Extraction of gold and silver involves leaching was \[C{N^ - }\] ion. Silver is later recovered by:
A.Displacement with \[Zn\]
D.Zone refining

Answer Verified Verified
Hint: During the extraction of metal, it was made to react with a species with which it can form a water-soluble complex. The impurities formed can be easily isolated by filtration. Further, the use of a more electropositive species acts as a good reducing agent to free the metal from the complex.

Complete answer:
The extraction of silver and gold is done by the cyanide process also known as Mac Arthur - Forrest process. During the extraction of silver and gold, the respective ore is leached with a dilute solution of \[NaCN\] or \[KCN\]. For example considers \[A{g_2}S\] is the ore of silver, and then the \[NaCN\] reacts with \[A{g_2}S\] to give a complex in presence of \[{O_2}\] which is expressed as:
\[A{g_2}S + 4NaCN\xrightarrow{{}}2Na[Ag{(CN)_2}] + N{a_2}S\].
\[Na[Ag{(CN)_2}]\] is soluble in water and is treated with \[Zn\]. The zinc metal is more electropositive in nature and can easily reduce the silver from the complex generating metallic silver during the reaction.
\[2Na[Ag{(CN)_2}] + Zn \to N{a_2}[Zn{(CN)_4}] + 2Ag \downarrow \]
The precipitated silver can be removed by filtration. In a similar way, the gold is extracted by leaching with \[C{N^ - }\] ion and then displacement with \[Zn\].
The impurities of gold and silver cannot be removed by liquation, distillation or zone refining. In this process, the molten metal cannot be purified by removing the impurities by draining away, selective boiling and condensation, or by separating the metal in solid-state.
So silver is later recovered by displacement with \[Zn\] of the intermediate complex formed by leaching with \[C{N^ - }\] ion.

So the correct answer is A.

Note: The electrochemical series has to be understood to give the possible reason for such displacement reaction. The more electropositive the metal, the greater is its reducing power to displace a metal from the solution.
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