Silver is electroplated on an electrode of surface area 10000 CM square by passage of 0.65 ampere current for 16m 40s .
given that density of silver Age 4 gram per mm calculate thickness of silver deposited.
Answers
Answered by
0
Answer:
0.1817 * 10^(-7) cm
Explanation:
Time, t= 16m 40s = (16*60) + 40 = 1000s
Current, I = 0.65 A
Let the weight of silver deposited be w grams.
Then, by Faraday's 1st Law Of Electrolysis,
w = z*I*t
w = (E/96500)*0.65*1000 (if you know Silver's electrochemical equivalent's value, you can put it in the formula directly or you can divide the chemical equivalent by 96500 to get it)
w = (108/96500)*0.65*1000 [E = {(Atomic Mass of Element/ Valency of Element)}]
w = 0.0011191*0.65*1000
w = 0.727 gram
Now, volume of Silver deposited = Mass/ Density
= 0.727 gm/ 4 gm per (mm)^3
= 0.1817 (mm)^3 = 0.1817 * 10^(-3) (cm)^3
Now, thickness of Ag deposited = Volume/ Surface Area
= 0.1817 * 10^(-3) (cm)^3 / 10000 (cm)^2
= 0.1817 * 10^(-7) cm
Similar questions