Animated Solution for Chemistry - Metallurgy: On heating, mixture of ${\text{Cu}}_2\text{O}$ and ${\text{Cu}}_2\text{S}$ will give

Imagine you are an ancient metallurgist, staring at a pile of rocks, knowing that trapped within them is a metal that will define an entire era of human civilization: Copper. Extracting copper from its ores is not just a mechanical process; it is a beautiful symphony of chemical reactions.

When we talk about the extraction of copper, especially from its sulphide ores like copper pyrites, we encounter a fascinating sequence of steps. After roasting and smelting, we are left with a molten mixture known as copper matte, which primarily contains cuprous sulphide (${\text{Cu}}_2\text{S}$) and some iron sulphide. The magic truly happens in the next step, inside a massive, pear-shaped vessel known as the Bessemer Converter.

A Metallurgical Marvel

The Bessemer converter is where the final transformation takes place. Hot air is blown through the molten matte. This air oxidizes the remaining iron sulphide to iron oxide, which is then removed as slag using silica. But what happens to the copper?

A portion of the cuprous sulphide (${\text{Cu}}_2\text{S}$) is oxidized by the blast of air to form cuprous oxide (${\text{Cu}}_2\text{O}$).

Now, we have a molten pool containing both ${\text{Cu}}_2\text{O}$ and unreacted ${\text{Cu}}_2\text{S}$. This is exactly the scenario presented in our problem!

Here is where nature shows its elegance. In many metallurgical processes, we need to add an external reducing agent, like carbon (coke) or carbon monoxide, to strip the oxygen away from the metal oxide. But copper is special. It is less electropositive, meaning it doesn't hold onto oxygen as tightly as metals like iron or aluminum do.

Because of this, we don't need to add any external reducing agent. The unreacted cuprous sulphide itself acts as the reducing agent for the cuprous oxide. They react with each other in a process beautifully termed as Auto-Reduction or Self-Reduction.

Why does this reaction proceed so spontaneously at high temperatures? The answer lies in Ellingham diagrams and Gibbs free energy. At the elevated temperatures inside the Bessemer converter (around $1200^{\circ }\text{C}$), the Gibbs free energy change ($\Delta G$) for the formation of sulfur dioxide is highly negative. Sulfur has a tremendous affinity for oxygen at these temperatures, much more so than copper.

This thermodynamic gradient acts as an unstoppable driving force. The system "wants" to form ${\text{SO}}_2$ gas because it represents a state of much lower energy and higher entropy (since a gas is being produced from condensed phases). The copper is merely a bystander that gets reduced as a consequence of sulfur's aggressive pursuit of oxygen.

Let's look at the exact chemical equation that governs this self-reduction:

Let's break down the oxidation states to truly understand the electron flow: - In ${\text{Cu}}_2\text{O}$, copper is in the $+1$ oxidation state. - In ${\text{Cu}}_2\text{S}$, copper is also in the $+1$ oxidation state, and sulfur is in the $-2$ state. - In the products, we get pure elemental copper ($\text{Cu}$), which has an oxidation state of $0$. - The sulfur is oxidized from $-2$ to $+4$ to form sulfur dioxide (${\text{SO}}_2$).

The sulfur in the sulphide ore sacrifices its electrons to reduce the copper ions down to their elemental metallic state. It is a perfect, self-contained redox reaction!

As the reaction proceeds, a massive amount of sulfur dioxide (${\text{SO}}_2$) gas is generated. While the copper is still molten, this gas bubbles through the liquid metal, trying to escape into the atmosphere.

As the Bessemer converter is tilted and the molten copper is poured out to cool, a fascinating physical phenomenon occurs. The solubility of ${\text{SO}}_2$ gas in copper drops drastically as the metal transitions from a liquid to a solid state. The trapped gas violently forces its way out of the solidifying surface.

This escaping gas leaves behind tiny craters and eruptions on the surface of the metal. The final solidified copper looks like it has blisters or boils all over it. Because of this distinct appearance, the product is famously called Blister Copper.

Blister copper is about $98\%$ pure. While it is a massive achievement, it still contains impurities like silver, gold, and traces of other metals, which are later removed through electrolytic refining to achieve $99.99\%$ purity for electrical applications.

Returning to our original question: On heating a mixture of ${\text{Cu}}_2\text{O}$ and ${\text{Cu}}_2\text{S}$, the auto-reduction process yields pure copper metal and sulfur dioxide gas.

Therefore, the correct option is (d) $\text{Cu}+{\text{SO}}_2$.

Never underestimate the power of self-reduction. Sometimes, the solution to a complex problem is already present within the system itself!