Question

What is the reducing agent in the reaction:
$2Na+2{{H}_{2}}O\to 2NaOH+{{H}_{2}}$ ?

Answer 1

A reaction where reduction and oxidation happens simultaneously is termed as redox reaction. A reducing agent is that which itself gets oxidized. While an oxidizing agent is the agent which itself gets reduced. A reducing agent has the ability to lose electrons, while the oxidizing agent gains electrons.

Complete answer:
We have been given a redox reaction, where oxidation and reduction is occurring at the same time. We have to find the reducing agent. A reducing agent is the species that gives electrons and gets oxidized itself. While, an oxidizing agent is the species that takes up the electrons and gets reduced. Reducing agent has an increase in its oxidation number, while an oxidizing agent has a decrease in the oxidation number.
In the equation, $2Na+2{{H}_{2}}O\to 2NaOH+{{H}_{2}}$, we will see the reduction and the oxidation halves to determine the reducing agent.
Reduction half, ${{H}_{2}}O(l)+{{e}^{-}}\to \dfrac{1}{2}{{H}_{2}}(g)\uparrow +H{{O}^{-}}$
Oxidation half, $Na(s)\to N{{a}^{+}}+{{e}^{-}}$
We can see from the equation that sodium loses electrons to form the cation and gets oxidized to NaOH in the products. Also, the oxidation state of Na (s) from the reactant is zero that changes to +1 in NaOH, which means an increase in the oxidation state that happens for reducing agents.
Hence, sodium (Na) is the reducing agent in the given equation.

Note:
The oxidation state of any element in its elementary form is taken as zero, for example, Na, Mg, ${{O}_{2}},{{S}_{8}},C{{l}_{2}}$. We can see that any reaction is a redox reaction by the fact that it has an increase as well as decrease in the oxidation number of the reacting species, when they form a product.