Question

Impure copper containing Fe, Au, and Ag as impurities is electrolytically refined. A current of 140 A for 482.5 s and decreased the mass of the anode by 22.26g and increased the mass of the cathode by 22.011g percentage of iron in impure copper is (given molar mass Fe = 55.55g/mol; molar mass of Cu = 63.54g/mol)
(A) 0.95
(B) 0.9
(C) 0.85
(D) 0.97

Answer 1

Hint: Faraday gave two laws that provide the quantitative aspect of electrolysis. They are as follows:
Faraday’s first law of electrolysis: The mass of a substance deposited or liberated at any electrode due to chemical reaction is directly proportional to the amount of electricity passed through the electrolyte. Let the mass of the substance deposited be ‘w’ gram when Q coulomb of electricity is passed, then we can write
     \[\begin{align}
  & W\propto Q \\
 & W=ZQ \\
\end{align}\]
Z is called the electrochemical equivalent of the substance deposited at the electrode. It is given as
     \[Z=\dfrac{\text{Equivalent mass of the substance}}{96500}\]
Faraday’s second law of electrolysis: When the same amount of electricity is passed through two different solutions, the masses of the substances deposited at the electrodes are directly proportional to the ratio of their equivalent masses. If the same amount of electricity is passed through $CuS{{O}_{4}}$ and $FeS{{O}_{4}}$ solutions, then we can write
     \[\dfrac{\text{Mass of Fe deposited}}{\text{Mass of Cu deposited}}=\dfrac{\text{Equivalent weight of Fe}}{\text{Equivalent weight of Cu}}\]

Complete answer:
We have been given that the mass of anode decreased by 22.26g
And, the mass of cathode increased by 22.011g
According to Faraday’s first law of electrolysis, we have
     \[W=ZQ\]
Amount of current passed through the electrolytic solution, I =140 A
Time for which 140 A of current was passed, t = 482.5 s
Therefore, the amount of electricity (in coulombs), Q = I$\times $ t = 140 A $\times $ 482.5 s = 67550 A s or 67550 coulombs.

Electrochemical equivalent of Cu, Z can be calculated as \[\]
\[Z=\dfrac{\text{Equivalent mass of Cu (E)}}{96500}\]
Molar mass of Cu is 63.54 g/mol. We can find the equivalent mass of Cu as
\[\begin{align}
  & \text{Equivalent mass of Cu = }\dfrac{\text{Molar mass of Cu}}{\text{Number of electrons gained}} \\
 & E=\dfrac{63.54\text{ }g/mol}{2eq/mol}=31.77g/eq \\
\end{align}\]
Substituting the equivalent mass of Cu in the equation for Z, we get
\[Z=\dfrac{31.77g/eq}{96500\text{ C/}eq}=\dfrac{31.77g}{96500\text{ C}}\]
Now, substituting the values of Z and Q in \[W=ZQ\], we can find the mass of Cu that should have been deposited at cathode.
\[\begin{align}
  & Q=67550\text{ C} \\
 & \text{Z= }\dfrac{31.77g}{96500\text{ C}} \\
 & W=ZQ=\dfrac{31.77g}{96500\text{ C}}\times 67550\text{ C} \\
 & W=22.239g \\
\end{align}\]

But the actual increase in the mass of Cu at cathode is 22.011g which is less than the mass of Cu that should have been deposited on the cathode, i.e. 22.239g. So the mass of Cu that remained in the solution is equal to impurity due to Fe present in the solution. Therefore, we can calculate the mass of Fe in the impure Cu as:
Impurity in Cu = 22.239g - 22.011g = 0.228g
According to Faraday’s second law of electrolysis, we can calculate the total amount of Fe present in the impure Cu. Let ‘${{W}_{Fe}}$’ be the mass of Fe that remained in the solution and ‘\[{{W}_{Cu}}\]’ be the mass of Cu passed into the solution which is 0.228g, then the ratio of these masses can be written as:
     \[\dfrac{{{W}_{Fe}}}{{{W}_{Cu}}}=\dfrac{\text{Equivalent mass of Fe}}{\text{Equivalent mass of Cu}}\]

We have calculated the equivalent mass of Cu to be 31.77 g/eq. Similarly, we can find the equivalent mass of Fe. Given, the molar mass of Fe is 55.55g/mol. Then, the equivalent mass of Fe will be:
     \[\begin{align}
  & \text{Equivalent weight of Fe = }\dfrac{\text{Molar mass of Fe}}{\text{Number of electrons gained}} \\
 & E=\dfrac{\text{55}\text{.55 }g/mol}{2eq/mol}=27.775g/eq \\
\end{align}\]
Substituting the equivalent mass of Cu and Fe in the equation of second law of electrolysis for Cu and Fe, we get
     \[\begin{align}
  & \dfrac{{{W}_{Fe}}}{{{W}_{Cu}}}=\dfrac{\text{Equivalent mass of Fe}}{\text{Equivalent mass of Cu}} \\
 & \dfrac{{{W}_{Fe}}}{{{W}_{Cu}}}=\dfrac{27.775g/eq}{31.77g/eq} \\
\end{align}\]
We know that ${{W}_{Cu}}$ = 0.228g. Therefore, ${{W}_{Fe}}$ will be
     \[\begin{align}
  & {{W}_{Fe}}=\dfrac{27.775g/eq}{31.77g/eq}\times {{W}_{Cu}} \\
 & {{W}_{Fe}}=\dfrac{27.775g/eq}{31.77g/eq}\times 0.228g \\
 & {{W}_{Fe}}=0.199g \\
\end{align}\]

Now, we have to calculate the percentage of Fe in the impure Cu. Mass of anode decreased is equal to the mass of the impure Cu which is equal to 22.26g. Therefore, the percentage of Fe in the impure Cu will be calculated as:
     \[\begin{align}
  & \dfrac{\text{Mass of Fe in impure Cu }}{\text{Mass of impure Cu}}\times 100 \\
 & \Rightarrow \dfrac{0.199g}{22.26g}\times 100=0.90 \\
\end{align}\]

So, the correct answer is “Option B”.

Note: In the electrolytic refining of Cu, anode is impure Cu which is oxidized to $C{{u}^{2+}}$. Pure Cu is obtained at cathode by reduction of $C{{u}^{2+}}$ to Cu. Carefully solve the question step by step to avoid any confusion or error.