Electrochemistry: Speed-Solving Techniques (10)

Question

Electrochemistry: speed-solving techniques for JEE+NEET

doubts.ai · Accepted Answer

Solution — Step by Step Q = I t = 5 	imes 30 	imes 60 = 9000 C. n e = Q / F = 9000 / 96500 0.0933 mol. Cu^{2+} + 2e^- → Cu. So 2 moles of electrons deposit 1 mole of Cu. Moles of Cu = 0.0933 / 2 = 0.0467 mol. Mass = 0.0467 63.5 \approx 2.96 g. Final answer: 2.96 g of Cu deposited. Why This Works Faraday's law: mass deposited = ItM{nF}, where n is the number of electrons per metal ion. For Cu^{2+}, n = 2. Plug in and you get the answer in one shot. This is the most predictable electrochemistry calculation in JEE/NEET. The only thing that changes between problems is the metal (and hence n and atomic mass). Alternative Method (Speed) Direct formula: m = ItM{nF} = 5 	imes 1800 	imes 63.5{2 96500 = 571500{193000} 2.96 g. One line. Done. Memorise m = ItM{nF} — Faraday's law in one line. For Cu, n = 2. For Ag, n = 1 (so silver deposits twice as much per coulomb on a per-mole basis). For Al, n = 3. Common Mistake Forgetting to convert minutes to seconds. 30 minutes is 1800 s, not 30 s. A factor-of-60 error gives a factor-of-60 wrong answer. Using the wrong n value. Cu^{2+} needs 2 electrons per atom, so n = 2. Cu^+ would need only 1. Read the formula in the problem carefully — CuSO 4 contains Cu^{2+}.

Question

Solution — Step by Step

Why This Works

Alternative Method (Speed)

Common Mistake

Try These Next