Calculate the volume in dm³ of hydrogen produced at s.t.p. during the electrolysis of dilute tetraoxosulphate (VI) acid, if 1.0 A was used for 15 minutes. [1F = 96,500 C, Volume of gas at s.t.p. = 22.4 dm³]

To calculate the volume of hydrogen produced, we will use Faraday's laws of electrolysis. Step 1: Calculate the total charge passed (Q). First, convert the time from minutes to seconds. t = 15 min × 60 s/min = 900 s Now, calculate the charge using the formula Q = I × t. Q = 1.0 A × 900 s = 900 C Step 2: Write the half-reaction for hydrogen production at the cathode. During the electrolysis of dilute tetraoxosulphate (VI) acid (H_2SO_4), hydrogen ions (H^+) are reduced at the cathode. 2H^+(aq) + 2e^- H_2(g) This reaction shows that 2 moles of electrons are required to produce 1 mole of hydrogen gas. Step 3: Calculate the moles of electrons transferred. Use Faraday's constant (F = 96,500 C/mol) to find the moles of electrons (n_e). n_e = (Q)/(F) = 900 C96,500 C/mol Step 4: Calculate the moles of hydrogen produced. From the cathode reaction, 2 moles of electrons produce 1 mole of H_2. Moles of H_2 = (1)/(2) × n_e = (1)/(2) × (900)/(96,500) mol Moles of H_2 = (450)/(96,500) mol ≈ 0.0046632 mol Step 5: Calculate the volume of hydrogen produced at s.t.p. At s.t.p., 1 mole of any gas occupies 22.4 dm^3. Volume of H_2 = Moles of H_2 × Molar volume at s.t.p. Volume of H_2 = (450)/(96,500) mol × 22.4 dm^3/mol Volume of H_2 = (10080)/(96500) dm^3 Volume of H_2 ≈ 0.10445 dm^3 Rounding to three significant figures: Volume of H_2 ≈ 0.104 dm^3 Send me the next one 📸