Question

Silver is deposited on a metallic vessel by passing a current of 0.2 A for 3 hrs. Calculate the weight of silver deposited. (At mass of silver= 108 amu ,F=96500 C?)

Answer 1

Q = it= 0.5 x 2 x 60 x 60 = 3600 C
${{Ag}^{+}}$ + ${{e}^{-}}$ ------> Ag
(1F =96500 C 108 g
96500 C of charge deposits 108 g Ag
.‘. 3600 C of charge will deposit Ag = 108x3600/96500 = 4.029 g
Thickness = Mass/AreaX Density
4.029/ 900 x 10.5
= 4.26 x${{10}^{-4}}$ cm

So Have You Got It

Answer 2

Answer :

2.95 × 10⁻⁴ cm

Explanation :

Given :

Current, I = 0.2 A

Time, t = 3 hrs

= (3 × 60 × 60) sec

= 10800 sec

As we know,

• Charge = Current × Time

Q = I × t

Putting values, we get

Q = (0.2)(10800)

= 2160 C

Now,

Mass of silver deposited by passing 96500 C charge = 108 g

Mass of silver deposited by passing 1 C charge = (108/96500) g

Mass of silver deposited by passing 2160 C charge = [ (108/96500) × 2160 ] g

= 2.417 g

Density of silver = 10.47 g/cm³ (given)

Mass of silver = 2.417 g

• Volume of silver = Mass of silver/Density of silver

Putting values, we get

Volume of silver = (2.417/10.47) cm³

= 0.236 cm³

We can also write volume as :

• Volume of silver = Area of silver × Length (Thickness) of silver

Thickness of silver = Volume of silver/Area of silver

Area of silver = 800 cm² (given)

Volume of silver = 0.236 cm³

Putting values, we get

Thickness of silver = (0.236/800) cm

= 2.95 × 10⁻⁴ cm