Question

Question #33342

Answer 1

Start with the unbalanced chemical equation

`NH_3 + O_2 -> NO_2 + H_2O`

Now, let's focus on assigning oxidation numbers for all atoms involved in the reaction

`N^(-3)H_3^(+1) + O_2^(0) -> N^(+4)O_2^(-2) + H_2^(+1)O^(-2)`

Notice that nitrogen went from an oxidation number (ON) of -3 on the reactants' side, to an ON of +4 on the products' side, while oxygen went from an ON of 0 on the reactants' side, to an ON of -2 on the products' side. The half-reactions look like this

`N^(-3) -> N^(+4) + 7e^(-)`
`O_2^(0) + 4e^(-) -> O_2^(2-)`

Since there are two oxygen atoms on the reactants' side, and each oxygen atom gained `2e^(-)`, oxygen gained a total number of `4e^(-)`. The total number of electrons transferred in this reaction is `4*7=28e^(-)`, which means that the oxidation half-reaction must be multiplied by 4, and the reduction half-reaction must be multiplied by 7.

This will produce these coefficients for the oxidized and reduced reactants

`4NH_3 + 7O_2 -> NO_2 + H_2O`

Now, balance nitrogen like you would normally, and you'll get

`4NH_3 + 7O_2 -> 4NO_2 + H_2O`

The hydrogen and oxygen are not balanced, since we have 12 H on the left side and only 2 on the right side, and 14 O on the left side and only 9 on the right. Let's balance hydrogen first.

In basic solution, for every H atom you need you must add `H_2O` on the side that needs it, and `OH^(-)` on the opposite side. We need 10 H atoms on the right side, so

`10OH^(-) + 4NH_3 + 7O_2 -> 4NO_2 + H_2O + 10H_2O`

Now the hydrogen is balanced. For every oxygen atom we need, `2OH^(-)` must be added to that side, and `H_2O` to the opposite side. We need 5 oxygen atoms on the right side, so

`5H_2O + 10OH^(-) + 4NH_3 + 7O_2 -> 4NO_2 + H_2O + 10H_2O + 10OH^(-)`

The balanced equation thus is

`4NH_3 + 7O_2 -> 4NO_2 + 6H_2O`

The `10OH^(-)` cancelled out, and we were left with 6 `H_2O` molecules on the right side.