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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To calculate the volume of oxygen evolved, we need to follow these steps: 1. Write the balanced half-reaction for the evolution of oxygen. 2. Calculate the total charge passed through the solution. 3. Determine the moles of electrons transferred. 4. Calculate the moles of oxygen produced using stoichiometry. 5. Use the ideal gas law to find the volume of oxygen at the given conditions. Step 1: Write the half-reaction for oxygen evolution. When acidified water undergoes electrolysis, oxygen gas is produced at the anode. The half-reaction is: 2H_2O(l) → O_2(g) + 4H^+(aq) + 4e^- This equation shows that 4 moles of electrons are required to produce 1 mole of oxygen gas. Step 2: Calculate the total charge passed (Q). Given current, I = 2.5 A Given time, t = 1.5 mins First, convert time to seconds: t = 1.5 mins × 60 s/min = 90 s Now, calculate the charge using the formula Q = I × t: Q = 2.5 A × 90 s = 225 C Step 3: Determine the moles of electrons transferred (n_e). Given Faraday's constant, F = 96500 C/mol e^- The moles of electrons transferred are: n_e = (Q)/(F) = 225 C96500 C/mol e^- ≈ 0.0023316 mol e^- Step 4: Calculate the moles of oxygen produced (n_O_2). From the half-reaction, 4 moles of electrons produce 1 mole of O_2. n_O_2 = n_e × 1 mol O_24 mol e^- = 0.0023316 mol e^- × (1)/(4) ≈ 0.0005829 mol O_2 Step 5: Use the ideal gas law to find the volume of oxygen (V). The ideal gas law is PV = nRT. We need to solve for V: V = (nRT)/(P). Given conditions: Temperature, T = 285 K Pressure, P = 0.91 × 10^5 Nm^-2 (which is Pascals, Pa) Moles of O_2, n = 0.0005829 mol The ideal gas constant, R = 8.314 J mol^-1 K^-1 (or Pa m^3 mol^-1 K^-1) Substitute the values into the ideal gas law: V = (0.0005829 mol) × (8.314 J mol^-1 K^-1) × (285 K)0.91 × 10^5 Pa V = (1.3807086)/(91000) m^3 V ≈ 0.0000151726 m^3 Convert the volume from cubic meters to cubic centimeters: 1 m^3 = (100 cm)^3 = 10^6 cm^3 V =