Question

I'm the redox reaction of sodium aluminum metal reacts with nitric acid it will form products of aluminum nitrate and molecular hydrogen. Write the balanced half reactions and identify each one as either oxidation or reduction. Find the final net balanced reaction from the half reactions.

Answer 1

Answer :

The balanced half reactions are:

Oxidation : $2Al\rightarrow 2Al^{3+}+6e^-$

Reduction : $6H^++6e^-\rightarrow 3H_2$

The net balanced reaction will be,

$2Al+6H^+\rightarrow 2Al^{3+}+3H_2$

or,

$2Al(s)+6HNO_3(aq)\rightarrow 2Al(NO_3)_3(aq)+3H_2(g)$

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

The given chemical reaction will be:

$2Al(s)+6HNO_3(aq)\rightarrow 2Al(NO_3)_3(aq)+3H_2(g)$

The oxidation-reduction half reaction will be :

Oxidation : $Al\rightarrow Al^{3+}+3e^-$

Reduction : $2H^++2e^-\rightarrow H_2$

To balance the electrons we are multiplying oxidation reaction by 2 and reduction reaction by 3, we get:

Oxidation : $2Al\rightarrow 2Al^{3+}+6e^-$

Reduction : $6H^++6e^-\rightarrow 3H_2$

The net balanced reaction will be,

$2Al+6H^+\rightarrow 2Al^{3+}+3H_2$

or,

$2Al(s)+6HNO_3(aq)\rightarrow 2Al(NO_3)_3(aq)+3H_2(g)$