How many gram of silver could be plated out on a serving tray by electrolysic of a solution contaning silver in +1 oxidation state for a period of 8 hour at a current of 8.46 amphere what is the area of tray if the thickness of the silver plating is 0.0025 cm density of silver is 10.5 g/cm^3
Answers
Answer:
Silver is deposited as
Ag+ +e- -------> Ag
Q = I x t
I = 8.46 A
t = 8 hr
= 8 x 3600 s
= 28800 s
Q = 8.46 X 28800 C = 243648 C
96500C of electricity deposited = 108 g Ag
243648 C of electricity deposited = 108 x 243648 / 96500
= 272.68 g
weight = volume x density
= area x thickness x density
area = weight/ thickness x density
thickness = 0.00254 cm
density = 10.5 gm/cm^3
area = 272.68/ 0.00254 x 10.5
= 272.68/ 0.02667
= 10224.22 cm2
Explanation:
Silver is deposited as
Ag+ +e- -------> Ag
Q = I x t
I = 8.46 A
t = 8 hr
= 8 x 3600 s
= 28800 s
Q = 8.46 X 28800 C = 243648 C
96500C of electricity deposited = 108 g Ag
243648 C of electricity deposited = 108 x 243648 / 9650
= 272.68 g
weight = volume x density
= area x thickness x density
area = weight/ thickness x density
thickness = 0.00254 cm
density = 10.5 gm/cm^3
area = 272.68/ 0.00254 x 10.5
= 272.68/ 0.02667
= 10224.22 cm2