Question

If 96500 coulombs of electricity liberates one gram equivalent of any substance, the time taken for a current of 0.15 amperes to deposit 20 mg of copper from a solution of copper sulphate is (Chemical equivalent of copper = 32)
A. 5 min 20 sec
B. 6 min 42 sec
C. 4 min 40 sec
D. 5 min 50 sec

Answer 1

Hint: Since the electricity is passed through this solution, so laws of electrolysis will be used here. These laws of electrolysis are used to obtain pure substances when electric current is passed and electrons are either gained or lost.
Formula Used: According to the formula of Faraday’s first law:
W=ZQ……(i)
‘W’ is the mass deposited or liberated
‘Q’ is amount of charge passed
‘Z’ is the electrochemical equivalent of the substance
And it is known that Q=It
Substituting the value of Q in equation (i), we get
W=ZIt……(ii)

Complete Step by Step Answer:
According to Faraday’s first law of electrolysis, during electrolysis the amount of substance that is released or deposited at cathode or anode is directly proportional to the amount of charge passed through the electrolytic solution.
Given W= 20 mg or
\[W = 20 \times {10^{ - 3}}g\]
I = 0.15 A
The value of ‘Z’ can be written as
\[Z = \dfrac{E}{{96500}}\]
Where ‘E’ is the chemical equivalent which is given as E=32
Substituting all the given values in equation (ii) and solving we get
\[20 \times {10^{ - 3}} = \dfrac{{32}}{{96500}} \times 0.15 \times t\]
Solving the above equation for the value of ‘t’, we get
t=402 seconds
or t = 6 minutes 42 seconds
Therefore, the time taken for a current of 0.15 amperes to deposit 20 mg of copper from a solution of copper sulphate is 6 minutes 42 seconds.
Hence, Option B is the correct answer.

Note: It was Michael Faraday who proposed the two laws of electrolysis for the first time in 1833. Electrolysis is a non-spontaneous reaction in which reduction and oxidation takes place by passing the electric current through an aqueous solution known as an electrolyte.