What mass of silver is deposited when a current of 2.6 A is passed through a solution of a silver salt for 40 minutes? [Ag= 108 ; F=96500C]

Analyze the question: This is a Faraday's law of electrolysis problem. We need the mass of silver deposited by a known current over a known time.

Concept:

• Charge: Q = I × t (t in seconds)
• Ag⁺ + e⁻ → Ag, so 1 mole of electrons (96500 C) deposits 1 mole of Ag (108 g)
• Mass = (Q × molar mass) / (96500 × n), where n = 1 for silver

Solution:

1. Convert the time to seconds: t = 40 × 60 = 2400 s

2. Calculate the charge: Q = 2.6 × 2400 = 6240 C

3. Calculate the mass of silver: Mass = (6240 × 108) / 96500 = 673,920 / 96,500 ≈ 6.98 g

Check: 96500 × 6.98 = 673,570, which is very close to 673,920, so the answer is about 6.983 g. This rounds to 6.98 g.

Answer: C. 6.98 g.

Tip: For Faraday's law questions, always follow this order: Q = It (convert minutes to seconds first), then moles of electrons = Q/96500, then use the ion's charge to find the moles of metal. For Ag⁺, the ratio is 1:1, which makes it easy. The most common mistake is forgetting to convert minutes to seconds.