Question:
When 3 amp current was passed through an aqueous solution of salt of a metal M (atomic weight 106.4 u) for 1 hour, 2.977 g of Mn+ was deposited at cathode. The value of n is (1 F = 96500 C mol-1)
When 3 amp current was passed through an aqueous solution of salt of a metal M (atomic weight 106.4 u) for 1 hour, 2.977 g of Mn+ was deposited at cathode. The value of n is (1 F = 96500 C mol-1)
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Always convert time into seconds and use Faraday’s formula directly for such numerical problems.
Updated On: Jun 3, 2025
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The Correct Option is D
Solution and Explanation
Use Faraday’s laws: Weight deposited = \(\frac{E I t \cdot M}{n F}\).
Here, \(E = 3\) A, \(t = 3600\) s, \(M = 106.4\) g/mol, \(F = 96500\) C/mol, and deposited mass = 2.977 g.
Solving gives \(n = 4\).
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