Question

Consider the following statements:
i.Atomic hydrogen is obtained by passing hydrogen through an electric arc.
ii.Hydrogen gas will not reduce heated aluminium oxide.
iii.Finely divided palladium absorbs large volumes of hydrogen gas.
iv.Pure nascent hydrogen is best obtained by reacting \[Na\] with \[{{C}_{2}}{{H}_{5}}OH.\]
Which of the above statement(s) is/are correct?
A.i. only
B.ii. only
C.i, ii and iii
D.ii, iii and iv

Answer 1

Hint: We know that a very high temperature is required for welding and cutting of metals. Burning of hydrogen in air produces a very hot flame. Dissociation of dihydrogen into atomic hydrogen by electric arc followed by recombination of atoms releases a large amount of heat.

Complete answer:
Hydrogen is a good reducing agent. When hydrogen is passed over many heated metallic oxides, they are reduced to the metals. Therefore, hydrogen reduces oxides of metal which are below in reactivity series. So that means the hydrogen can reduce the oxides of only those metals which are less reactive than hydrogen itself. Moreover, the metal oxide likes Sodium Oxide, Copper Oxide which cannot be reduced by hydrogen.
Some compounds can be both reducing and oxidizing agents. Hydrogen is one of the elements that can be both reducing and oxidizing. Hydrogen gas is a reducing agent when it reacts with nonmetals and an oxidizing agent when it reacts with metals. Electrochemical series describes the arrangement of elements in order of their increasing electrode potential values. The series that has been established by measuring the Potential of various electrodes versus standard hydrogen electrodes (SHE).
The normal diatomic hydrogen \[\left( {{H}_{2}} \right)\] is passed through an electric arc, which separates it into atomic hydrogen \[\left( H \right).\]
Standard electrode potential of aluminium is \[-1.66V\], which is why it acts as a strong reducing agent.
Finely divided palladium can be used as a catalyst and at room temperature it can absorb up to \[900\] times its own volume of hydrogen.
\[2{{C}_{2}}{{H}_{5}}OH+2Na\to 2{{C}_{2}}{{H}_{5}}ONa+{{H}_{2}}\]
Therefore, the correct answer is option C.

Note:
Remember that electrochemical series is a series of chemical elements arranged in order of their standard electrode potentials. So from top to bottom in electrochemical series, reducing potential increases and oxidizing potential decreases. From the series, we can see that the reducing potential of hydrogen is less than that of Aluminium.