Answer: Dissolve only the metal being refined (e.g., Cu), not impurities above it (Ag, Au) or those below that fall off without dissolving.
- A Dissolve only the metal being refined (e.g., Cu), not impurities above it (Ag, Au) or those below that fall off without dissolving
- B Dissolve highly every metal present in the impure anode simultaneously and largely indiscriminately as frequently observed in practice
- C Deliberately deposit most of the impurity metals directly onto the cathode right alongside the pure metal in many documented cases
- D Prevent any and all dissolution of the impure anode from occurring throughout the entire electrolysis according to conventional understanding
Correct answer: A. Dissolve only the metal being refined (e.g., Cu), not impurities above it (Ag, Au) or those below that fall off without dissolving
Explanation: Anode potential is set between Cu (dissolves, E° -0.34V vs SHE as cathode/+0.34 as anode) and noble metals (Ag, Au, Pt) which remain as anode mud; less noble impurities (Ni, Fe) dissolve but don't plate at cathode (Cu is preferentially plated).
An Ellingham diagram plots ΔG° of oxide formation against temperature for different metals; whichever line is LOWER (more negative ΔG°) at a given temperature reduces the oxide of any metal whose line sits above it - the basis of carbon reduction (Fe, Zn) vs electrolytic reduction (Al, Mg, Na) decisions.
Concept context
The science of extracting metals from ores and refining them for use. Covers concentration methods, reduction techniques, refining processes, and the thermodynamic principles that govern metal extraction.