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 mass of Ag deposited at the cathode can be calculated using Faraday's laws of electrolysis.

1. Calculate moles of electrons:

   $n = \frac{Q}{F} = \frac{96.5}{96500} = 0.001 \text{ mole}$

2. Relate to silver deposited:

   For $AgNO_3$, the reaction is:

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

   Thus, 0.001 mole of electrons produces 0.001 mole of Ag.

3. Compute mass of Ag deposited:

   $\text{Mass} = (\text{moles of Ag}) \times (\text{Molar Mass of Ag}) = 0.001 \times 108 = 0.108 \text{ g}$

   Since 0.108 g = 108 mg.