Question

The quantity of electricity needed to deposit $ 127.08 $ g of copper is
(A) 1 faraday
(B) 4 coulomb
(C) 4 faraday
(D) 1 ampere

Answer 1

Hint: To answer this question, you must recall the Faraday’s laws of electrolysis. Michael Faraday gave two laws of electrolysis to give an explanation of the quantitative aspects of electrolysis. Using the two laws, we can easily determine and express the magnitudes of electrolytic effects, i.e. the amount of the substance deposited and the quantity of electric current passed through the electrolytic reaction mixture.

Formula used: $ E = \dfrac{M}{n} $
Where, $ E $ denotes the equivalent mass of the substance deposited/ liberated/ precipitated
 $ M $ denotes the molar mass of the substance deposited/ liberated/ precipitated
And $ n $ represents the n- factor of the substance deposited/ liberated/ precipitated.

Complete step by step solution
It is known from the definition of 1 faraday that it is the amount of charge that is required to liberate/ deposit/ precipitate 1 gram equivalent of the substance. 1 gram equivalent of a substance contains mass of substance equal to that of the equivalent weight of the substance.
We can find the equivalent weight of copper using the formula $ E = \dfrac{M}{n} $ . The n- factor for copper is 2. So, we get, $ E = \dfrac{{63.5}}{2} = 31.75{\text{ g}} $
Thus, we can say that $ 31.75{\text{ g}} $ of copper requires 1 faraday charge.
So, $ 127.08 $ g of copper requires, $ \dfrac{{127.08}}{{31.75}}{\text{ Faraday}} $
 $ {\text{Charge}} = 4{\text{ Faraday}} $
The correct answer is C.

Note
We know that, n- factor of a substance can be defined as the combining capacity or valency of the substance. For instance, if we consider an acid, so the n- factor of the acid is its basicity. Similarly in case of electrolysis, the n- factor represents the number of electrons transferred per molecule of the substance.