The standard reduction potentials at $$298$$ K for the following half cells are given below :
$$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$$, $$E° = 1.33$$ V
$$Fe^{3+}(aq) + 3e^- \rightarrow Fe$$, $$E° = -0.04$$ V
$$Ni^{2+}(aq) + 2e^- \rightarrow Ni$$, $$E° = -0.25$$ V
$$Ag^+(aq) + e^- \rightarrow Ag$$, $$E° = 0.80$$ V
$$Au^{3+}(aq) + 3e^- \rightarrow Au$$, $$E° = 1.40$$ V
Consider the given electrochemical reactions, The number of metal(s) which will be oxidized by $$Cr_2O_7^{2-}$$, in aqueous solution is ______

$$Cr_2O_7^{2-}$$ has $$E° = 1.33$$ V (as oxidizing agent). It can oxidize metals whose reduction potential is less than 1.33 V.

The metals and their reduction potentials:

- Fe: $$Fe^{3+}/Fe$$ has $$E° = -0.04$$ V (< 1.33 V) — Can be oxidized ✓

- Ni: $$Ni^{2+}/Ni$$ has $$E° = -0.25$$ V (< 1.33 V) — Can be oxidized ✓

- Ag: $$Ag^+/Ag$$ has $$E° = 0.80$$ V (< 1.33 V) — Can be oxidized ✓

- Au: $$Au^{3+}/Au$$ has $$E° = 1.40$$ V (> 1.33 V) — Cannot be oxidized ✗

Number of metals oxidized by $$Cr_2O_7^{2-}$$ = 3 (Fe, Ni, Ag).

The answer is $$\boxed{3}$$.