Question

In the refining of Ag metal through electrolysis, what will be the weight of 250 g Ag anode if 10 ampere current is passed for 3 hours? Purity of silver is 80% by weight in gram is:
(A) 88
(B) 99
(C) 64
(D) 108

Answer 1

Attempt this question by remembering Faraday's law for electrolysis and apply them. The process in which a chemical reaction usually occurs through the induction of electrical energy that is referred as an electrolytic cell and this occurrence of non-spontaneous reactions under the effect of electric energy is called as electrolysis

Formula used: We will require the following formulas:
Q = It
W = ZQ
where,
Q = Quantity of electricity passed
I = current in amperes
t = time
W = mass deposited or refined
Z = Electrochemical equivalent

Complete answer:
Let us begin with the study about Faraday’s law for electrolysis as follows:-
Faraday’s first law: This law states that the amount of chemical reaction that occurs at any electrode during electrolysis under the effect of electrical energy is directly proportional to the quantity of electricity passed through the electrolyte. Mathematically it can be represented as follows:-
Q = It
W = ZQ
where,
Q = Quantity of electricity passed
I = current in amperes
t = time
W = mass deposited or refined
Z = Electrochemical equivalent
-The values provided in the question are as follows:-
I = current in amperes = 10 A
t = time = 3 hours = 3(3600) seconds
-Calculation of quantity of electricity passed:-
$\begin{aligned} & \Rightarrow Q=It \\\ & \Rightarrow Q=10A\times 3\times 3600seconds \\\ & \Rightarrow Q=108000C \\\ \end{aligned}$
-Calculation of electrochemical equivalent:-
As we know that Z = $\dfrac{\text{molar mass}}{\text{n-factor }\times \text{ 96500C/mol}}$
The reaction occurred is: $A{{g}^{+}}+{{e}^{-}}\to Ag$
n-factor = 1
Molar mass of Ag = 108g/mol
So Z = $\dfrac{\text{108g/mol}}{\text{1 }\times \text{ 96500C/mol}}$
-Calculation of mass deposited:-
W = ZQ
$\begin{aligned} & W=\dfrac{\text{108g/mol}}{\text{1 }\times \text{ 96500C/mol}}\times 108000C \\\ & W=120.8704g \\\ \end{aligned}$
Since the percentage purity is given as 80%
Hence the mass of Ag deposited = $120.8704\times \dfrac{100}{80}=151.1g\simeq 151g$
Now the weight of Ag anode is = 250g – 151g = 99

Thus the correct option is: (B) 99 .

Note:
-Remember that the second law of Faraday states that “when the same quantity or amount of electricity is passed through solutions of different electrolytes, the amounts of the substances deposited or liberated at the electrodes are directly proportional to their chemical equivalents”.
-Also n-factor for a redox reaction is equal to the total number of electrons transferred.