Question

The standard reduction potential at 298 K for the following half cells are given below  
NO − 3 3 − + 4H+ + 3e- → NO(g) +2H2O E0 = 0.97V 
\(V^{2+} (aq) + 2e^-→V\)           E⁰ = -1.19V 
\(Fe^{3+}(aq) + 3e^- → Fe\)        E⁰ = -0.04V 
\(Ag^+ (aq) + e^- → Ag(s)\)      E⁰ = 0.80V 
\(Au^{3+}(aq)+3e^-→Au(s)\) E⁰ = 1.40V 
The number of metal(s) which will be oxidized by  \(N0^-_3\) Oin aqueous solution is______.

Answer 1

Correct Answer: 3

Nitrate Ion Oxidation Analysis 

For \( \text{NO}_3^- \) to oxidize a metal, \( \text{NO}_3^- \) must itself be reduced. The given reduction half-reaction is:

\( \text{NO}_3^- + 4\text{H}^+ + 3e^- \rightarrow \text{NO} + 2\text{H}_2\text{O} \qquad E^0 = 0.97 \text{ V} \)

For a metal to be oxidized by \( \text{NO}_3^- \), the overall cell potential (\( E^0_{cell} \)) for the redox reaction must be positive. The cell potential is calculated as:

\( E^0_{cell} = E^0_{reduction} - E^0_{oxidation} \)

In our case, \( E^0_{reduction} \) is 0.97 V (for the \( \text{NO}_3^- \) reduction). For the cell potential to be positive, \( E^0_{oxidation} \) must be less than 0.97 V.

Since we are given standard reduction potentials, we need to reverse the sign to obtain the oxidation potentials for each metal:

• V: \( E^0_{oxidation} = -(-1.19 \text{ V}) = 1.19 \text{ V} \)
• Fe: \( E^0_{oxidation} = -(-0.04 \text{ V}) = 0.04 \text{ V} \)
• Ag: \( E^0_{oxidation} = -(0.80 \text{ V}) = -0.80 \text{ V} \)
• Au: \( E^0_{oxidation} = -(1.40 \text{ V}) = -1.40 \text{ V} \)

Comparing these values to 0.97 V, we find that Fe, Ag, and Au have oxidation potentials less than 0.97 V. Therefore, these three metals can be oxidized by \( \text{NO}_3^- \) in aqueous solution.

Conclusion:

Vanadium (V) has an oxidation potential greater than 0.97 V, so it will not be oxidized by \( \text{NO}_3^- \). So, the number of metal(s) which will be oxidized by \( \text{NO}_3^- \)  in aqueous solution  is 3.
 

Answer 2

In the given half reactions, metals V, Fe, and Ag will be oxidized by NO\(_3^-\) because their reduction potentials are less than 0.97 V. Thus, three metals will be oxidized by NO\(_3^-\).