}}{\text{Moles of Q}}$$ $$n = \frac{7.2124 \text{ mol}}{2.4 \text{ mol}} \approx 3.005$$ Rounding to the nearest whole number, the charge on metal Q is 3. The charge on metal Q is $\boxed{3}$ ii) The volume of oxygen gas produced at standard temperature and pressure Step 1: Identify the reaction at the anode. Since molten oxide of metal Q is electrolyzed, oxide ions ($\text{O}^{2-}$) are oxidized at the anode to produce oxygen gas. $$2\text{O}^{2-} \to \text{O}_2(g) + 4\text{e}^-$$ This equation shows that 4 moles of electrons are required to produce 1 mole of oxygen gas. Step 2: Use the moles of electrons passed (calculated in part i). $$\text{Moles of electrons} \approx 7.2124 \text{ mol}$$ Step 3: Calculate the moles of oxygen gas produced. $$\text{Moles of O}_2 = \frac{\text{Moles of electrons}}{4}$$ $$\text{Moles of O}_2 = \frac{7.2124 \text{ mol}}{4} \approx 1.8031 \text{ mol}$$ Step 4: Calculate the volume of oxygen gas at standard temperature and pressure (STP). $$\text{Volume of O}_2 = \text{Moles of O}_2 \times \text{Molar gas volume at STP}$$ $$\text{Volume of O}_2 = 1.8031 \text{ mol} \times 22.4 \text{ dm}^3\text{/mol}$$ $$\text{Volume of O}_2 \approx 40.389 \text{ dm}^3$$ Rounding to three significant figures: The volume of oxygen gas produced is $\boxed{40.4 \text{ dm}^3}$ 3 done, 2 left today. You're making progress.