Question

In the Lane's process for the manufacture of hydrogen the reagents used in oxidation-reduction process are respectively:
(A) $Fe/{{H}_{2}}O$ and water gas.
(B) $F{{e}_{3}}{{O}_{4}}$ and ${{H}_{2}}$
(C) ${{O}_{2}}$ and ${{H}_{2}}$
(D) $Fe/{{H}_{2}}O$ and $F{{e}_{3}}{{O}_{4}}$

Answer 1

We know that the lane’s process, di-hydrogen is manufactured. This is done by the oxidation of iron using superheated steam. Here to verify the reagents that are too hydrogen reagent used in the Redox reaction (oxidation-reduction) process we will pass it through heat of around $1023K\text{ }to\text{ }1073K.$

Complete answer:
Lane’s process of manufacturing dihydrogen takes place in two steps. Let us see more about these steps. The first step of this process is the oxidation stage. In this stage, iron is oxidized by using steam which is superheated at temperatures nearly about $1023K\text{ }to\text{ }1073K.$ Iron is oxidized to iron (III) oxide. The chemical equation for this process is given below.
$3Fe+4{{H}_{2}}O\xrightarrow[1023K-1073K]{\Delta }F{{e}_{3}}{{O}_{4}}+4{{H}_{2}}\uparrow$
The second step of this process is the reduction stage. In this stage, the iron (III) oxide formed in the last stage is reduced using water-gas. Water-gas is the combination of two gases – carbon monoxide $\left( CO \right)$ and dihydrogen $({{H}_{2}}).$ The water-gas reacts with iron (iii) oxide and reduces it to metallic iron. This step is known as vivification. The chemical equations for both of these processes are given below.
$F{{e}_{3}}{{O}_{4}}+4{{H}_{2}}\to 3Fe+4{{H}_{2}}O$
The iron formed in the reduction stage is again recycled for use in the oxidation stage. In this way, this process continues and hydrogen is efficiently produced at an industrial level.
Therefore, the correct answer is option A.

Note:
Don’t confuse between water gas and producer gas. The water-gas contains only carbon monoxide and dihydrogen while the producer gas contains carbon dioxide and nitrogen along with the components of water gas.