Problem 49 - 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) \(\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{g})+3 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\ell)\) (b) \(\mathrm{Si}(\mathrm{s})+2 \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{SiCl}_{4}(\ell)\)

Step-by-Step Solution

Answer

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

Step 1: Determine Oxidation States in Reaction (a)

In the reaction \( \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{g})+3 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\ell) \), the oxidation state of carbon in \( \mathrm{C}_{2}\mathrm{H}_{4} \) is -2, as each hydrogen contributes +1 and there are four hydrogens. The oxidation state of oxygen in \( \mathrm{O}_{2} \) is 0. In the products, carbon in \( \mathrm{CO}_{2} \) is +4 and oxygen in \( \mathrm{H}_{2}\mathrm{O} \) is -2.

Step 2: Identify Changes and Assign Oxidized and Reduced Reactant in Reaction (a)

The carbon atoms in \( \mathrm{C}_{2}\mathrm{H}_{4} \) are oxidized from -2 to +4 as they lose electrons. Meanwhile, the oxygen in \( \mathrm{O}_{2} \) is reduced from 0 to -2 as it gains electrons.

Step 3: Determine Oxidizing and Reducing Agents in Reaction (a)

Since \( \mathrm{C}_{2}\mathrm{H}_{4} \) is oxidized, it acts as the reducing agent, while \( \mathrm{O}_{2} \) is reduced and thus acts as the oxidizing agent.

Step 4: Determine Oxidation States in Reaction (b)

In the reaction \( \mathrm{Si}(\mathrm{s})+2 \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{SiCl}_{4}(\ell) \), silicon initially has an oxidation state of 0 and chlorine in \( \mathrm{Cl}_{2} \) is also 0. In \( \mathrm{SiCl}_{4} \), silicon has an oxidation state of +4 and each chlorine has an oxidation state of -1.

Step 5: Identify Changes and Assign Oxidized and Reduced Reactant in Reaction (b)

Silicon is oxidized from 0 to +4, losing electrons, while chlorine is reduced from 0 to -1, gaining electrons.

Step 6: Determine Oxidizing and Reducing Agents in Reaction (b)

Silicon acts as the reducing agent because it is oxidized, while \( \mathrm{Cl}_{2} \) acts as the oxidizing agent as it is reduced.

Key Concepts

Oxidation States

In any redox (reduction-oxidation) reaction, understanding oxidation states is crucial. An oxidation state represents the theoretical charge an atom would have if all bonds were ionic. It helps us track how electrons are transferred between atoms.
In general, oxidation involves an increase in oxidation state, while reduction involves a decrease. Let's look at an example from the mentioned reactions.

• For reaction (a) \(\mathrm{C}_2\mathrm{H}_4 + 3\mathrm{O}_2 \rightarrow 2\mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O}\), the carbon in \(\mathrm{C}_2\mathrm{H}_4\) starts with an oxidation state of -2.
• Each hydrogen contributes a +1 oxidation state.
• For oxygen, \(\mathrm{O}_2\) starts at 0 and becomes -2 in \(\mathrm{H}_2\mathrm{O}\) and \(\mathrm{CO}_2\).
• For reaction (b) \(\mathrm{Si} + 2\mathrm{Cl}_2 \rightarrow \mathrm{SiCl}_4\), silicon starts at an oxidation state of 0 and changes to +4.

Learning to calculate these states helps in determining the flow of electrons during reaction processes.

Oxidizing Agent

An oxidizing agent is a substance that causes another substance to lose electrons and thereby be oxidized. It acts by accepting electrons itself. In the process, the oxidizing agent is reduced as it gains electrons.
This is clearly seen in our example reactions.

• In reaction (a), \(\mathrm{O}_2\) acts as the oxidizing agent. It starts with an oxidation state of 0 and gets reduced to -2 in \(\mathrm{H}_2\mathrm{O}\) and \(\mathrm{CO}_2\), indicating it has gained electrons.
• For reaction (b), \(\mathrm{Cl}_2\) is the oxidizing agent. Initially at 0, it reaches -1 in \(\mathrm{SiCl}_4\).

Identifying the oxidizing agent involves recognizing the substance being reduced during the reaction.

Reducing Agent

A reducing agent is a substance that donates electrons to another compound, facilitating its reduction. As it donates electrons, it undergoes oxidation itself.This can be seen in our sample reactions:

• In reaction (a), \(\mathrm{C}_{2}\mathrm{H}_{4}\) is the reducing agent because it donates electrons and oxidizes from -2 to +4.
• Moreover, in reaction (b), silicon \(\mathrm{Si}\) is the reducing agent.

Identifying the reducing agent involves understanding which substance is losing electrons, becoming oxidized in the process.