The mass of zinc produced by the electrolysis of zine sulphate solution with a steady current of 0.015 A for 15 minutes is _________ × 10⁻⁴ g.
(Atomic mass of zinc =65.4 amu)

Solution

Chemistry ma’am · 4 steps

Find the charge first, then remember each Zn²⁺ ion needs two electrons to become zinc.

1. Charge passed. Convert the time to seconds: 15 min = 900 s. Then Q = It = 0.015 × 900 = 13.5 C.
2. Electrons per atom. At the cathode, Zn²⁺ + 2e⁻ → Zn. So 2 moles of electrons, that is 2 × 96500 C, deposit one mole of zinc, which is 65.4 g.

   m = (M Q)/nF = (65.4 × 13.5)/(2 × 96500) g

3. Put in numbers. The top is 65.4 × 13.5 = 882.9 and the bottom is 193000. So m = 4.57 × 10⁻³ g = 45.7 × 10⁻⁴ g.
4. So. The mass of zinc deposited is about 45.7 × 10⁻⁴ g, which rounds to 46.

The common slip: 91. Forgetting that zinc needs 2 electrons per ion (taking n = 1) doubles the mass and gives 91, not 46.

The idea to keep. In any Faraday's law question, write the electrode half-reaction first. The number of electrons in it is n, and the mass deposited is MQ/nF.

Answer from the NTA final answer key. Solution by PrepWiser.