Silver is electro-deposited on a metallic vessel of surface area 800 cm' by passing a current of 0-2 ampere for 3 ho
the thickness of silver deposited given that its density is 10:47 g cm-. (At. mass of Ag = 107.92). (H.P.
[Ans. 2-8
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Answers
Answer :
2.87 × 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 96487 C charge = 107.92 g
Mass of silver deposited by passing 1 C charge = (107.92/96487) g
Mass of silver deposited by passing 2160 C charge = [ (107.92/96487) × 2160 ] g
= 2.415 g
Density of silver = 10.47 g/cm³ (given)
Mass of silver = 2.415 g
- Volume of silver = Mass of silver/Density of silver
Putting values, we get
Volume of silver = (2.415/10.47) cm³
= 0.23 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 = 0.23 cm³
Putting values, we get
Thickness of silver = (0.23/800) cm
= 2.87 × 10⁻⁴ cm