Question

Silver is electrodeposited on a metallic vessel of surface area 800 cm² by passing current of 0.2 amp for 3 hours. Calculate the thickness of silver deposited. 

(Density of silver = 10.47 g cm⁻³, atomic mass of silver = 107.92 amu)

Answer 1

Given: Surface area = 800 cm²

Current (I) = 0.2 A

Time (t) = 3 hours = 3 × 3600 = 10800 s

Atomic mass of Ag = 107.92 g/mol

Density of Ag = 10.47 g/cm³

Faraday’s constant (F) = 96500 C/mol

Electrode reaction:

\[\ce{Ag+ + e− -> Ag}\]

⇒ 1 mol e⁻ deposits 1 mol Ag

Q  = I × t

= 0.2 × 10800

= 2160 C

Moles of Ag deposited (n) = `Q/F`

= `2160/96500`

= 0.02237 mol

Mass of Ag deposited (m) = n × M

= 0.02237 × 107.92

≈ 2.414 g

Volume of Ag deposited (V) = `"mass"/"density"`

= `0.2305/800`

= 2.88 × 10⁻¹⁴ cm