In the electrolytic refining of blister copper, the total number of main impurities, from the following, removed as anode mud is _________.
Pb, Sb, Se, Te, Ru, Ag, Au and Pt

During electrolytic refining of copper

• Anode : impure blister copper.
• Cathode : thin, pure copper sheets.
• Electrolyte : acidified $$\text{CuSO}_4$$ solution.

At the anode, copper is oxidised : $$$\text{Cu (s)} \rightarrow \text{Cu}^{2+} + 2e^-$$$

The behaviour of each impurity depends on its standard reduction potential $$E^{\circ}$$ relative to that of copper ($$E^{\circ}_{\text{Cu}^{2+}/\text{Cu}} = +0.34 \text{ V}$$).

1. Impurities that are more electropositive than copper (lower $$E^{\circ}$$ values) dissolve into the solution as ions. Examples : $$\text{Fe}, \text{Zn}, \text{Pb}^{2+}$$, etc.

2. Impurities that are less electropositive / noble (higher $$E^{\circ}$$ values) do not get oxidised. They are insoluble in the electrolyte and simply fall off the anode as a heavy deposit called anode mud.

The common constituents of anode mud are : $$\text{Ag}, \text{Au}, \text{Pt}, \text{Sb}, \text{Se}, \text{Te}$$ (and traces of $$\text{Sn}, \text{Bi}$$ etc.).

Let us examine the eight elements given in the question one by one.

Case 1: $$\text{Pb (Lead)}$$
$$E^{\circ}_{\text{Pb}^{2+}/\text{Pb}} = -0.13 \text{ V}$$ is much lower than $$+0.34 \text{ V}$$, so lead is more electropositive than copper. It is, therefore, oxidised to $$\text{Pb}^{2+}$$. Lead does not remain as anode mud.

Case 2: $$\text{Sb (Antimony)}$$
A typical reduction potential for $$\text{Sb}^{3+}/\text{Sb}$$ is about $$+0.20 \text{ V}$$, nobler than Pb but still less than Cu. Antimony does not dissolve appreciably; it settles as part of the mud.

Case 3: $$\text{Se (Selenium)}$$
Selenium is present in copper ores as $$\text{Cu}_2\text{Se}$$. Being much less electropositive than copper, it remains undissolved and appears in the mud.

Case 4: $$\text{Te (Tellurium)}$$ behaves analogously to selenium and also enters the anode mud.

Case 5: $$\text{Ru (Ruthenium)}$$
The standard potential $$E^{\circ}_{\text{Ru}^{3+}/\text{Ru}} \approx +0.46 \text{ V}$$ is higher than that of copper, so ruthenium will not be oxidised and hence is not an important impurity in commercial blister copper. It is therefore not counted among the main constituents of anode mud.

Case 6: $$\text{Ag (Silver)}$$
$$E^{\circ}_{\text{Ag}^+/\text{Ag}} = +0.80 \text{ V}$$. Silver is far nobler than copper and appears entirely in the anode mud.

Case 7: $$\text{Au (Gold)}$$
$$E^{\circ}_{\text{Au}^{3+}/\text{Au}} = +1.50 \text{ V}$$. Gold is also collected in the mud.

Case 8: $$\text{Pt (Platinum)}$$ is the most noble of all; it too stays as insoluble anode mud.

Collecting the results:

• Remain as anode mud : $$\text{Sb}, \text{Se}, \text{Te}, \text{Ag}, \text{Au}, \text{Pt}$$  → 6 elements.
• Do not remain as mud : $$\text{Pb}, \text{Ru}$$

Hence, the total number of main impurities that are removed as anode mud is 6.

Final answer : $$6$$