Question

For each unbalanced equation given below write unbalanced half-reactions. identify the species oxidized and the species reduced. identify the oxidizing and reducing agents. (a) \(\mathrm{Ag}(s)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{Ag}^{+}(a q)+\mathrm{NO}(g)\) (b) \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{O}_{2}(g)\)

Short answer

Question: In each of the following reactions, identify the species oxidized, the species reduced, the oxidizing agent, and the reducing agent. (a) \(\mathrm{Ag}(s) + \mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{Ag}^+(a q) + \mathrm{NO}(g)\) (b) \(\mathrm{CO}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g) + \mathrm{O}_{2}(g)\) Answer: (a) In reaction (a), the species oxidized is \(\mathrm{Ag}(s)\), the species reduced is \(\mathrm{NO}_{3}^{-}(a q)\), the oxidizing agent is \(\mathrm{NO}_{3}^{-}(a q)\), and the reducing agent is \(\mathrm{Ag}(s)\). (b) In reaction (b), the species oxidized is \(\mathrm{H}_{2} \mathrm{O}(l)\), the species reduced is \(\mathrm{CO}_{2}(g)\), the oxidizing agent is \(\mathrm{CO}_{2}(g)\), and the reducing agent is \(\mathrm{H}_{2} \mathrm{O}(l)\).

Step-by-step solution

(a) Identifying the species with changing oxidation states

In the given equation, \(\mathrm{Ag}(s)\) and \(\mathrm{NO}_{3}^{-}(a q)\) are the reactants, and \(\mathrm{Ag}^+(a q)\) and \(\mathrm{NO}(g)\) are the products. We observe that the oxidation state of \(\mathrm{Ag}\) changes from \(0\) in \(\mathrm{Ag}(s)\) to \(+1\) in \(\mathrm{Ag}^+(a q)\), and the oxidation state of nitrogen changes from \(+5\) in \(\mathrm{NO}_{3}^{-}\) to \(+2\) in \(\mathrm{NO}(g)\). These are the species with changing oxidation states.

(a) Writing unbalanced half-reactions

The unbalanced half-reactions can be written as follows: Oxidation: \(\mathrm{Ag}(s) \longrightarrow \mathrm{Ag}^+(a q) + e^-\) Reduction: \(\mathrm{NO}_{3}^{-}(a q) + 3e^- \longrightarrow \mathrm{NO}(g)\)

(a) Identifying the species oxidized, reduced, oxidizing agents, and reducing agents

Since \(\mathrm{Ag}(s)\) loses electrons and increases its oxidation state from \(0\) to \(+1\), it is oxidized and acts as the reducing agent. On the other hand, \(\mathrm{NO}_{3}^{-}(a q)\) gains electrons and decreases its oxidation state from \(+5\) to \(+2\), so it is reduced and acts as the oxidizing agent.

(b) Identifying the species with changing oxidation states

In the given equation, \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) are the reactants, and \(\mathrm{C}_{2} \mathrm{H}_{4}(g)\) and \(\mathrm{O}_{2}(g)\) are the products. We observe that the oxidation state of carbon changes from \(+4\) in \(\mathrm{CO}_{2}\) to \(-2\) in \(\mathrm{C}_{2} \mathrm{H}_{4}\), and the oxidation state of oxygen changes from \(-2\) in \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) to \(0\) in \(\mathrm{O}_{2}\). These are the species with changing oxidation states.

(b) Writing unbalanced half-reactions

The unbalanced half-reactions can be written as follows: Oxidation: \(2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{O}_{2}(g) + 4e^- + 4H^+\) Reduction: \(\mathrm{CO}_{2}(g) + 8e^- + 8H^+ \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g) + 2 \mathrm{H}_{2} \mathrm{O}(l)\)

(b) Identifying the species oxidized, reduced, oxidizing agents, and reducing agents

Since \(\mathrm{H}_{2} \mathrm{O}(l)\) loses electrons and the oxidation state of oxygen increases from \(-2\) to \(0\), \(\mathrm{H}_{2} \mathrm{O}(l)\) is oxidized and acts as the reducing agent. On the other hand, \(\mathrm{CO}_{2}(g)\) gains electrons and the oxidation state of carbon decreases from \(+4\) to \(-2\), so \(\mathrm{CO}_{2}(g)\) is reduced and acts as the oxidizing agent.