Question

Anthracene can be electrolytically oxidised to anthraquinone. What mass of anthraquinone can be produced by the passage of 1 ampere current for an hour at 100% efficiency?
   





A. A. 3.654 g
B. 2 g
C. 12 g
D. 1.2932 g

Answer 1

Hint: In this question, anthracene is getting oxidised to anthraquinone electrolytically. Therefore we will use the Faraday’s laws of electrolysis in order to determine the amount of anthraquinone produced.
Complete step by step solution:
In this question, we will use Faraday’s first law of electrolysis which states that the amount of a substance that gets deposited on an electrode during electrolysis is directly proportional to the amount of charge passed through the electrolytic solution i.e.
$ { mass\quad of\quad substance\quad deposited\quad \propto \quad Amount\quad of\quad charge }$
$ \Rightarrow m=ZQ$
Where ‘m’ is the mass of the substance deposited, ‘Q’ is the amount of the charge passed through the solution and ‘Z’ is the proportionality constant; known as the electro-chemical equivalent of the substance. ‘Z’ will be equal to the amount of the substance deposited when a charge of 1 coulomb is passed through the solution.
Let us examine the reaction given in the question:








In order to find the total amount of charge required to convert 1 mole of anthracene to 1 mole of anthraquinone, we need to find the valency factor which is the total change in oxidation number that the atoms in a species undergo per molecule of the species.
In the above reaction, the carbon atoms labelled as ‘1’ and ‘2’ are undergoing a change in their oxidation state. The oxidation number of carbon ‘1’ in anthracene is -1 whereas the oxidation number of the carbon ‘1’’ in anthraquinone is +2. The total change in oxidation number is +3 i.e.[+2-(-1)]=+3. Since the total number of carbon atoms that are undergoing the change in oxidation number in anthracene is two, therefore a total of 6 electrons ($2\times (3)=6$) are required to convert 1 molecule of anthracene into 1 molecule of anthraquinone. Therefore a total of 6 moles of electrons will be required to produce 1 mole of anthraquinone from 1 mole of anthracene. 6 moles of electrons will carry 6 Farad of electric charge (1F=96500 C/mol). The Molar mass of anthraquinone is 208 g/mol. Therefore by using the Faraday’s first law of electrolysis:
$ \begin{matrix} 208 g/mol \\ Molar\quad\ mass\ of\ anthraquinone \end{matrix}=Z\times 6\times 96500\quad C/mol$
$ \Rightarrow Z=\cfrac { 208 g/mol }{ 6\times 96500 C/mol } $
Now, when a current of 1 ampere is passed through anthracene for 1 hour, it produces a total charge of:
$ Q=I\times t$ ; where ‘Q’ is the total charge, ‘I’ is the current passed and ‘t’ is the time for which the current is passed.
$ \Rightarrow Q=1A\times 60\times 60s$
Therefore:
$ \begin{matrix} m \\ Mass\ of\ Anthraquinone \end{matrix}=\begin{matrix} \cfrac { 208 g/mol }{ 6\times 96500 C/mol } \\ (Z) \end{matrix}\times \begin{matrix} 1A \\ I \end{matrix}\times \begin{matrix} 60\times 60s \\ t \end{matrix}$
$\begin{matrix} m \\ Mass\ of\ Anthraquinone \end{matrix}=1.2932 g$

Therefore, the correct answer is (D) 1.2932 g.

Note: Always remember that the valency factor in this case will be 6 and not 3 since a total of 2 carbon atoms in anthracene are undergoing a change in their oxidation state from -1 in anthracene to +2 in anthraquinone.