Question

The final step in the smelting of CuFeS \(_{2}\) is $$ \mathrm{Cu}_{2} \mathrm{~S}(s)+2 \mathrm{Cu}_{2} \mathrm{O}(s) \longrightarrow 6 \mathrm{Cu}(I)+\mathrm{SO}_{2}(g) $$ (a) Give the oxidation states of copper in \(\mathrm{Cu}_{2} \mathrm{~S}, \mathrm{Cu}_{2} \mathrm{O},\) and \(\mathrm{Cu} .\) (b) What are the oxidizing and reducing agents in this reaction?

Short answer

Oxidation states: Cu in \(\text{Cu}_2\text{S}\) and \(\text{Cu}_2\text{O}\) is +1, in \(\text{Cu}\) is 0. Oxidizing agent: \(\text{Cu}_2\text{O}\), Reducing agent: \(\text{Cu}_2\text{S}\).

Step-by-step solution

Determine Oxidation States in \(\text{Cu}_2\text{S}\)

In \(\text{Cu}_2\text{S}\), sulfur typically has an oxidation state of -2. Let the oxidation state of copper be \(x\). Therefore, the equation can be written as \(2x + (-2) = 0\). Solving for \(x\), we get \(x = +1\). Thus, the oxidation state of Cu in \(\text{Cu}_2\text{S}\) is +1.

Determine Oxidation States in \(\text{Cu}_2\text{O}\)

In \(\text{Cu}_2\text{O}\), oxygen typically has an oxidation state of -2. Let the oxidation state of copper be \(x\). Therefore, the equation can be written as \(2x + (-2) = 0\). Solving for \(x\), we get \(x = +1\). Thus, the oxidation state of Cu in \(\text{Cu}_2\text{O}\) is +1.

Determine Oxidation State in \(\text{Cu}\)

In elemental copper (Cu), the oxidation state is 0 because it is not combined with any other element.

Identify Oxidizing and Reducing Agents

In the given reaction, \(\text{Cu}_2\text{S}(s) + 2\text{Cu}_2\text{O}(s) \rightarrow 6\text{Cu}(I) +\text{SO}_2(g)\), copper is reduced because its oxidation state decreases from +1 in \(\text{Cu}_2\text{S}\) and \(\text{Cu}_2\text{O}\) to 0 in \(\text{Cu}\). Thus, \(\text{Cu}_2\text{O}\) acts as the oxidizing agent (it gains electrons), and \(\text{Cu}_2\text{S}\) acts as the reducing agent (it loses electrons).

Key concepts

Copper Smelting

Copper smelting is a process of extracting copper metal from its ores. The process involves several steps, including the final step of reacting copper(I) sulfide [\(\text{Cu}_2\text{S}\)] with copper(I) oxide [\(\text{Cu}_2\text{O}\)].
In this reaction, the copper is separated from its compounds, resulting in pure copper and sulfur dioxide gas [\(\text{SO}_2\)].
The basic chemical equation for this smelting step is:
\[\text{Cu}_2\text{S}(s) + 2\text{Cu}_2\text{O}(s) \rightarrow 6\text{Cu}(s) + \text{SO}_2(g)\]
This is a redox reaction where copper changes its oxidation states, and it's vital to understand these changes to grasp the chemistry involved in smelting.

Oxidizing Agents

In any redox (reduction-oxidation) reaction, an oxidizing agent plays a crucial role. It's the substance that accepts electrons and gets reduced during the reaction. In the context of copper smelting, \(\text{Cu}_2\text{O}\) acts as the oxidizing agent.
Here's why: Copper in \(\text{Cu}_2\text{O}\) has an oxidation state of +1. During the reaction, it is converted to copper metal (\(\text{Cu}\)), which has an oxidation state of 0. This means that \(\text{Cu}_2\text{O}\) gains electrons and is thus reduced. So, we can confidently say that \(\text{Cu}_2\text{O}\) is the oxidizing agent.
The role of oxidizing agents is essential in driving the redox reactions forward, facilitating various industrial processes, including copper smelting.

Reducing Agents

Just as oxidizing agents are critical, reducing agents are equally important in redox reactions. A reducing agent donates electrons and gets oxidized. In the copper smelting reaction, \(\text{Cu}_2\text{S}\) acts as the reducing agent.
In \(\text{Cu}_2\text{S}\), copper has an oxidation state of +1. During the reaction, it is converted to copper metal (\(\text{Cu}\)), in which the oxidation state is 0. This indicates that \(\text{Cu}_2\text{S}\) loses electrons and gets oxidized, making it the reducing agent.
The function of reducing agents is to supply the necessary electrons for the reduction of another substance, making these agents indispensable in redox chemistry and processes like copper smelting.

Redox Reactions

Redox reactions are chemical processes in which the oxidation state of atoms changes due to the transfer of electrons. These reactions consist of two half-reactions: oxidation and reduction.
In the context of copper smelting:

• Oxidation: Copper in \(\text{Cu}_2\text{S}\) loses electrons and gets oxidized from +1 to 0.
• Reduction: Copper in \(\text{Cu}_2\text{O}\) gains electrons and gets reduced from +1 to 0.

The overall equation \(\text{Cu}_2\text{S}(s) + 2\text{Cu}_2\text{O}(s) \rightarrow 6\text{Cu}(s) + \text{SO}_2(g)\) showcases how electrons are transferred between substances, highlighting the redox nature of the reaction.
Understanding these fundamental aspects of redox reactions is key to grasping the broader application in industrial processes like copper smelting.