Question

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.

Short answer

In conclusion, nitric acid can oxidize all the given metals (iron, copper, rhodium, zinc, lead, and tin) to +2 ions, as their standard reduction potentials are lower than the standard reduction potential of nitric acid.

Step-by-step solution

Find the standard reduction potentials of the metals and nitric acid

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

Compare the standard reduction potentials

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.

Key concepts

standard reduction potential

The standard reduction potential is a crucial concept in understanding redox reactions. This potential measures a substance's tendency to gain electrons, also known as being reduced. We express it in volts (V), representing the energy change involved in the electron transfer.
To determine whether a metal can be oxidized by an oxidizing agent like nitric acid, we compare their standard reduction potentials. Standard reduction potentials are found in tables listing various substances.
The higher the reduction potential, the stronger the tendency to gain electrons and undergo reduction. Conversely, a lower or negative reduction potential indicates a substance more likely to lose electrons (i.e., be oxidized).
### Practical Example For example, consider nitric acid with a standard reduction potential of +0.96V. Any metal with a lower reduction potential can donate electrons to nitric acid and be oxidized. Hence, nitric acid can oxidize metals like iron, copper, rhodium, zinc, lead, and tin, all of which have lower potential values than nitric acid.

oxidizing agents

Oxidizing agents play an essential role in redox (reduction-oxidation) reactions. These substances accept electrons from other compounds (making them reduced) and in turn oxidize the other compounds by taking their electrons.
The higher the standard reduction potential of an oxidizing agent, the stronger it is because it has a greater ability to accept electrons. Nitric acid is a potent oxidizing agent due to its high standard reduction potential.
### Core Characteristics - **Electron Acceptor:** Gains electrons in the reaction. - **Promotes Oxidation:** Causes another substance to lose electrons. - **Example in Action:** In our examples, nitric acid oxidizes the given metals by accepting electrons, resulting in the oxidation of these metals to +2 ions.

metal oxidation

Metal oxidation involves a metal losing electrons, effectively transforming from a neutral state to a positively charged ion. This process is essential in chemical reactions, particularly in forming compounds.
When a metal undergoes oxidation, it typically increases its oxidation state. In our scenario with the metals and nitric acid, each metal forms a +2 ion after losing two electrons.
### Characteristics of Oxidized Metals - **Increased Oxidation State:** The metal ion formed is more positively charged. - **Electron Loss:** Metals in the elemental state (like Fe, Cu) become ions by losing electrons (e.g., Fe to Fe²⁺). - **Driven by Oxidizing Agent:** Here, nitric acid drives this electron loss due to its high reduction potential.

half-reaction

Half-reactions are a way to dissect complex oxidation-reduction reactions into two simpler components, each depicting the reduction or oxidation process separately.
In redox reactions, one half-reaction represents the reduction (electron gain), while the other demonstrates oxidation (electron loss). This simplification allows easy identification of electron transfer in reactions.
### Example Explained For example, let's consider a zinc half-reaction: - Reduction: Zn²⁺(aq) + 2e⁻ → Zn(s)
This represents the reduction process when zinc ions gain electrons to form solid zinc. - Oxidation: Zn(s) → Zn²⁺(aq) + 2e⁻
Conversely, this shows zinc oxidizing by losing electrons to become zinc ions. The analysis of half-reactions thus enables better understanding and solving of complex oxidation-reduction problems.