Why are sulfide ores converted to oxides before reduction rather than being reduced directly?

Direct reduction of metal sulfides with carbon is thermodynamically unfavorable because carbon disulfide ($CS_2$) is not as thermodynamically stable as carbon dioxide ($CO_2$) or carbon monoxide ($CO$). The standard Gibbs free energy change $\Delta G^\circ$ for converting $MS$ to $M$ using carbon is positive, whereas $\Delta G^\circ$ for reducing metal oxides $MO$ with $C$ or $CO$ is strongly negative at operational temperatures.

How does Zone Refining yield ultra-pure semiconductors like Silicon ($Si$) and Germanium ($Ge$)?

Zone Refining operates on the principle of fractional crystallization: impurities are significantly more soluble in the liquid molten state than in the solid state of the metal. A moving circular heater melts a narrow zone of an impure rod. As the heater advances, pure metal crystallizes out behind it while impurities stay in the melt and are pushed to one end, which is ultimately cut off.

What is the exact function of cryolite ($Na_3AlF_6$) in the Hall-Héroult electrolytic extraction of Aluminum?

Pure alumina ($Al_2O_3$) has a very high melting point ($\sim 2323\text{ K}$) and exhibits poor electrical conductivity. Adding cryolite ($Na_3AlF_6$) along with fluorspar ($CaF_2$) lowers the operating melting point to roughly $1140\text{ K}$ and dramatically boosts the electrical conductivity of the molten bath, making extraction energy-efficient.

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Question #1NEET 2023

In the extraction of copper from its sulfide ore ( $Cu_2S$ ), the metal is finally obtained by the reduction of cuprous oxide ( $Cu_2O$ ) with which of the following species?

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Step-by-Step Solution & Concept

Step 1

Identify the partial oxidation phase during the roasting of copper glance ore: $2Cu_2S + 3O_2 \rightarrow 2Cu_2O + 2SO_2$ .

Step 2

Recognize that the remaining unreacted copper(I) sulfide ( $Cu_2S$ ) acts directly as an auto-reducing agent for the newly formed $Cu_2O$ .

Step 3

Write the auto-reduction equation: $2Cu_2O + Cu_2S \rightarrow 6Cu + SO_2$ , which forms blister copper.

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Question #2JEE Main 2024

According to the Ellingham diagram, at what minimum temperature range does carbon ( $C$ ) act as a thermodynamically superior reducing agent compared to carbon monoxide ( $CO$ ) for reducing metal oxides?

Solution Locked

Step-by-Step Solution & Concept

Step 1

Examine the standard free energy curves for $2C + O_2 \rightarrow 2CO$ (negative slope, $\Delta S^\circ > 0$ ) and $2CO + O_2 \rightarrow 2CO_2$ (positive slope, $\Delta S^\circ < 0$ ).

Step 2

Find the intersection temperature of these two curves, which occurs at $T \approx 710^\circ\text{C}$ ( $983\text{ K}$ ).

Step 3

Conclude that above $710^\circ\text{C}$ , the line for $C \rightarrow CO$ lies lower than $CO \rightarrow CO_2$ , rendering $\Delta G^\circ$ more negative and making carbon ( $C$ ) the more effective reducing agent.

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Mastering Metallurgy: Principles, Ellingham Diagrams & Refining | ExamiFly

Chapter Overview

Core Subtopics & Syllabus Breakdown

Key Formulas & High-Yield Equations

Gibbs Free Energy Criteria for Reduction

Ellingham Diagram Slope Equation

Equilibrium Constant Relation in Pyrometallurgy

Van 't Hoff Thermal Dependency

Faraday's First Law of Electrometallurgy

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