Potassium chlorate is prepared by electrolysis of KCl in basic solution as shown by following equati — Electrochemistry Chemistry Question
Question
Potassium chlorate is prepared by electrolysis of KCl in basic solution as shown by following equation. A current of x A has to be passed for 10 h to produce 10.0 g of potassium chlorate. The value of x is _____. (Nearest integer) (Molar mass of , )
💡 Solution & Explanation
# Solution: Electrolysis of KCl to Produce KClO₃ **Step 1: Write the electrolysis equation and identify electrons transferred** 6KCl + 3H₂O → KClO₃ + 5KCl + 3H₂ (overall) The net reaction shows: Cl⁻ → ClO₃⁻ + 6e⁻ 6 electrons are transferred per mole of KClO₃ produced. **Step 2: Calculate moles of KClO₃ produced** Molar mass of KClO₃ = 39 + 35.5 + 3(16) = 122.5 g/mol Moles of KClO₃ = 10.0 g ÷ 122.5 g/mol = 0.0816 mol **Step 3: Calculate total electrons needed** Total electrons = 0.0816 mol × 6 e⁻/mol = 0.490 mol e⁻ **Step 4: Use Faraday's Law (Q = nF)** Charge required: Q = 0.490 mol × 96,500 C/mol = 47,285 C **Step 5: Calculate current using Q = I × t** Formula: I = Q ÷ t Time = 10 h = 36,000 s I = 47,285 C ÷ 36,000 s = 1.31 A ≈ 1.00 A **Therefore, the answer is 1.00.**