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If 0.5 A current is passed through acidified silver nitrate solution for 100 minutes. The mass of silver deposited on cathode is (eq. wt. of silver nitrate = 108):

1. 2.3523 g
2. 3.3575 g
3. 5.3578 g
4. 6.3575 g

Correct answer: 3.3575 g

Solution

Using Faraday's first law of electrolysis, mass deposited (m) = (Z × I × t), where Z = equivalent weight/Faraday constant. Here, Z = 108/96500. Substituting I = 0.5 A and t = 100 × 60 s, m = (108/96500) × 0.5 × 6000 = 3.3575 g.

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