The Ostwald process for the commercial production of nitric acid involves the following three steps: $$4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ $$2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)$$ $$3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g)$$ a. Which reactions in the Ostwald process are oxidation–reduction reactions? b. Identify each oxidizing agent and reducing agent.

We will assign oxidation states to all elements in the reactions. To do this, remember the following rules: 1. The oxidation state of an element in its elemental state is 0 (e.g., O₂ or N₂). 2. The more electronegative element is assigned its typical oxidation state (e.g., -2 for oxygen in most compounds, except peroxides). 3. The remaining elements will have the oxidation state that preserves the overall charge neutrality. Here are the oxidation states of elements in each reaction: \( 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \) (N: -3, H: +1, O: 0) -----> (N: +2, O: -2, H: +1, O: -2) \( 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \) (N: +2, O: -2, O: 0) -----> (N: +4, O: -2, O: -2) \( 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \) (N: +4, O: -2, O: -2, +1, H: +1, O: -2) -----> (H: +1, N: +5, O: -2, O: -2, O: -2, N: +2, O: -2)

The first two reactions in the Ostwald process involve changes in oxidation states for elements and are therefore oxidation-reduction reactions. The third reaction does not have any changes in oxidation states, so it is not an oxidation-reduction reaction. \( 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \) - Oxidation-reduction reaction \( 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \) - Oxidation-reduction reaction \( 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \) - Not an oxidation-reduction reaction The answer to part a is that the first two reactions are oxidation-reduction reactions.

In the first reaction, the oxidation state of nitrogen increases from -3 to +2, which means ammonia (NH₃) is being oxidized. Thus, it is the reducing agent. Oxygen's oxidation state decreases from 0 to -2, indicating that molecular oxygen (O₂) is being reduced. Therefore, oxygen is the oxidizing agent. In the second reaction, the oxidation state of nitrogen increases from +2 to +4, which means nitric oxide (NO) is being oxidized. Thus, it is the reducing agent. Oxygen's oxidation state decreases from 0 to -2, indicating that molecular oxygen (O₂) is being reduced, and it is the oxidizing agent. The answer to part b is: - In the first reaction, NH₃ is the reducing agent, and O₂ is the oxidizing agent. - In the second reaction, NO is the reducing agent, and O₂ is the oxidizing agent.