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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Here are the solutions for section G and the chemistry problem. G. Fill in the blanks with suitable verbs from the brackets. 1. The army attacked the enemy camp at night. 2. The river flowed across the town. Chemistry Problem The reaction at the cathode of a lead-acid battery during discharge, starting with PbO_2 and consuming 2e^-, to produce PbSO_4 and H_2O is: Step 1: Identify the given components and the type of reaction. The problem shows PbO_2 at the cathode, 2e^- consumed, and PbSO_4 + 2H_2O as products. This is a reduction half-reaction in a lead-acid battery. Step 2: Write the unbalanced half-reaction with the given species. PbO_2 + 2e^- → PbSO_4 + 2H_2O Step 3: Balance the atoms other than oxygen and hydrogen. Lead (Pb) is already balanced (1 on each side). Sulfur (S) is present in PbSO_4 on the product side, so we need SO_4^2- on the reactant side. PbO_2 + SO_4^2- + 2e^- → PbSO_4 + 2H_2O Step 4: Balance oxygen atoms by adding water molecules (if needed) and hydrogen atoms by adding H^+ ions (since it's an acidic environment). Reactant side has 2 oxygen from PbO_2 and 4 oxygen from SO_4^2- (total 6 oxygen). Product side has 4 oxygen from PbSO_4 and 2 oxygen from 2H_2O (total 6 oxygen). Oxygen is balanced. Product side has 4 hydrogen atoms from 2H_2O. Add 4H^+ to the reactant side. PbO_2(s) + SO_4^2-(aq) + 4H^+(aq) + 2e^- → PbSO_4(s) + 2H_2O(l) Step 5: Verify the charge balance. Reactant side charge: (0) + (-2) + (+4) + (-2) = 0 Product side charge: (0) + (0) = 0 The charges are balanced. The complete balanced half-reaction is: PbO_2(s) + SO_4^2-(aq) + 4H^+(aq) + 2e^- → PbSO_4(s) + 2H_2O(l) What's next?