To determine which pairs of metals can be oxidized by the nitrate ion \((NO_3^-) \) in an aqueous solution, we need to compare the standard reduction potentials \(E^\circ\) of the given metal ions with that of the nitrate ion.
### Step-by-Step Solution:
1. **Identify the Standard Reduction Potential of Nitrate Ion:**
The standard reduction potential \(E^\circ\) for the reduction of the nitrate ion \(NO_3^-\) is given as \(+0.96 \, V\).
2. **List the Standard Reduction Potentials of the Given Metal Ions:**
- \(V^{2+} + 2e^- \rightarrow V\), \(E^\circ = -1.19 \, V\)
- \(Fe^{3+} + 3e^- \rightarrow Fe\), \(E^\circ = -0.04 \, V\)
- \(Au^{3+} + 3e^- \rightarrow Au\), \(E^\circ = +1.40 \, V\)
- \(Hg^{2+} + 2e^- \rightarrow Hg\), \(E^\circ = +0.86 \, V\)
3. **Determine Which Metals Can Be Oxidized by Nitrate Ion:**
For a metal to be oxidized by \(NO_3^-\), its reduction potential must be lower than that of \(NO_3^-\) (i.e., \(E^\circ < +0.96 \, V\)).
- For \(V^{2+}\): \(-1.19 \, V < +0.96 \, V\) (can be oxidized)
- For \(Fe^{3+}\): \(-0.04 \, V < +0.96 \, V\) (can be oxidized)
- For \(Au^{3+}\): \(+1.40 \, V > +0.96 \, V\) (cannot be oxidized)
- For \(Hg^{2+}\): \(+0.86 \, V < +0.96 \, V\) (can be oxidized)
4. **Identify the Pairs of Metals:**
The metals that can be oxidized by \(NO_3^-\) are:
- \(V\) and \(Hg\)
- \(V\) and \(Fe\)
- \(Fe\) and \(Hg\)
5. **Conclusion:**
The pairs of metals that can be oxidized by the nitrate ion in aqueous solution are:
- \(Hg\) and \(Fe\)
- \(V\) and \(Fe\)
- \(V\) and \(Hg\)
### Final Answer:
The pairs of metals that can be oxidized by \(NO_3^-\) are:
- \(Hg\) and \(Fe\)
- \(V\) and \(Fe\)
- \(V\) and \(Hg\)
