Problem 39 - Chemistry and Chemical Reactivity

Question

In the following reactions, decide which reactant is oxidized and which is reduced. Designate the oxidizing agent and the reducing agent. (a) \(\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{g})+3 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (b) \(\mathrm{Si}(\mathrm{s})+2 \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{SiCl}_{4}(\ell)\)

Step-by-Step Solution

Answer

(a) \(\mathrm{C}_2\mathrm{H}_4\) is oxidized, reducing agent; \(\mathrm{O}_2\) is reduced, oxidizing agent. (b) \(\mathrm{Si}\) is oxidized, reducing agent; \(\mathrm{Cl}_2\) is reduced, oxidizing agent.

Step 1: Define Oxidation and Reduction

Oxidation refers to the loss of electrons by a molecule, atom, or ion, whereas reduction refers to the gain of electrons. Thus, in a chemical reaction, the substance that loses electrons is oxidized and the one that gains electrons is reduced.

Step 2: Determine Oxidation States for Reaction (a)

In the reaction \( \mathrm{C}_2\mathrm{H}_4 + 3 \mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2 + 2 \mathrm{H}_2\mathrm{O} \), assign oxidation states: Carbon in \( \mathrm{C}_2\mathrm{H}_4 \) is \( -2 \), and in \( \mathrm{CO}_2 \) it is \( +4 \). Oxygen in \( \mathrm{O}_2 \) is \( 0 \) and in \( \mathrm{CO}_2 \) and \( \mathrm{H}_2\mathrm{O} \) it is \( -2 \).

Step 3: Identify the Oxidized Species (Reaction a)

Carbon goes from \(-2\) to \(+4\), meaning it loses electrons and is oxidized.

Step 4: Identify the Reduced Species and Agents (Reaction a)

Oxygen goes from \(0\) to \(-2\), meaning it gains electrons and is reduced. Hence, \(\mathrm{C}_2\mathrm{H}_4\) is the reducing agent while \(\mathrm{O}_2\) is the oxidizing agent.

Step 5: Determine Oxidation States for Reaction (b)

In the reaction \( \mathrm{Si} + 2 \mathrm{Cl}_2 \rightarrow \mathrm{SiCl}_4 \), assign oxidation states: Silicon starts at \( 0 \) and goes to \(+4\) in \( \mathrm{SiCl}_4 \). Chlorine starts at \( 0 \) and goes to \(-1\) in \( \mathrm{SiCl}_4 \).

Step 6: Identify the Oxidized Species (Reaction b)

Silicon increases its oxidation state from \(0\) to \(+4\), thus it is oxidized.

Step 7: Identify the Reduced Species and Agents (Reaction b)

Chlorine decreases from \(0\) to \(-1\), thus it is reduced. \(\mathrm{Si}\) is the reducing agent and \(\mathrm{Cl}_2\) is the oxidizing agent.

Key Concepts

Oxidizing Agent

In a chemical reaction, an oxidizing agent is the substance that accepts electrons. This means it promotes oxidation by gaining electrons itself.
Oxidizing agents are crucial because they make it possible for other substances to lose electrons and become oxidized.
- In the reaction where ethylene (\( \mathrm{C}_2\mathrm{H}_4 \)) reacts with oxygen (\( \mathrm{O}_2 \)), oxygen serves as the oxidizing agent. Oxygen gains electrons, changing its oxidation state from \( 0 \) to \( -2 \) and helping carbon to lose electrons.- Similarly, in the silicon (\( \mathrm{Si} \)) and chlorine (\( \mathrm{Cl}_2 \)) reaction, chlorine acts as the oxidizing agent. It accepts electrons from silicon, reducing its own oxidation state from \( 0 \) to \( -1 \).Remember, an oxidizing agent is always reduced in the reaction process, which means it takes part in a reduction itself.

Reducing Agent

A reducing agent is the substance that donates electrons to another species in a chemical reaction. This means it aids another substance in its reduction, while the reducing agent itself gets oxidized.
- In the combustion of ethylene (\( \mathrm{C}_2\mathrm{H}_4 \)), ethylene serves as the reducing agent. It donates electrons to oxygen, increasing its oxidation state from \( -2 \) to \( +4 \).- In the reaction between silicon and chlorine, silicon acts as the reducing agent. It gives away electrons to chlorine, as its oxidation state escalates from \( 0 \) to \( +4 \).Reducing agents, therefore, are central to the oxidation process as they allow another species to undergo reduction by losing electrons.

Oxidation States

The oxidation state is a concept that helps in the understanding of electron transfer in reactions. It indicates the degree of oxidation of an atom in a compound, reflecting its ability to lose or gain electrons.
- In the ethylene and oxygen reaction, carbon's oxidation state changes from \( -2 \) in \( \mathrm{C}_2\mathrm{H}_4 \) to \( +4 \) in \( \mathrm{CO}_2 \). This indicates a loss of electrons (oxidation).
- Oxygen goes from an oxidation state of \( 0 \) in \( \mathrm{O}_2 \) to \( -2 \) in both \( \mathrm{CO}_2 \) and \( \mathrm{H}_2\mathrm{O} \), evidencing the gain of electrons (reduction).For the second reaction involving silicon and chlorine:- Silicon’s oxidation state changes from \( 0 \) in elemental silicon to \( +4 \) in \( \mathrm{SiCl}_4 \), demonstrating oxidation through electron loss.- Chlorine shows a reduction by moving from an oxidation state of \( 0 \) to \( -1 \) in \( \mathrm{SiCl}_4 \).By tracking these changes, one can identify which species are oxidized or reduced, recognizing the electron flow in the reactions.