Question

The catalyst used in the manufacture of hydrogen by Bosch’s process is:
(A)- $F{{e}_{2}}{{O}_{3}}$
(B)- $C{{r}_{2}}{{O}_{3}}$
(C)- both A and B
(D)- Cu

Answer 1

Bosch’s process involves water gas and steam for the production of hydrogen commercially. Water gas is a mixture of carbon monoxide (CO) and hydrogen (${{H}_{2}}$). Catalyst used in Bosch’s process separates CO from ${{H}_{2}}$ by oxidizing CO to $C{{O}_{2}}$.

Complete answer:
Bosch’s process is used for the industrial or large scale production of hydrogen. production of hydrogen occurs in the following steps:
First step involves the production of water gas from coke, a substance high in carbon content prepared by heating coal in absence of air, and steam.
Steam is passed over a hot-red coke at a temperature of about 1200$^{o}C$ to produce an equivolume mixture of carbon monoxide and hydrogen gas which is commonly known as water gas. The chemical equation for the production of water gas is given as
$\begin{aligned} & C+{{H}_{2}}O\xrightarrow{{{1200}^{o}}C}CO+{{H}_{2}} \\\ & \text{ water gas} \\\ \end{aligned}$
In second step, water gas is mixed with excess of steam and passed over a catalyst which is heated ferric oxide ($F{{e}_{2}}{{O}_{3}}$) and chromic oxide ($C{{r}_{2}}{{O}_{3}}$). This results in the formation of carbon dioxide ($C{{O}_{2}}$) and hydrogen gas (${{H}_{2}}$).
$\begin{aligned} & CO+{{H}_{2}}+{{H}_{2}}O\xrightarrow{F{{e}_{2}}{{O}_{3}}/C{{r}_{2}}{{O}_{3}}}C{{O}_{2}}+2{{H}_{2}} \\\ & \text{water gas} \\\ \end{aligned}$
This step is necessary for the separation of carbon monoxide (CO) from water gas ($CO+{{H}_{2}}$) to obtain hydrogen gas. Ferric oxide ($F{{e}_{2}}{{O}_{3}}$) and chromic oxide ($C{{r}_{2}}{{O}_{3}}$) used as catalyst separate CO by oxidizing into carbon dioxide ($C{{O}_{2}}$).
Finally, the mixture of $C{{O}_{2}}$ and ${{H}_{2}}$ is passed through cold water under pressure of 30 atmosphere or through a solution of potassium hydroxide (KOH). $C{{O}_{2}}$ being soluble in water (and KOH) dissolves whereas ${{H}_{2}}$ does not dissolve and thus, is left behind.
$\begin{aligned} & {{H}_{2}}O+C{{O}_{2}}\to {{H}_{2}}C{{O}_{3}} \\\ & \text{ carbonic acid} \\\ \end{aligned}$
$2KOH+C{{O}_{2}}\to {{K}_{2}}C{{O}_{3}}+{{H}_{2}}O$
$F{{e}_{2}}{{O}_{3}}$ acts the catalyst and $C{{r}_{2}}{{O}_{3}}$ acts more like its promoter. Sometimes, we also represent the catalyst used in Bosch’s process as $F{{e}_{2}}C{{r}_{2}}{{O}_{6}}$. ($F{{e}_{2}}{{O}_{3}}+C{{r}_{2}}{{O}_{3}}\to F{{e}_{2}}C{{r}_{2}}{{O}_{6}}$)

Therefore, the correct option is (C).

Note:
Though C{{r}_{2}}O{{ & }_{3}} acts as the promoter to catalyst $F{{e}_{2}}{{O}_{3}}$, they are always used together in Bosch’s process. The catalyst should be able to remove CO from water gas mixture. Both $F{{e}_{2}}{{O}_{3}}$ and $C{{r}_{2}}{{O}_{3}}$ are good oxidant agents and oxidize CO into $C{{O}_{2}}$.