Question

${\text{NaOH}}$ can also be prepared by electrolysis of aqueous ${\text{NaCl}}$. Amount of ${\text{NaOH}}$ formed when ${\text{0}}{\text{.445L}}$ of ${\text{NaC}}{{\text{l}}_{{\text{(aq)}}}}$ is electrolysed for ${\text{137s}}$ with a current of ${\text{1}}{\text{.08A}}$, is:
A) ${\text{0}}{\text{.09g}}$
B) ${\text{0}}{\text{.12g}}$
C) ${\text{0}}{\text{.06g}}$
D) Amount of ${\text{NaCl}}$ not been given

Answer 1

We have to know that acid reacts with the base means to give salt and water. According to this discussion, Sodium hydroxide reacts with hydrochloric acid to give a product of sodium chloride as salt and water. Salt reacts with water to give acid and base. According to the second discussion, sodium chloride as salt reacts with water to give the product of Sodium hydroxide and hydrochloric acid. It is a reversible reaction.
${\text{HCl + NaOH}} \rightleftarrows {\text{NaCl + }}{{\text{H}}_{\text{2}}}{\text{O}}$

Formula used:
In faraday’s law,
${\text{m = ZIt}}$
${\text{Z = }}\dfrac{{\text{E}}}{{\text{F}}}$
Here, The mass of the substance liberated at electrode is ${\text{m}}$ and the electrochemical equivalent of the substance liberated at electrode is ${\text{Z}}$ and the equivalent mass of the substance liberated at electrode is ${\text{E}}$ and the faraday’s constant is ${\text{F}}$ and the current of the substance is ${\text{I}}$ and the time of the substance is ${\text{t}}$.

Complete step by step answer:
$NaOH$ can also be prepared by electrolysis of aqueous $NaCl$.
The required current ${\text{I}}$, is ${\text{1}}{\text{.08A}}$.
The required time ${\text{t}}$, is ${\text{137s}}$.
In Faraday’s law, The mass of the substance liberated at an electrode during electrolysis is directly proportional to the quantity of charge passed through the cell.
$m \propto Q$
${\text{Q = It}}$
$m \propto It$
${\text{m = ZIt}}$
${\text{Z = }}\dfrac{{\text{E}}}{{\text{F}}}$
${\text{m = }}\dfrac{{{\text{EIt}}}}{{\text{F}}}$
Here, The mass of the substance liberated at the electrode is ${\text{m}}$.
The electro chemical equivalent of the substance liberated at the electrode is ${\text{Z}}$.
The equivalent mass of the substance liberated at the electrode is ${\text{E}}$.
The faraday’s constant is ${\text{F}}$.
The current of the substance is ${\text{I}}$.
The time of the substance is ${\text{t}}$.
The charge of the substance is ${\text{Q}}$.
The electro chemical equivalent of the ${\text{NaOH}}$ liberated at electrode ${\text{Z}}$ is equal to the ratio between the equivalent mass of the ${\text{NaOH}}$ liberated at electrode is ${\text{E}}$ to the faraday’s constant is ${\text{F}}$.
The equivalent mass of the ${\text{NaOH}}$ liberated at electrode ${\text{E}}$ is $40$.
The faraday’s constant ${\text{F}}$ is $96500$.
Amount of ${\text{NaOH}}$ formed when ${\text{0}}{\text{.445L}}$ of ${\text{NaC}}{{\text{l}}_{{\text{(aq)}}}}$ is electrolysed for ${\text{137s}}$ with a current of${\text{1}}{\text{.08A}}$ is calculated as,
${\text{m = }}\dfrac{{{\text{EIt}}}}{{\text{F}}}$
We apply known values,
${\text{m = }}\dfrac{{40 \times {\text{1}}{\text{.08}} \times {\text{137}}}}{{96500}}$
On simplification we get,
$\Rightarrow m = 0.06$
According to the above calculation, ${\text{NaOH}}$ can also be prepared by electrolysis of aqueous ${\text{NaCl}}$. Amount of ${\text{NaOH}}$ formed when ${\text{0}}{\text{.445L}}$ of ${\text{NaC}}{{\text{l}}_{{\text{(aq)}}}}$ is electrolysed for ${\text{137s}}$ with a current of ${\text{1}}{\text{.08A}}$, is ${\text{0}}{\text{.06g}}$.
Hence, option (C) is correct.

Note:
As we know that the SI unit of current is ampere. The symbol of the ampere is ${\text{A}}$. The SI unit of time is seconds. We have to remember that the symbol of time is ’${\text{s}}$’. The SI unit of charge is coulomb. The symbol of coulomb is ${\text{C}}$.