What reaction will take place if a silver coin is put in dilute `HNO_(3)` ?

Answer

To determine the reaction that occurs when a silver coin is placed in dilute nitric acid (HNO₃), we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this scenario are:
- Silver (Ag) from the silver coin
- Dilute nitric acid (HNO₃)

### Step 2: Understand the Nature of the Reaction
Dilute HNO₃ acts as an oxidizing agent. This means it can oxidize silver (Ag) while itself being reduced.

### Step 3: Write the Oxidation and Reduction Half-Reactions
1. **Oxidation Half-Reaction**: Silver (Ag) is oxidized to silver nitrate (AgNO₃).
\[ \text{Ag} \rightarrow \text{Ag}^+ + e^- \]
2. **Reduction Half-Reaction**: Nitrogen in HNO₃ is reduced. In dilute HNO₃, nitrogen is typically reduced from +5 oxidation state in HNO₃ to +2 in nitric oxide (NO).
\[ \text{2HNO}_3 + 4e^- \rightarrow \text{2NO} + 2H_2O + 2H^+ \]

### Step 4: Combine the Half-Reactions
To combine the half-reactions, we need to balance the electrons. The oxidation of 3 moles of silver gives 3 moles of electrons, and we need to balance this with the reduction of HNO₃.

The balanced overall reaction can be written as:
\[ 3 \text{Ag} + 4 \text{HNO}_3 \rightarrow 3 \text{AgNO}_3 + 2 \text{NO} + 2 \text{H}_2O \]

### Step 5: Confirm the Balancing of Atoms and Charges
- **Silver (Ag)**: 3 on both sides
- **Nitrogen (N)**: 4 on both sides
- **Oxygen (O)**: 12 on both sides (from 4 HNO₃ and products)
- **Hydrogen (H)**: 8 on both sides (from 4 HNO₃ and products)

### Conclusion
The reaction that takes place when a silver coin is put in dilute HNO₃ is:
\[ 3 \text{Ag} + 4 \text{HNO}_3 \rightarrow 3 \text{AgNO}_3 + 2 \text{NO} + 2 \text{H}_2O \]