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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Step 1: Write the half-reactions for the electrolysis of NaOH solution using graphite electrodes. At the cathode (reduction): Water is reduced to hydrogen gas and hydroxide ions. 2H_2O(l) + 2e^- H_2(g) + 2OH^-(aq) At the anode (oxidation): Hydroxide ions are oxidized to oxygen gas and water. 4OH^-(aq) O_2(g) + 2H_2O(l) + 4e^- Step 2: Determine the moles of electrons passed. The quantity of electricity released is 0.5 \, F. One Faraday (1 \, F) corresponds to 1 mole of electrons. Therefore, moles of electrons passed = 0.5 \, mol e^-. Step 3: Calculate the moles of H_2 produced. From the cathode reaction, 2 moles of electrons produce 1 mole of H_2. Moles of H_2 = 0.5 \, mol e^- × 1 \, mol H_22 \, mol e^- = 0.25 \, mol H_2 Step 4: Calculate the moles of O_2 produced. From the anode reaction, 4 moles of electrons produce 1 mole of O_2. Moles of O_2 = 0.5 \, mol e^- × 1 \, mol O_24 \, mol e^- = 0.125 \, mol O_2 Step 5: Convert moles of gases to volume at STP. At STP (Standard Temperature and Pressure), 1 mole of any ideal gas occupies 22.4 \, L. Volume of H_2: Volume of H_2 = 0.25 \, mol × 22.4 \, L/mol = 5.6 \, L Volume of O_2: Volume of O_2 = 0.125 \, mol × 22.4 \, L/mol = 2.8 \, L The volume of H_2 released is 5.6 L and the volume of O_2 released is 2.8 L.