Question

The mass of benzene in g that would be required to produce a current of one ampere (Assume current efficiency 50%) for three hour from the following data:
${C_6}{H_6} + \dfrac{{15}}{2}{O_2} \to 6C{O_2} + 2{H_2}O$
(A) 0.58
(B) 0.38
(C) 0.24
(D) 0.29

Answer 1

Faraday was a scientist who gave two faraday laws for electrolysis and to understand the relationship between the amount of chemical reaction and the electricity passed during electrolysis. On using Faraday's second law $\dfrac{{weight}}{{equivalent \ weight}} = \dfrac{Q}{{96500}}$ we can calculate the required mass of the compound.

Complete step by step answer:
As we know that Benzene is an organic chemical compound that is a hydrocarbon in which six carbon atoms with hydrogen atoms attached to them are arranged in the shape of a ring. It has a molecular formula of ${C_6}{H_6}$ . It has a sweet smell and is used for production of plastics, synthetics and is used to make lubricants, dyes, drugs and pesticides.
Faraday was a scientist who gave two laws for electrolysis which are:
First law: This law states that the amount of chemical reaction at an electrode during the electrolysis process is directly proportional to the electricity passed through the electrolyte.
Second law- This law states that the amount of substances that are liberated when electricity is passed through the electrolytic solution is directly proportional to their equivalent chemical weights. According to this law
$\dfrac{{weight}}{{equivalent \ weight}} = \dfrac{Q}{{96500}}$
Thus, quantity of electricity is calculated as :
$Q = It$
where $Q$ is the quantity of electricity in coulombs and $I$ Is the current in amperes and $t$ is time in seconds.
We know that the charge on one electron is $1.6021 \times {10^{ - 19}}C$
Therefore, the charge on one mol of electrons
$\Rightarrow C = {N_A} \times 1.6021 \times {10^{ - 19}}C$
Where ${N_A}$ is Avogadro number
$\Rightarrow C = 6.022 \times {10^{23}} \times 1.6021 \times {10^{ - 19}}$
$\Rightarrow C = 96482Cmo{l^{ - 1}}$
For the given question we shall use Faraday’s second law
$\dfrac{{weight}}{{equivalent \ weight}} = \dfrac{Q}{{96500}}$
$eq.weight = \dfrac{{molecular \ weight}}{{nfactor}}$
And we know, $Q = It$
Given, ${C_6}{H_6} + \dfrac{{15}}{2}{O_2} \to 6C{O_2} + 2{H_2}O$
Molecular weight of benzene = $78g$
n factor = $\dfrac{{15}}{2}$
$\Rightarrow \dfrac{w}{{\dfrac{{78}}{{7.5}}}} = \dfrac{{I \times t}}{{96500}}$
$t = 3 \times 60 \times 60 \ seconds$
$\Rightarrow \dfrac{w}{{\dfrac{{78}}{{7.5}}}} = \dfrac{{1 \times 3 \times 60 \times 60}}{{96500}}$
$\Rightarrow w = 0.58g$
Therefore, the required mass of benzene is $0.58g$ and the correct option is, ‘(A) 0.58’.

Note: Benzene is quite dangerous for our skin and it may cause irritation and burning sensation after prolonged contact with the human skin. It is insoluble in water as it does not form hydrogen bonds with water whereas it is soluble in organic solvents. It burns with a sooty smell and shows resonance therefore, it is very stable.