Question

In the reaction, \[NaH\,+\,{{H}_{2}}O\,\to \,NaOH\,+\,{{H}_{2}}\]
A.) \[{{H}^{-}}\] is oxidized
B.) \[{{H}^{+}}\] is reduced
C.) both A and B
D.) none of the above

Answer 1

Hint: Sodium hydride ignites spontaneously at room temperature when it comes in contact with moist air and reacts violently with water, liberating hydrogen, which is also a highly flammable gas as it gets ignited by the heat of the reaction.

Complete step by step solution:

\[\overset{+1}{\mathop{Na}}\,\overset{-1}{\mathop{H}}\,\,+\,\overset{+1}{\mathop{{{H}_{2}}}}\,\overset{-2}{\mathop{O}}\,\,\to \,\overset{+1}{\mathop{Na}}\,\overset{-2}{\mathop{O}}\,\overset{+1}{\mathop{H}}\,\,+\,\overset{0}{\mathop{{{H}_{2}}}}\,\]

This is an acid-base reaction, where the water acts as an acid or proton donor and the hydride anion NaH acts as the base or proton acceptor. In the above reaction, the oxidation state of hydrogen in NaH changes from -1 to +1 and hydrogen in water changes from +1 to 0. Since the oxidation numbers of hydrogen atoms change when final products are formed, this reaction is also an example of redox reaction. Let’s elaborate:
Left side of the equation
In NaH, the oxidation number of hydrogen is -1; which makes it a reducing agent.
And in $H_2O$, the oxidation number of hydrogen is +1; which makes it an oxidizing agent.
Right side of the equation
In NaOH the oxidation number of hydrogen is +1.
In \[{{H}_{2}}\] the oxidation number of hydrogen is 0.
Oxidation numbers of sodium and oxygen atoms have not changed at all.

Therefore, the correct answer is (c).

Note: This irreversible reaction occurs very fast and forms a basic solution of NaOH. To determine the reducing agent and oxidizing agent remember this acronym OILRIG (Oxidation Is Loss of electrons Reduction Is Gain of electrons)