The standard reduction potentials at $298 \mathrm{~K}$ for the following half cells are given below:

$$\mathrm{NO}_{3}^{-}+4 \mathrm{H}^{+}+3 \mathrm{e}^{-} \rightarrow \mathrm{NO}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O} \quad \mathrm{E}^{\theta}=0.97 \mathrm{~V}$$

$$\mathrm{V}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{V} \quad\quad\quad \mathrm{E}^{\theta}=-1.19 \mathrm{~V}$$

$$\mathrm{Fe}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} \quad\quad\quad \mathrm{E}^{\theta}=-0.04 \mathrm{~V}$$

$$\mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Ag}(\mathrm{s}) \quad\quad\quad \mathrm{E}^{\theta}=0.80 \mathrm{~V}$$

$$\mathrm{Au}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Au}(\mathrm{s}) \quad\quad\quad \mathrm{E}^{\theta}=1.40 \mathrm{~V}$$

The number of metal(s) which will be oxidized by $\mathrm{NO}_{3}^{-}$ in aqueous solution is __________.

<p>To determine the metals that can be oxidized by $NO_3^-$, we need to find the metals that have a standard reduction potential (SRP) lower than 0.97 V (since $NO_3^-$ has an SRP of 0.97 V).</p> <p>From the given data, we can find the metals that satisfy this condition:</p> <ol> <li>$V^{2+} (\text{aq}) + 2e^- \rightarrow V$, $E^\theta = -1.19 \, \text{V}$ (less than 0.97 V, hence can be oxidized)</li><br/> <li>$Fe^{3+} (\text{aq}) + 3e^- \rightarrow Fe$, $E^\theta = -0.04 \, \text{V}$ (less than 0.97 V, hence can be oxidized)</li><br/> <li>$Ag^+ (\text{aq}) + e^- \rightarrow Ag (\text{s})$, $E^\theta = 0.80 \, \text{V}$ (less than 0.97 V, hence can be oxidized)</li><br/> <li>$Au^{3+} (\text{aq}) + 3e^- \rightarrow Au (\text{s})$, $E^\theta = 1.40 \, \text{V}$ (greater than 0.97 V, hence cannot be oxidized)</li> </ol> <p>Therefore, $V$, $Fe$, and $Ag$ are the metals that can be oxidized by $NO_3^-$.<br/><br/> The total number of such metals is 3.</p>