Question

When a piece of sodium metal is dropped in water, hydrogen gas evolved because:
A.Sodium is reduced and acts as an oxidizing agent.
B.Water is oxidized and acts as a reducing agent.
C.Sodium loses electrons and is oxidized while water is reduced.
D.Water loses electrons and is oxidized to hydrogen.

Answer 1

Hint: Determine the oxidation number of each species in the reaction. An increase in oxidation number means that the element loses its electrons and gets oxidized. A decrease in oxidation number means that the element gains electrons and gets reduced.

Complete step by step answer:
Step 1: Write the balanced chemical equation when a piece of sodium metal is dropped in water as follows:
When sodium metal is dropped in water, hydrogen gas is evolved. Thus, the reaction is,
${\text{2Na}} + 2{{\text{H}}_2}{\text{O}} \to {{\text{H}}_2} + 2{\text{NaOH}}$.
In the reaction, sodium hydroxide is formed and hydrogen gas is evolved.
Step 2: Determine the oxidation numbers of sodium and hydrogen atoms in the reactant and product side as follows:
The reaction is,
${\text{2Na}} + 2{{\text{H}}_2}{\text{O}} \to {{\text{H}}_2} + 2{\text{NaOH}}$
Oxidation number of sodium: The oxidation number of any element in its elemental form is always zero. Thus, the oxidation number of sodium is $0$.
Oxidation number of hydrogen in water: The net oxidation number on water is $0$. The oxidation number of oxygen is $ - 2$. Let the oxidation number of hydrogen be $x$. Thus,
$
  \left( {2 \times x} \right) + \left( { - 2} \right) = 0 \\
  2x = 2 \\
  x = + 1 \\
 $
Thus, the oxidation number of hydrogen in water: is $ + 1$.
Oxidation number of ${{\text{H}}_2}$: The oxidation number of any element in its elemental form is always zero. Thus, the oxidation number of ${{\text{H}}_2}$ is $0$.
Oxidation number of sodium${\text{Na}}$ in ${\text{NaOH}}$: The net oxidation number on ${\text{NaOH}}$ is $0$. The oxidation number of oxygen is $ - 2$. The oxidation number of hydrogen is $ + 1$. Let the oxidation number of sodium be $x$. Thus,
$
  \left( x \right) + \left( { - 2} \right) + \left( { + 1} \right) = 0 \\
  x = 2 - 1 \\
  x = + 1 \\
 $
Thus, the oxidation number of sodium in ${\text{NaOH}}$: is $ + 1$.
Thus,
$
  {\text{ 0 + 1 0 + 1}} \\
  {\text{2Na}} + 2{{\text{H}}_2}{\text{O}} \to {{\text{H}}_2} + 2{\text{NaOH}} \\
 $
Step 3: Determine the species getting oxidized and reduced as follows:
The oxidation number of sodium changes from $0$ to $ + 1$. Thus, the oxidation number of sodium increases. Thus, sodium metal loses an electron and gets oxidized.
The oxidation number of hydrogen changes from $ + 1$ to $0$. Thus, the oxidation number of hydrogen decreases. Thus, the hydrogen atom gains an electron and gets reduced.
Thus, sodium loses electrons and is oxidized and water is reduced.

So, the correct answer is Option C .

Note:
A chemical species which loses its electrons and itself gets oxidized is known as a reducing agent.
A chemical species which gains electrons and itself gets reduced is known as an oxidizing agent.
Thus,
Sodium metal loses electrons and gets oxidized. Thus, sodium metal is a reducing agent.
Water gains electrons and gets reduced. Thus, water is an oxidizing agent.