Question

In the refining of silver by electrolytic method what will be the weight of 100 g Ag anode if 5 ampere current is passed for 2 hours? Purity of silver is 95% by weight in gram is:
 (A) 57
(B) 32
(C) 60
(D) 13

Answer 1

Hint: We will attempt this question by using Faraday's law for electrolysis. The in which a chemical reaction occurs through the induction of electrical energy is referred to as an electrolytic cell and this process of carrying out non-spontaneous reactions under the influence of electric energy is known as electrolysis.
 Formula used:
We will require the following formulas in this question:-
$\begin{align}
  & Q=It \\
 & W=ZQ \\
\end{align}$
where,
Q = Quantity of electricity passed
I = current in amperes
t = time
W = mass deposited or refined
Z = Electrochemical equivalent

Complete answer:
Let us first study about Faraday’s law for electrolysis which will be required to solve this question as follows:-
It states that the amount of chemical reaction which occurs at any electrode during electrolysis under the influence of electrical energy is proportional to the quantity of electricity passed through the electrolyte. Mathematically it is written as:-
$\begin{align}
  & Q=It \\
 & W=ZQ \\
\end{align}$
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 = 5 A
t = time = 2 hours = 2(3600) seconds
 -Calculation of quantity of electricity passed:-
$\begin{align}
  & \Rightarrow Q=It \\
 & \Rightarrow Q=5A\times 2\times 3600\sec onds \\
 & \Rightarrow Q=36000C \\
\end{align}$
 -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
$\begin{align}
  & W=\dfrac{\text{108g/mol}}{\text{1 }\times \text{ 96500C/mol}}\times 36000C \\
 & W=40.2901g \\
\end{align}$
 Since the %purity is given as 95%
Hence the mass of Ag deposited = $40.2901\times \dfrac{100}{95}=42.41g$
Now the weight of Ag anode is = 100g – 42.41g = 57.59g

Thus the correct option is: (A) 57

Note:
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”.
-To solve these questions, we must acquire the knowledge of both laws given by Faraday.