JEE Main 2023 — Electrochemistry Question with Solution

To determine the metals that can be oxidized by $N{O}_3^{-}$, we need to find the metals that have a standard reduction potential (SRP) lower than 0.97 V (since $N{O}_3^{-}$ has an SRP of 0.97 V).

From the given data, we can find the metals that satisfy this condition:

1. $V^{2+}(\text{aq})+2e^{-}\to V$, $E^{\theta }=-1.19\text{V}$ (less than 0.97 V, hence can be oxidized)


2. $F{e}^{3+}(\text{aq})+3e^{-}\to Fe$, $E^{\theta }=-0.04\text{V}$ (less than 0.97 V, hence can be oxidized)


3. $A{g}^{+}(\text{aq})+e^{-}\to Ag(\text{s})$, $E^{\theta }=0.80\text{V}$ (less than 0.97 V, hence can be oxidized)


4. $A{u}^{3+}(\text{aq})+3e^{-}\to Au(\text{s})$, $E^{\theta }=1.40\text{V}$ (greater than 0.97 V, hence cannot be oxidized)

Therefore, $V$, $Fe$, and $Ag$ are the metals that can be oxidized by $N{O}_3^{-}$.

The total number of such metals is 3.