Silver is electrodeposited on a metallic vessel of surface area 800 cm² by passing a current of 0.20 amp for 3.0 hrs. The thickness of silver deposited ___.

Given its density is 10.47g/cc [Atomic weight of Ag = 108]

2.88 × 10^–4 cm

The correct answer is option 3. 2.88 × 10^–4 cm.

Let the thickness of the silver deposited be \(x\) cm.

Weight of silver deposited = density x V = \(10.47 × 800x\)g

We know, from Faraday's law we know that

\(W = Zit\)

\(10.47 × 800x = \frac{108}{96500} × 0.2 × 3 × 3600\)

\(x = \frac{108 × 0.2 × 3 × 3600}{96500 ×10.47 ×800}\)

or, \(x = \frac{233280}{808284000}\)

or, \(x = 2.88 × 10^{− 4} cm\)