Question

What is the value of current (in A) that has to be supplied to an electrolytic cell to deposit 108g of Ag in $ {10^4} $ second, current efficiency = 77.2%.

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 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 valued given to us are: $ M = 108g,t = {10^4}s $
We’ll have to find the value of Z first. The formula for finding Z is: $ Z = \dfrac{{Equivalent{\text{ }}weight(g/mol)}}{{96500(C/mol)}} $
The equivalent weight of Ag is: $ Eq.wt = \dfrac{{M.{M_{Ag}}}}{{n - factor}} $
The reaction that occurs during deposition is: $ A{g^ + } + {e^ - } \to A{g_{(s)}}(n - factor = 1) $
The molecular weight of Ag is 108g/mol, the equivalent weight will be $ = \dfrac{{108}}{1} = 108g/mol $
Substituting the values to find out the value of Z; $ Z = \dfrac{{108g/mol}}{{96500C/mol}} $
Finding the current (I) in amperes from equation (1): $ I = \dfrac{M}{{Z \times t}} = \dfrac{{108g}}{{\dfrac{{108}}{{96500}}g/C \times {{10}^4}s}} = 96500 \times {10^{ - 4}}A = 9.6500A $
We re given that the current efficiency is 77.2%, hence the Current required (A) will be:
 $ I = 9.6500 \times \dfrac{{77.2}}{{100}} = 7.34A $
The required answer is therefore, 7.34 Amperes.

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}} $ .