Question

Dihydrogen of high purity $\left(>99.95\%\right)$ is obtained through

• A. The electrolysis of acidified water using $\mathrm{Pt}$ electrodes.
• B. The reaction of $\mathrm{Zn}$ with dilute $\mathrm{HCl}$.
• C. The electrolysis of brine solution.
• D. The electrolysis of warm $\mathrm{Ba}{\left(\mathrm{OH}\right)}_2$.

Answer 1

The correct option is D

The electrolysis of warm $\mathrm{Ba}{\left(\mathrm{OH}\right)}_2$.

Explanation for the correct option:

D) The electrolysis of warm $\mathbf{Ba}{\left(\mathrm{OH}\right)}_{\mathbf{2}}$:

1. In the laboratory, high purity compounds can be obtained only with electrolysis.
2. Similarly, highly purified dihydrogen is also obtained by electrolyzing water using a small amount of acid or base.
3. Here, $\mathrm{Ba}{\left(\mathrm{OH}\right)}_2$ is a base. Hence, electrolyzing water in the presence of a base $\mathrm{Ba}{\left(\mathrm{OH}\right)}_2$ gives us the Highly purified $\left(>99.95\%\right)$ Dihydrogen $\left({\mathrm{H}}_2\right)$.
4. This highly purified Dihydrogen can be collected at the cathode and the dioxygen $\left({\mathrm{O}}_2\right)$ can be collected at the anode.

$2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\underset{\mathrm{Ba}{\left(\mathrm{OH}\right)}_2}{\overset{\mathrm{Electrolysis}}{\to }}\underset{\left(\mathrm{Cathode}\right)}{2{\mathrm{H}}_2\left(\mathrm{g}\right)}+\underset{\left(\mathrm{Anode}\right)}{{\mathrm{O}}_2\left(\mathrm{g}\right)}$

Explanation for the incorrect options:

A) The electrolysis of acidified water using $\mathbf{Pt}$ electrodes:

The highly purified dihydrogen is obtained by electrolyzing the water using a small amount of acid or base in the presence of Nickel $\left(\mathrm{Ni}\right)$ electrodes, not Platinum $\left(\mathrm{Pt}\right)$electrodes.

B) The reaction of $\mathbf{Zn}$ with dilute $\mathbf{HCl}$:

By reacting $\mathrm{Zn}$ with dilute $\mathrm{HCl}$, dihydrogen can be obtained but not in a highly pure form.

C) The electrolysis of brine solution:

By electrolyzing the Brine solution, along with the Chlorine $\left({\mathrm{Cl}}_2\right)$, Sodium ions $\left({\mathrm{Na}}^{+}\right)$ and Hydroxide ions $\left({\mathrm{OH}}^{-}\right)$, Dihydrogen is also obtained. But, as a byproduct, not as a highly purified form.

Hence, option D is correct.