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)

• A. 2
• B. 1
• C. 3
• D. 4

Hint:

Always convert time into seconds and use Faraday’s formula directly for such numerical problems.

Answer 1

The Correct Option is D

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 \(n = 4\).