Question
Question: How much time is required to deposit \[1\times {{10}^{-3}}cm\] thick layer of silver(density is \[1....
How much time is required to deposit 1×10−3cm thick layer of silver(density is 1.05g/cm3) on a surface of area 100cm2 by passing a current of 5A through the AgNO3 solution ?
A. 125s
B. 115s
C. 18.7s
D. 27.25s
Solution
According to Faraday’s first law the amount of substance produced at an electrode during electrolysis is directly proportional to the quantity of electricity passed though the electrolyte.
Hence, we can write as:
W∝Q⇒Q=it
W∝it
⇒W=Zit
W = Deposited weight
Q = Quantity of electricity passed
I = Current required to electrolysis
T = Time required for the process
Z = Electro-chemical equivalent of the substance
Complete step by step answer:
Here we have the electrolysis of silver nitrate solution(AgNO3)
So we have to find the mass of silver deposited from the question by using density and volume
We have known that;
Mass=Density×Volume
Mass=Density×surface.area×thickness
We have density, thickness and surface area as 1.05g/cm3,1×10−3cm and 100cm2 respectively. So we could write as;
Mass=1.05×100×1×10−3
⇒Mass=0.105g
So, we have to find the required time to deposit 0.105g of silver by passing 5A using the equation of Faraday’s first law. According Faraday’s first law
W=Zit
From this equation we have to find the required time. Hence we could rewrite the equation as;
t=ZiW
Z=96500Equivalent.weight, here the equivalent weight of silver is 108
W=0.105g
⇒i=5A
Substitute these values in the above equation.
So we get,
⇒t=5×1080.105×96500
⇒t=18.7s
Hence we get the time required is 18.7s to deposit 0.105g by passing 5A of current through the silver nitrate solution
So, the correct answer is Option C.
Note: Also, Other than Faraday’s first law we have Faraday’s second law which states that Masses of different substances liberated or dissolved by the same amount of electricity passed is directly proportional to their chemical equivalent. Hence we have to be damn sure about the question where law has to be used in the questions to solve it.