Question

Electrolysis of dilute aqueous NaCl solution was carried out by passing 10mA current. This time required to liberate 0.01moles of ${H_2}$ gas at the cathode is:
A.$9.65 \times {10^4}\sec $
B.$19.3 \times {10^4}\sec $
C.$28.95 \times {10^4}\sec $
D.$38.6 \times {10^4}\sec $

Answer 1

Hint: We have to calculate the time taken for the liberation of hydrogen gas at the cathode using the amount of hydrogen released (in moles), current passed and Faraday’s constant. The amount of hydrogen released in moles is calculated using the electrons transferred during the process multiplied by the moles of hydrogen released. We have to multiply the amount of hydrogen released, current passed and Faraday’s constant to get time taken for the liberation of hydrogen.

Complete step by step answer:
Given data contains,
Current passed is 10mA.
Moles of hydrogen released are 0.01moles.
In the electrolysis of aqueous solution of sodium chloride, as the reduction potential of sodium ion is less than the hydrogen ion, the reaction that takes place at the cathode could be given as,
$2{H^ + } + 2{e^ - } \to {H_2}$
0.01moles of hydrogen would require $2 \times 0.01 = 0.02$ moles of electrons (or) $0.02$ faradays.
We know the formula to calculate the amount of Faradays passed,
Amount of Faradays passed=$\dfrac{{I\left( A \right) \times t\left( s \right)}}{{96500C \cdot mo{l^{ - 1}}}}$
On rearranging the equation to calculate the time taken we get,
Let us now substitute the values of current, moles of hydrogen released in the equation.
\[t = {\text{Amount of Faradays passed}} \times 96500 \times I\]
$0.02 = \dfrac{{\dfrac{{10mA}}{{1000mA/A}} \times t\left( s \right)}}{{96500C \cdot mo{l^{ - 1}}}}$
$ \Rightarrow t = 0.02 \times 96500 \times 0.01$
On multiplying we get,
$ \Rightarrow t = 19.3 \times {10^4}\sec $
The time required to liberate 0.01moles of ${H_2}$ gas at the cathode is $19.3 \times {10^4}\sec $.
Therefore, the option (B) is correct.

Note:
We must know that electrolysis is one of the many methods which are widely used in industry, it is used not only for extraction of metals but also to reduce the impurities in metal. This process is used in the medical field as well like for permanent removal of hair, etc. Even though such an application has some side effects but for long run it is good.