In an electrolysis, a current of 200 A was passed through molten oxide of metal Q for 58 minutes and 64.8 g of the metal deposited. Determine:

Here's the solution to question 4: 4. In an electrolysis, a current of 200 A was passed through molten oxide of metal Q for 58 minutes and 64.8 g of the metal deposited. Determine: Q = 27, 1F = 96500 C, molar gas volume stp = 22.4 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 min × 60 s/min = 3480 s Now, calculate the charge: Q_total = I × t Q_total = 200 A × 3480 s = 696000 C Step 2: Calculate the moles of metal Q deposited. Moles of Q = Mass of QMolar mass of Q Moles of Q = 64.8 g27 g/mol = 2.4 mol Step 3: Calculate the moles of electrons passed. Moles of electrons = Q_totalFaraday constant Moles of electrons = 696000 C96500 C/mol ≈ 7.2124 mol Step 4: Determine the charge on metal Q (n). The deposition of metal Q from its oxide involves the reduction of Q^n+ ions: Q^n+ + ne^- 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 = Moles of electronsMoles of Q n = 7.2124 mol2.4 mol ≈ 3.005 Rounding to the nearest whole number, the charge on metal Q is 3. The charge on metal Q is 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 (O^2-) are oxidized at the anode to produce oxygen gas. 2O^2- O_2(g) + 4e^- 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). Moles of electrons ≈ 7.2124 mol Step 3: Calculate the moles of oxygen gas produced. Moles of O_2 = Moles of electrons4 Moles of O_2 = 7.2124 mol4 ≈ 1.8031 mol Step 4: Calculate the volume of oxygen gas at standard temperature and pressure (STP). Volume of O_2 = Moles of O_2 × Molar gas volume at STP Volume of O_2 = 1.8031 mol × 22.4 dm^3/mol Volume of O_2 ≈ 40.389 dm^3 Rounding to three significant figures: The volume of oxygen gas produced is 40.4 dm^3 3 done, 2 left today. You're making progress.