Question

The mass (m) of a substance deposited on cathode or anode during electrolysis is given as:
[where, Q= Charge, M=Molar mass, n=n-factor, F= Faraday’s constant]
A) M= Qm/Fn
B) m= QM/Fn
C) m= QMFn
D) Q= mFn/M

Answer 1

Hint: Try to recall that Faraday proposed two laws of electrolysis to explain the quantitative aspects of electrolysis. Also, the amount of substance deposited is proportional to the number of moles of electrons passed. Now by using this you can easily answer the given question.

Complete step by step answer:
It is known to you that the quantity of electric charge passed through an electrolyte and the amount of substance deposited at the electrodes was presented by the Faraday in the form of the law of electrolysis and named these laws as the first law of electrolysis and second law of electrolysis.
Now, the first law of electrolysis: According to this law, the amount of a substance deposited or liberated at an electrode is directly proportional to the amount of charge passed (utilized) through the solution i.e. $W\alpha Q$.
or, $W = ZQ$ where, W= weight liberated or deposited, Q= charge in coulomb and Z= electrochemical equivalent.
The second law of electrolysis: According to this law, during electrolysis when the same amount of charge is passed through different electrolyte solutions connected in series then the weight of substance dissolved or deposited at cathode or anode is in the ratio of their equivalent weights i.e. $\dfrac{{{w_1}}}{{{w_2}}} = \dfrac{{{E_1}}}{{{E_2}}}$.
So, by combining both laws we can say that, the amount of chemical change occurred i.e., the moles of substances deposited or liberated is proportional to the number of moles of electrons exchanged during the oxidation-reduction reactions that occur i.e.,

$$\dfrac{Q}{F} = \dfrac{m}{M} \times n \\\ m = \dfrac{{MQ}}{{Fn}} \\\ Q = \dfrac{{Fmn}}{M} \\\ M = \dfrac{{Fmn}}{Q} \\\$$

Where, M= Molar mass, Q= Charge, n= n-factor, F=Faraday’s constant, m= mass of substance deposited.

Therefore, from above we can conclude that option D is the correct option for the given question.

Note: It should be remembered to you that One Faraday is the charge required to liberate or deposit a one-gram equivalent of a substance at the corresponding electrode.
Also, you should remember that Faraday’s law of electrolysis establishes a relationship between the amount of substance deposited or liberated and the quantity of charge passed through the solution.