This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Hey bazuunelson — ready when you are. 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: i) Charge on metal Q ii) The volume of oxygen gas produced at standard temperature and pressure (STP). (Q = 27, 1 F = 96500 C, molar gas volume at STP = 22.4 dm^3) Step 1: Calculate the total charge passed. The current I = 200 A and time t = 58 minutes. Convert time to seconds: t = 58 min × 60 s/min = 3480 s The total charge Q_total is given by Q_total = I × t. Q_total = 200 A × 3480 s = 696000 C Step 2: Calculate the moles of metal Q deposited. The molar mass of metal Q is given as 27 g/mol. The mass of metal Q deposited is 64.8 g. Moles of Q = Mass of QMolar mass of Q = 64.8 g27 g/mol = 2.4 mol Step 3: Determine the charge on metal Q. According to Faraday's laws of electrolysis, the charge required to deposit one mole of a metal is n × F, where n is the charge on the metal ion and F is Faraday's constant (96500 C/mol). The total charge passed is 696000 C and 2.4 mol of metal Q was deposited. So, 2.4 mol × n × F = Q_total. 2.4 mol × n × 96500 C/mol = 696000 C n = 696000 C2.4 mol × 96500 C/mol = (696000)/(231600) ≈ 3.005 The charge on metal Q is approximately 3. Therefore, the metal ion is Q^3+. i) Charge on metal Q: The charge on metal Q is 3 Step 4: Write the half-reaction for the deposition of metal Q. Q^3+(l) + 3e^- Q(s) Step 5: Write the half-reaction for the production of oxygen gas. In molten oxide electrolysis, the oxide ions (O^2-) are oxidized at the anode to produce oxygen gas. 2O^2-(l) O_2(g) + 4e^- Step 6: Calculate the moles of electrons transferred. From Step 3, we know that 2.4 mol of Q^3+ ions required 3 moles of electrons per mole of Q. Moles of electrons = 2.4 mol Q × 3 mol e^-1 mol Q = 7.2 mol e^- Alternatively, using the total charge: Moles of electrons = Q_totalF = 696000 C96500 C/mol e^- ≈ 7.21 mol e^- We will use 7.2 mol e^- for consistency with the calculated charge on Q. Step 7: Calculate the moles of oxygen gas produced. From the half-reaction for oxygen production, 4 mol of electrons produce 1 mol of O_2. Moles of O_2 = 7.2 mol e^- × 1 mol O_24 mol e^- = 1.8 mol O_2 Step 8: Calculate the volume of oxygen gas produced at STP. At STP, 1 mol of any gas occupies 22.4 dm^3. Volume of O_2 = 1.8 mol × 22.4 dm^3/mol = 40.32 dm^3 ii) The volume of oxygen gas produced at standard temperature and pressure (STP): The volume of oxygen gas produced is 40.32 dm^3 Send me the next one 📸