Question

How would I solve this

Answer 1

In the redox reaction
`\( \text{PbO} + \text{NH}_3 \rightarrow \text{N}_2 + \text{H}_2\text{O} + \text{Pb} \)`, the oxidation half-reaction involves lead reduction, while the reduction half-reaction involves ammonia oxidation. In aqueous solution, the net ionic form is
`\( \text{PbO} + 6\text{NH}_3 \rightarrow \text{Pb}(\text{NH}_3)_6^(2+) + 2\text{OH}^- \).`

The given redox equation is:

`\[ \text{(a) } \text{PbO}(s) + \text{NH}_3(g) \rightarrow \text{N}_2(g) + \text{H}_2\text{O}(l) + \text{Pb}(s) \]`

To determine the oxidation and reduction half-reactions, let's assign oxidation numbers.

1. **Oxidation Half-Reaction:**

`\[ \text{Pb}^(4+) + 4\text{e}^- \rightarrow \text{Pb}^(2+) \]`

This is because lead (
`\(\text{Pb}\)`) is reduced from an oxidation state of +4 to +2.

2. **Reduction Half-Reaction:**

`\[ \text{NH}_3 + 3\text{e}^- \rightarrow \text{N}_2 + 3\text{H}^+ \]`

This is because ammonia (
`\(\text{NH}_3\)`) is oxidized from an oxidation state of -3 to 0.

**Net Ionic Form in Aqueous Solution:**

`\[ \text{PbO}(s) + 6\text{NH}_3(aq) \rightarrow \text{Pb}(\text{NH}_3)_6^(2+) + 2\text{OH}^- \]`

In this form, we include states-of-matter for reactants and products in aqueous solution, considering that ammonia (
`\(\text{NH}_3\)`) readily dissolves in water.

Note: The balanced coefficients for ammonia and hydroxide ions are selected to ensure charge balance in the net ionic form.