Question

Calculate the mass of Ag deposited at cathode when 96.5 coulomb of electricity is passed through a solution of $AgNO_3$.

(Molar Mass of Ag = 108 g $mol^{-1}$)

• A. 5.4 mg
• B. 16.2 mg
• C. 21.2 mg
• D. 108 mg

Answer 1

Answer: (D)

108 mg

Answer 2

The deposition of silver occurs via the reaction:

$Ag^+ + e^- \rightarrow Ag$

Using Faraday's law:

$Mass\,deposited = \frac{Molar\,mass \times Q}{F}$

Here,

Molar mass of Ag = 108 g/mol, Q = 96.5 C, F = 96500 C/mol.

So,

$Mass = \frac{108 \times 96.5}{96500} = \frac{10422}{96500} = 0.108 \,g = 108 \,mg$