Answer
385.5k+ views
Hint: For extracting the metal we need to react the given metal with a species with which it tends to form a complex which is water soluble. With the process of filtration we can isolate the impurities present by this method. The use of electropositive metal can be done as a reducing agent so that the metal can be freed from the complex.
Complete step by step answer:
The extraction of gold and silver is done majorly by the Macarthur Forrest process which is also known as the cyanide process. For extracting the gold and silver, the ore is leached by the dilute solution of sodium cyanide or the potassium cyanide. If we take the ore of silver that is $${\text{A}}{{\text{g}}_{\text{2}}}{\text{S}}$$ and react it with the sodium cyanide in the presence of oxygen the formation of the complex takes place. Look at the following reaction:
$${\text{A}}{{\text{g}}_{\text{2}}}{\text{S + 4NaCN}} \to {\text{2Na[Ag(CN}}{{\text{)}}_2}] + {\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}$$
The complex formed is a soluble complex in water which is further treated with zinc. As the metal of zinc is more electropositive in nature and tends to reduce the silver easily from the complex which generates the metallic silver in the reaction. $${\text{2Na[Ag(CN}}{{\text{)}}_2}] + Zn \to N{a_2}[{\text{Zn(CN}}{{\text{)}}_{\text{4}}}{\text{] + 2Ag}}$$
$${\text{2Na[Ag(CN}}{{\text{)}}_2}] + Zn \to N{a_2}[{\text{Zn(CN}}{{\text{)}}_{\text{4}}}{\text{] + 2Ag}}$$
The silver which is precipitated can be removed by the filtration process. In the same way the gold can be extracted through the cyanide process.
So, the correct answer is Option A.
Note: The impurities which are present in the silver or gold cannot be removed through the process of liquation, filtration or zone refining. We cannot purify the molten metal by the impurities through selective boiling or condensation. The metal which has a high electropositive character has high reducing power also to displace a metal from the solution.
Complete step by step answer:
The extraction of gold and silver is done majorly by the Macarthur Forrest process which is also known as the cyanide process. For extracting the gold and silver, the ore is leached by the dilute solution of sodium cyanide or the potassium cyanide. If we take the ore of silver that is $${\text{A}}{{\text{g}}_{\text{2}}}{\text{S}}$$ and react it with the sodium cyanide in the presence of oxygen the formation of the complex takes place. Look at the following reaction:
$${\text{A}}{{\text{g}}_{\text{2}}}{\text{S + 4NaCN}} \to {\text{2Na[Ag(CN}}{{\text{)}}_2}] + {\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}$$
The complex formed is a soluble complex in water which is further treated with zinc. As the metal of zinc is more electropositive in nature and tends to reduce the silver easily from the complex which generates the metallic silver in the reaction. $${\text{2Na[Ag(CN}}{{\text{)}}_2}] + Zn \to N{a_2}[{\text{Zn(CN}}{{\text{)}}_{\text{4}}}{\text{] + 2Ag}}$$
$${\text{2Na[Ag(CN}}{{\text{)}}_2}] + Zn \to N{a_2}[{\text{Zn(CN}}{{\text{)}}_{\text{4}}}{\text{] + 2Ag}}$$
The silver which is precipitated can be removed by the filtration process. In the same way the gold can be extracted through the cyanide process.
So, the correct answer is Option A.
Note: The impurities which are present in the silver or gold cannot be removed through the process of liquation, filtration or zone refining. We cannot purify the molten metal by the impurities through selective boiling or condensation. The metal which has a high electropositive character has high reducing power also to displace a metal from the solution.
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