Chapter 20, Problem 60A

Interpretation Introduction

Interpretation: The chemical equation, $N_2{O}_4+N_2{H}_4\to N_2+H_{2}O$ is to be balanced.

Concept introduction: The chemical reaction that involves both oxidation and reduction is the redox reaction. Redox reactions can be balanced by following the half-reaction method. That is,

• First, the given redox reaction is divided into half reactions one oxidation and the other reduction.
• Based on mass and charge the half-reactions are balanced.
• The electrons on both sides of the reaction are equalized.
• Then, add the half-reactions together and the common one is eliminated to get the balanced equation.
• The addition of water molecules and H+ will also help in the balancing of equations.

Answer to Problem 60A

The balanced equation is, $N_2{O}_4+2N_2{H}_4\to 3N_2+4H_{2}O$

Explanation of Solution

The reaction, $N_2{O}_4+N_2{H}_4\to N_2+H_{2}O$ is a redox reaction. The half-reactions present in the reaction are,

Oxidation: $N_2{O}_4+2e^{-}\to N_2$

Reduction: $N_2{H}_4+6e^{-}\to N_2+2e^{-}$

The electron count is the same so this equation but the oxygen count is not the same it can be balanced by the addition of water.

The addition of water and H+ will make the reaction as,

Reduction: $N_2{O}_4+8H^{+}\to N_2+4H_{2}O$

Oxidation: $N_2{H}_4\to N_2+4H^{+}$

Now the charge can be balanced by adding electrons. So, the equation becomes,

Reduction: $N_2{O}_4+8e^{-}\to N_2+4H_{2}O$

Oxidation: $N_2{H}_4\to N_2+4H^{+}+4e^{-}$

The electron count can be made equal by multiplying the oxidation equation by 2.

Reduction: $N_2{O}_4+8H^{+}+8e^{-}\to N_2+4H_{2}O$

Oxidation: $N_2{H}_4\to 2N_2+8H^{+}+8e^{-}$

The addition of two reactions will give the balanced equation as,

  $N_2{O}_4+2N_2{H}_4\to 3N_2+4H_{2}O$