Question

What weight of copper will be deposited by passing 2 faradays of electricity through a cupric salt
(A) 2.0
(B) 3.175
(C) 63.5
(D) 127.0

Answer 1

Hint: The answer is found by using the combined formula deduced from Faraday’s law of electrolysis. First, we need to find the equivalent weight of Cu, substitute it in the formula and this will give us the weight of copper that gets deposited.

Complete step by step solution:
-We are given the case of electrolysis. It can be simply explained as the process
of carrying out non-spontaneous reactions under the influence of electric energy.
- As we know, Michael Faraday was the first scientist who described the
quantitative aspects of the laws of electrolysis. He proposed two laws in order to explain the
electrolysis namely Faraday's first law of electrolysis and second law of electrolysis.
- Faraday's first law of electrolysis states that during the electrolysis the amount of
chemical reaction which occurs at any electrode under the influence of electrical energy is
proportional to the quantity of electricity passed through the electrolyte.
- Faraday's second law of electrolysis states that during the electrolysis, when the
same quantity of electricity is passed through the electrolytic solution, the number of different
substances liberated are proportional to their chemical equivalent weights.
-By combining these laws of electrolysis, we can understand that the amount of
electricity needed for oxidation-reduction depends on the stoichiometry of the electrode reaction.

It can be mathematically expressed as
$\dfrac{W}{Equivalent\quad weight}=\dfrac{I\times t}{96500}$
We need to figure out the amount of copper (W) that will get deposited by passing 2 Faradays of
electricity. We can write the reduction reaction as follows,
$Cu_{(aq)}^{2+}+\quad 2{{e}^{-}}\quad \to \quad C{{u}_{(s)}}$

It’s a reduction reaction since electrons are getting added. By looking at the reaction,
two moles of electrons are required to deposit one mole Cu. It can be also said like 2 faradays of electricity will deposit one mole Cu.
The Equivalent weight of Cu =$\dfrac{Atomic\quad weight}{2}$
= $\dfrac{63.5}{2}$
= 31.75 g
By substituting this value in the above formula of Faraday's law, we can obtain as follows,
$\dfrac{W}{31.75}=\dfrac{2\times 96500}{96500}$
Thus W =$\dfrac{2\times 96500}{96500}\times 31.75$
= 63.5 g

Therefore, the weight of copper that will get deposited on passing 2 faradays of electricity is 63.5g. That#39;s the option (C).

Note: The answer can also be found in a simpler way. From the written redox reaction, we know that 2 moles of electrons will deposit one mole Cu. That is 2 Faraday of electricity will deposit one mole Cu and one mole Cu contains 63.5 g. ( $Number\quad of\quad moles=\quad\dfrac{Given\quad mass(W)}{Molar\quad mass}$)