WAEC 2021 Chemistry · Question 42

How many coulombs of electricity would liberate 1.08g of Ag from a solution of silver salt? [Ag = 108.0; 1F = 96500 C]

Options

A.96500 C
B.9650 C
C.965 C
D.9.65 C

Answer: C. 965 C

Explanation written with AI and checked against the answer key

Analyze the question: We need the quantity of electricity (in coulombs) that deposits 1.08 g of silver from a silver salt solution. Concept: Faraday's laws of electrolysis. One mole of electrons (1 Faraday = 96500 C) deposits one mole of a metal if the metal ion carries a +1 charge. For silver: Ag⁺ + e⁻ → Ag So 1 F (96500 C) liberates 1 mole of Ag = 108 g. Solution: Step 1: Find the moles of Ag liberated. Moles = mass ÷ molar mass = 1.08 ÷ 108 = 0.01 mol Step 2: Each mole of Ag needs 1 mole of electrons, so the charge needed is: Q = 0.01 × 96500 = 965 C Alternatively, by proportion: 108 g of Ag needs 96500 C 1.08 g of Ag needs (1.08 ÷ 108) × 96500 = 965 C Answer: C. 965 C Tip: First check the ion's charge (Ag⁺ means 1 F per mole). Then use the ratio: Charge = (mass given ÷ molar mass) × charge on ion × 96500. Here 1.08 is exactly 1/100 of 108, so the answer is 96500 ÷ 100 = 965 C. Spotting such simple ratios saves time in the exam.

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