Potassium chlorate is prepared by the electrolysis of KCl in basic solution If only 60% of the curre — Electrochemistry Chemistry Question
Question
Potassium chlorate is prepared by the electrolysis of KCl in basic solution If only 60% of the current is utilized in the reaction, the time (rounded to the nearest hour) required to produce 10g of KClO using a current of 2A is …………… (Given: F = 96,500 C mol ; molar mass of KClO = 122g mol ) 3 -1 3 -1
💡 Solution & Explanation
**Step 1: Write the electrolysis reaction and determine electrons transferred** In basic solution, KCl is oxidized to KClO₃: Cl⁻ → ClO₃⁻ + 6e⁻ 6 moles of electrons are required to produce 1 mole of KClO₃. **Step 2: Calculate moles of KClO₃ needed** Moles of KClO₃ = 10 g ÷ 122 g/mol = 0.0820 mol **Step 3: Calculate total electrons (moles) required** Moles of electrons = 0.0820 mol × 6 = 0.490 mol **Step 4: Account for current efficiency (60%)** Since only 60% of current is utilized: Effective moles of electrons = 0.490 ÷ 0.60 = 0.817 mol **Step 5: Convert moles of electrons to charge (coulombs)** Charge required = 0.817 mol × 96,500 C/mol = 78,840 C **Step 6: Calculate time using Q = I × t** Time = Q ÷ I = 78,840 C ÷ 2 A = 39,420 seconds Convert to hours: 39,420 s ÷ 3,600 s/hr = 10.95 hrs ≈ **11 hours** Therefore, the answer is 11.00.