Balancing the Reaction ( P_4 HNO_3 rightarrow H_3PO_4 NO_2 H_2O ) Using the Oxidation Method

In this article, we will detail the step-by-step process of balancing the chemical reaction between ( P_4 ) and ( HNO_3 ) to form ( H_3PO_4 ), ( NO_2 ), and ( H_2O ) using the oxidation method. This method is particularly useful in understanding reductive and oxidative states of elements in the reaction.

Step 1: Assign Oxidation States

To begin with, we need to assign oxidation states to the elements involved in the reaction.

( P_4 ): Elemental form, oxidation state 0. ( N ) in ( HNO_3 ): 5 oxidation state. ( P ) in ( H_3PO_4 ): 5 oxidation state. ( N ) in ( NO_2 ): 4 oxidation state. ( H ) in ( H_3PO_4 ): 1 oxidation state. ( O ) in ( H_3PO_4 ), ( NO_2 ), and ( H_2O ): -2 oxidation state.

Step 2: Identify Oxidation and Reduction

Next, we identify the oxidation and reduction processes taking place in the reaction.

Oxidation: Phosphorus is oxidized from 0 in ( P_4 ) to 5 in ( H_3PO_4 ). Reduction: Nitrogen is reduced from 5 in ( HNO_3 ) to 4 in ( NO_2 ).

Step 3: Write Half-Reactions

We write half-reactions for both oxidation and reduction processes. These half-reactions will help us to balance the overall reaction later.

Oxidation Half-Reaction:

( n cdot P_4 rightarrow 4 cdot H_3PO_4 )

Reduction Half-Reaction:

( n cdot 4 cdot HNO_3 rightarrow 4 cdot NO_2 2 cdot H_2O )

Step 4: Balance Each Half-Reaction

In this step, we balance the elements, charge, and water in each half-reaction.

Oxidation Half-Reaction:

( n cdot P_4 rightarrow 4 cdot H_3PO_4 ) ( n ) phosphorus atoms on both sides. Hydrogen and oxygen need to be balanced later.

Reduction Half-Reaction:

( n cdot 4 cdot HNO_3 rightarrow 4 cdot NO_2 2 cdot H_2O ) 4 nitrogen atoms on both sides. Balance hydrogen: 4 H from ( 4 cdot HNO_3 ) gives 2 H2O.

Step 5: Combine Half-Reactions

We now combine the half-reactions to form the overall reaction.

( n cdot P_4 n cdot 4 cdot HNO_3 rightarrow 4 cdot H_3PO_4 4 cdot NO_2 2 cdot H_2O )

Step 6: Balance the Overall Reaction

Finally, we check if the reaction is balanced in terms of mass and charge.

Phosphorus: 4 on both sides. Nitrogen: 4 on both sides. Hydrogen: 12 from ( 4 cdot HNO_3 ) equals 12 in ( 4 cdot H_3PO_4 ) and ( 2 cdot H_2O ). Oxygen: 4 times 3 12 from ( 4 cdot HNO_3 ) equals 4 times 4 16 from ( 4 cdot H_3PO_4 ) and 4 from ( 4 cdot NO_2 ) and 2 from ( 2 cdot H_2O ).

The balanced equation is:

( P_4 10 cdot HNO_3 rightarrow 4 cdot H_3PO_4 10 cdot NO_2 2 cdot H_2O )

This balanced equation ensures that all elements and charges are balanced in the reaction. By using the oxidation method, we can effectively balance complex reactions involving multiple half-reactions.

Keywords: Balancing Chemical Equations, Oxidation Method, Phosphoric Acid, Nitrogen Dioxide, Cathodic Principle