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Refining Process Against Impurities MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

  • Electrolytic Refining is considered one the most difficult concept.

  • Vapour Phase Refining is considered one of the most asked concept.

  • 11 Questions around this concept.

Solve by difficulty

Aluminium is extracted by the electrolysis of

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

Which method of purification is represented by the following equation :

\mathrm{T i(s)+2 I_{2}(g) \stackrel{523 k}{\longrightarrow} T i I_{4}(g) \stackrel{1700 k}{\longrightarrow} T i(s)+2 I_{2}(g)}

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Liquation

In this method a low melting metal like tin can be made to flow on a sloping surface. In this way it is separated from higher melting impurities.

Distillation Method

This is very useful for low boiling metals like zinc and mercury. The impure metal is evaporated to obtain the pure metal as distillate.

Electrolytic Refining

In this method, the impure metal is made to act as anode. A strip of the same metal in pure form is used as cathode. They are put in a suitable electrolytic bath containing soluble salt of the same metal. The more basic metal remains in the solution and the less basic ones go to the anode mud. This process is also explained using the concept of electrode potential, over potential, and Gibbs energy which you have seen in previous sections. The reactions are:

\begin{array}{ll}{\text {Anode: }} & {\mathrm{M} \rightarrow \mathrm{M}^{\mathrm{n}+}+\mathrm{ne}^{-}} \\ {\text {Cathode: }} & {\mathrm{M}^{\mathrm{n}+}+\mathrm{ne}^{-} \rightarrow \mathrm{M}}\end{array}

Copper is refined using an electrolytic method. Anodes are of impure copper and pure copper strips are taken as cathode. The electrolyte is acidified solution of copper sulphate and the net result of electrolysis is the transfer of copper in pure form from the anode to the cathode:

\begin{array}{ll}{\text { Anode: }} & {\mathrm{Cu} \rightarrow \mathrm{Cu}^{2+}+2 \mathrm{e}^{-}} \\ {\text {Cathode: }} & {\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}}\end{array}

Impurities from the blister copper deposit as anode mud which contains antimony, selenium, tellurium, silver, gold and platinum; recovery of these elements may meet the cost of refining. Zinc may also be refined this way.

Zone Refining Process

A movable heater is fitted around a rod of impure metal. The heater is slowly moved across the rod. The metal melts at the point of heating and as the heater moves on from one end of the rod to the other end, the pure metal crystallised while the impurities pass on the adjacent melted zone.

Vapour Phase Refining

In this method, the metal is converted into its volatile compound and collected elsewhere.

It is then decomposed to give pure metal. So, the two requirements are:

(i) the metal should form a volatile compound with an available reagent,

(ii) the volatile compound should be easily decomposable so that the recovery is easy.

The following examples will illustrate this technique.

Mond Process for Refining Nickel: In this process, nickel is heated in a stream of carbon monoxide forming a volatile complex, nickel tetracarbonyl:

\mathrm{Ni}+4 \mathrm{CO} \stackrel{330-350 \mathrm{~K}}{\longrightarrow} \mathrm{Ni}(\mathrm{CO})_{4}

The carbonyl is subjected to higher temperature so that it is decomposed giving the pure metal:

\mathrm{Ni}(\mathrm{CO})_{4} \stackrel{450-470 \mathrm{~K}}{\longrightarrow} \mathrm{Ni}+4 \mathrm{CO}

Van Arkel's Method for Refining Zirconium or Titanium: This method is very useful for removing all the oxygen and nitrogen present in the form of impurity in certain metals like Zr and Ti. The crude metal is heated in an evacuated vessel with iodine. The metal iodide being more covalent, volatilises:

\mathrm{Zr}+2 \mathrm{I}_{2} \rightarrow \mathrm{ZrI}_{4}

The metal iodide is decomposed on a tungsten filament, electrically heated to about 1800K. The pure metal is thus deposited on the filament.

\mathrm{ZrI}_{4} \rightarrow \mathrm{Zr}+2 \mathrm{I}_{2}

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