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In electro-refining of copper, some gold is deposited as:
(A) Cathode mud
(B) Electrolyte
(C) Anode mud
(D) None of these

Answer Verified Verified
Hint: The Refining itself means to remove impurities or undesirable substances from a substance. So, electro-refining is a process to remove impurities using electricity. Basically, it's a process of refining in metals using electrolysis.

Complete step by step answer:
Here we will take an example of electrolytic copper refining to understand the process more clearly. Copper is typically mined from its coal referred to as copper. It is about 98 to 99 percent pure. However, the electro-refining process can easily make it 99.95% pure which makes it a good product to be used in electrical components.

A block of impure copper is taken as an anode or positive electrode. Copper sulphate which is acidified with vitriol is employed as a graphite-coated electrolyte alongside pure copper tubes, as a cathode or negative electrode. In this phase of electrolysis copper sulphate divides into a positive ion of copper ($C{{u}^{2+}}$) and a negative ion of sulphate ($S{{O}_{4}}^{-}$). The positive copper ion ($C{{u}^{2+}}$) or cations travel towards the negative electrode made from pure copper where it absorbs the electrons from the cathode. Cu atoms are deposited on the cathode’s graphite layer.

The cathode is coated with graphite in the process of electrolytic metal processing or merely electro grinding so that the concentrated material can be easily removed. This is one of the most growing electrolysis procedures.

Electrolytic Refining of Copper


However, anode’s metallic impurities are also mixed with $S{{O}_{4}}$, forming metallic sulphate in the electrolyte solution and dissolving. In this above process, some impurities get dissolved in the solution while some deposits as anode mud below the anode electrode. Impurities which dissolve within the solution are Fe, Ni, Zn while impurities like Au, Ag and Pt deposits as anode mud below the anode electrode.
So, we can say that gold (Au) is deposited as Anode mud below the anode electrode.

Therefore, among the options if we check then option (C)Anode mud is the correct answer of the question.


Note:
- ${{H}^{+}}$from the electrolyte isn't reduced to ${{H}_{2}}$ (g) at the cathode.
-$C{{u}^{2+}}$lies below ${{H}^{+}}$within the table of ordinary reduction potentials.
-$C{{u}^{2+}}$is a stronger oxidant than ${{H}^{+}}$.
-$C{{u}^{2+}}$is more easily reduced than ${{H}^{+}}$.
-Keep in mind the impurities like Fe, Ni, Zn get dissolved within the solution but impurities like Au, Ag and Pt are deposited as anode mud below the anode.

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