The standard reduction potentials at 295 K for the following half cells are given below:
NO$$_3^-$$ + 4H$$^+$$ + 3e$$^-$$ $$\rightarrow$$ NO(g) + 2H$$_2$$O   E° = 0.97 V
V$$^{2+}$$(aq) + 2e$$^-$$ $$\rightarrow$$ V(s)   E° = -1.19 V
Fe$$^{3+}$$(aq) + 3e$$^-$$ $$\rightarrow$$ Fe(s)   E° = -0.04 V
Ag$$^+$$(aq) + e$$^-$$ $$\rightarrow$$ Ag(s)   E° = 0.80 V
Au$$^{3+}$$(aq) + 3e$$^-$$ $$\rightarrow$$ Au(s)   E° = 1.40 V
The number of metal(s) which will be oxidised by NO$$_3^-$$ in aqueous solution is ______.

For a redox reaction to be spontaneous under standard conditions, the standard cell potential must be positive:
$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \gt 0 \; -(1)$$

The nitrate ion in acidic medium is supplied as the oxidising agent:
$$NO_3^- + 4H^+ + 3e^- \rightarrow NO(g) + 2H_2O$$ with $$E^\circ = 0.97\;V$$

When $$NO_3^-$$ is reduced, it acts as the cathode. The metal that gets oxidised will therefore act as the anode. Let the general half-reaction for the metal be
$$M^{n+} + ne^- \rightarrow M(s)$$ with reduction potential $$E^\circ_{(M^{n+}/M)}$$

Putting this into $$(1)$$: $$E^\circ_{\text{cell}} = 0.97\;V - E^\circ_{(M^{n+}/M)}$$

For spontaneity $$E^\circ_{\text{cell}} \gt 0$$, so
$$0.97\;V - E^\circ_{(M^{n+}/M)} \gt 0$$ $$\Longrightarrow E^\circ_{(M^{n+}/M)} \lt 0.97\;V \; -(2)$$

Compare the given metal reduction potentials with $$0.97\;V$$:

• $$V^{2+} + 2e^- \rightarrow V(s)$$, $$E^\circ = -1.19\;V$$
• $$Fe^{3+} + 3e^- \rightarrow Fe(s)$$, $$E^\circ = -0.04\;V$$
• $$Ag^+ + e^- \rightarrow Ag(s)$$, $$E^\circ = 0.80\;V$$
• $$Au^{3+} + 3e^- \rightarrow Au(s)$$, $$E^\circ = 1.40\;V$$

According to inequality $$(2)$$, the metals satisfying $$E^\circ_{(M^{n+}/M)} \lt 0.97\;V$$ are V, Fe and Ag. Au does not satisfy the condition because $$1.40\;V \gt 0.97\;V$$.

Hence, the number of metals that will be oxidised by $$NO_3^-$$ in aqueous acidic solution is $$3$$.