During the process of electrolytic refining of copper, some metals present as impurity settle as 'anode mud'. These are

In electrolytic refining, a thick block of impure copper is made the anode while a thin, high-purity sheet of copper is the cathode. The electrolyte is usually an aqueous solution of acidified $$CuSO_4$$.

• At the cathode: $$Cu^{2+} + 2e^- \rightarrow Cu$$ (pure copper is deposited).
• At the anode: $$Cu \rightarrow Cu^{2+} + 2e^-$$ (impure copper dissolves).

The fate of each impurity depends on its standard reduction potential $$E^\circ$$.

Case 1: Metals more electropositive than copper (e.g., $$Fe, Ni, Zn, Pb$$) have lower $$E^\circ$$ values than $$Cu^{2+}/Cu$$. Hence, when the anode dissolves, these metals also pass into the solution as ions because oxidation of these metals is easier than that of copper. They do not deposit at the cathode under normal refining conditions and therefore remain in the electrolyte.

Case 2: Metals more electronegative than copper (e.g., $$Ag, Au, Pt$$) have higher $$E^\circ$$ values than $$Cu^{2+}/Cu$$. They are less readily oxidised, so they do not form ions when the anode dissolves. Instead, they simply fall off the anode as insoluble particles. These heavy, unoxidised particles collect beneath the anode as the dark, granular sludge called “anode mud” or “anode slime.”

Therefore, among the given choices, the metals that settle as anode mud during copper refining are $$Ag$$ and $$Au$$.

Checking each option:
Option A: Sn is more electropositive than Cu and dissolves; Ag becomes mud — partially correct, so discard.
Option B: Pb and Zn dissolve into the electrolyte, so no anode mud.
Option C: Ag and Au both form anode mud — correct.
Option D: Fe and Ni dissolve into the solution — not anode mud.

Option C which is: Ag and Au