Question

What is the charge if 7.375 g of metallic tin is deposited by a passage of 24125C through a solution containing the ion? (atomic mass of Sn = 118)
A) +1
B) +3
C) +2
D) +4

Answer 1

Hint: To find the answer to this problem, we’ll use Faraday's Laws of Electrolysis. As we have studied earlier, there are two laws of electrolysis stated by Faraday. We have to recall Faraday's First Law of electrolysis to find the appropriate answer to this question.

Complete Step By Step Answer:
Faraday has stated two laws of electrolysis. The first law states that the mass of an element liberated/deposited during electrolysis is directly proportional to the amount of electricity passed through the solution, during electrolysis. Mathematically we can represent this law as:
Mass Deposited $ (M) \propto $ Current $ (I) \times $ Time (t)
I.e. $ M \propto It $
Removing the proportionality sign, $ M = Z \times I \times t $ ( $ I \times t = Q $ ) --- (1)
Where, M is the mass of the deposit, I is the current passed(A), t is the time for which the electricity is passed, Z is the electrochemical equivalent and Q is the Charge required.
The electrochemical equivalent is the amount of substance deposited at the electrode when 1 coulomb of charge is passed for 1 second. It has the unit of gram/coulomb.
The valued given to us are: $ M = 118g/mol,m = 7.375g,Q = 24125C $
Note that no. of gram equivalent $ n = \dfrac{Q}{F} $ (where Q is the charge and F is Faraday) -- (1)
The no. of gram equivalent is equal to $ n = \dfrac{{Mass}}{{Molar{\text{ }}Mass}} \times {n_f} $ -- (2)
From (1) and (2) we get that $ \dfrac{Q}{F} = \dfrac{{Mass}}{{Molar{\text{ }}Mass}} \times {n_f} $
Substituting the respective values we get, $ \dfrac{{24125}}{{96500}} = \dfrac{{7.375}}{{118}} \times {n_f} $
The n-factor is equal to the charge on the ion.
Hence n-factor $ {n_f} = \dfrac{{24125 \times 118}}{{96500 \times 7.375}} $
 $ {n_f} = 4 $
Hence the charge on Sn ion is +4.
The correct option is (D).

Note:
The faraday’s second law states that the mass of elements liberated by a constant quantity of electricity, through different electrolytes, will be proportional to the chemical equivalents to the ions undergoing the reaction. The equation can be given as:
Mass deposited, $ M \propto E $ (where E is the equivalent mass )
E can be given as: $ E = \dfrac{{Mass{\text{ }}of{\text{ }}1mol{\text{ }}of{\text{ }}ion}}{{Ch\arg e{\text{ }}on{\text{ }}the{\text{ }}ion}} $ .