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How would I solve this

How would I solve this-example-1
User Golobitch
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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.

User Cyperpunk
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