Metallurgy — Practice Quiz

CO becomes a more effective reducing agent at higher temperatures because its ΔG becomes increasingly negative

Step 1: In the Ellingham diagram, the y-axis shows ΔG° for oxide formation and the x-axis shows temperature. A line with a negative slope means ΔG° becomes more negative as temperature increases. Step 2: For the reaction 2C + O₂ → 2CO, two moles of CO gas are formed from one mole of O₂ gas — this means disorder (entropy) increases, so ΔS is positive. Step 3: Since ΔG = ΔH - TΔS, a positive ΔS makes the term (-TΔS) increasingly negative as T increases, causing ΔG to decrease (become more negative). Step 4: This means CO formation is thermodynamically more favourable at higher temperatures — CO

Aluminium (Al)

Step 1: In the Ellingham diagram, for carbon to reduce a metal oxide, the line for CO or CO₂ formation must lie BELOW the line for that metal oxide formation at the given temperature. Step 2: The Ellingham line for Al₂O₃ formation lies very LOW on the diagram (very negative ΔG°), indicating that Al₂O₃ is extremely stable. Step 3: The carbon line (for CO formation) does not go below the Al₂O₃ line at any practically achievable temperature in a furnace. This means carbon cannot thermodynamically reduce Al₂O₃. Step 4: For Fe, Zn, and Pb oxides, the CO line does intersect and fall below their resp

Cryolite acts as the electrolyte and dissolves alumina, lowering its melting point for electrolysis

Step 1: Pure alumina (Al₂O₃) has a very high melting point (~2345 K), making its direct electrolysis extremely energy-intensive and impractical. Step 2: In the Hall-Heroult process, purified alumina (20%) is dissolved in molten cryolite (Na₃AlF₆). About 10% CaCl₂ is also added. Step 3: This mixture melts at a much lower temperature (~1270 K), making the electrolysis more energy-efficient and practical. The molten mixture conducts electricity and serves as the electrolyte. Step 4: The electrolytic cell uses a carbon-lined iron tank as the cathode. Carbon blocks immersed in the melt act as the a

Impure metal dissolves into the solution and less electropositive impurities collect as anode mud

Step 1: In electrolytic refining, the impure metal forms the anode and pure metal forms the cathode. The electrolyte is an aqueous solution of a salt of the metal being refined. Step 2: At the anode (impure metal), oxidation occurs — metal atoms lose electrons and dissolve as metal ions into the electrolyte: M(s) → M^n+(aq) + ne⁻. Step 3: At the cathode (pure metal), reduction occurs — metal ions from solution gain electrons and deposit as pure metal: M^n+(aq) + ne⁻ → M(s). Step 4: Impurities that are LESS electropositive than the metal being refined (e.g., Ag, Au, Pt in copper refining) do NO