Question

In each of the following elements:-Assign oxidation numbers.-Identify the oxidizing and reducing agents.-Identify the change in oxidation number.NH4NO2 —> N2 + 2H2O

Answer 1

Given equation:

`NH_4NO_2\rightarrow N_2+2H_2O`

The oxidation number of the elements are assigned below:

`N^(-3)4H^+N^(+3)2O^(-2)\rightarrow N^0_2+2H_2O`

The oxidation number of N in NH₄⁺ is -3

The oxidation number of N in NO₂⁻ is +3

The oxidation number of N in N₂ is 0

The oxidation number of H is + 1

The oxidation number of O is -2

Oxidizing and reducing agents.

NH₄NO₂ is the oxidizing agent and is also the reducing agent.

Hence, the reaction is a comproportionation or synproportionation - a chemical reaction where two reactants containing the same element but with different oxidation numbers, form a compound having an intermediate oxidation number.

Change in oxidation number.

The oxidation number of N changes from -3 in NH₄⁺ to 0 in N₂

Also the oxidation number of N changes from +3 in NO₂⁻ to 0 in N₂