Nitric acid is a powerful oxidizing agent. Using standard reduction potentials, predict whether the following metals can be oxidized to +2 ions by nitric acid: (a) iron, (b) copper, (c) rhodium, (d) zinc, (e) lead, (f) tin.

We need to look up the standard reduction potentials for the half-reactions of the metals and nitric acid in a standard reduction potential table. Here are the relevant half-reactions and their potentials: - Iron (Fe): \[Fe^{2+}(aq) + 2e^{-} \rightarrow Fe(s)\] Eº = -0.44 V - Copper (Cu): \[Cu^{2+}(aq) + 2e^{-} \rightarrow Cu(s)\] Eº = +0.34 V - Rhodium (Rh): \[Rh^{2+}(aq) + 2e^{-} \rightarrow Rh(s)\] Eº = -0.65 V - Zinc (Zn): \[Zn^{2+}(aq) + 2e^{-} \rightarrow Zn(s)\] Eº = -0.76 V - Lead (Pb): \[Pb^{2+}(aq) + 2e^{-} \rightarrow Pb(s)\] Eº = -0.13 V - Tin (Sn): \[Sn^{2+}(aq) + 2e^{-} \rightarrow Sn(s) \] Eº = -0.14 V - Nitric acid (HNO3): \[NO_{3}^{-}(aq) + 4H^{+}(aq) + 3e^{-} \rightarrow NO(g) + 2H_{2}O(l)\] Eº = +0.96 V

Since nitric acid has a standard reduction potential of +0.96V, it can oxidize metals that have a lower standard reduction potential. We will compare the nitric acid's standard reduction potential with the standard reduction potentials of each metal: (a) Iron (Fe): Eº(Fe) = -0.44 V is lower than Eº(HNO3) = +0.96 V, so iron can be oxidized by nitric acid. (b) Copper (Cu): Eº(Cu) = +0.34 V is lower than Eº(HNO3) = +0.96 V, so copper can be oxidized by nitric acid. (c) Rhodium (Rh): Eº(Rh) = -0.65 V is lower than Eº(HNO3) = +0.96 V, so rhodium can be oxidized by nitric acid. (d) Zinc (Zn): Eº(Zn) = -0.76 V is lower than Eº(HNO3) = +0.96 V, so zinc can be oxidized by nitric acid. (e) Lead (Pb): Eº(Pb) = -0.13 V is lower than Eº(HNO3) = +0.96 V, so lead can be oxidized by nitric acid. (f) Tin (Sn): Eº(Sn) = -0.14 V is lower than Eº(HNO3) = +0.96 V, so tin can be oxidized by nitric acid. In conclusion, nitric acid can oxidize all the given metals to +2 ions.