Question

In the cyanide process for extraction of gold and silver from ores, the cyanide solution acts as a:
(A) Reducing agent to reduce the gold and silver compounds in the ores into metallic states.
(B) Leaching agent to bring the gold and silver into the solution as cyanide complexes and thus separate these metals from the ores.
(C) Leaching agent to dissolve all the other constituents of the ores leaving the gold and silver as metals.
(D) Leaching agent to bring the ores into solution

Answer 1

To solve this question, we must first understand the cyanide process or the gold cyanidation process. Then we need to assess the extraction mechanism the components used in that and then only we can conclude the correct answer.

Complete step-by-step answer: Before we move forward with the solution of this given question, let us first understand some basic concepts: Cyanide Process: Also known as Mac Arthur - Forrest process. Silver and gold are extracted by the cyanide process. After the preliminary crushing and concentration by froth floatation process, the ore (crushed auriferous rocks in the case of gold) is leached with dilute $\left( {0.4{\text{ }} - {\text{ }}7\% } \right)$ solution of sodium cyanide made alkaline by adding lime kept agitated by a current of air. Silver (or gold) pass into solution as argentocyanide (or aurocyanide).It is based on the fact that gold/silver dissolves in dilute solution of sodium cyanide in presence of atmospheric oxygen with the formation of complex cyanide.
Step 1: Gold is mixed with the sodium cyanide which leads to the following Elsener’s equation and reaction. This is done to make soluble gold.
This method of making soluble gold is known as leaching. In the process of leaching, a dilute form of sodium cyanide is added into the ore containing the gold. Since gold is soluble after the leaching process, it is free to move through the membrane while the rest of ore cannot pass through the membrane.
Step 2: The reaction of the process are as follows:
$A{g_2}S + 4NaCN \to 2Na\left[ {Ag{{\left( {CN} \right)}_2}} \right] + N{a_2}S$
Silver is obtained by reaction with $Zn$ :
$2Na\left[ {Ag{{\left( {CN} \right)}_2}} \right]\,\,\, + \,\,Zn\,\, \to \,\,N{a_2}\left[ {Zn{{\left( {CN} \right)}_4}} \right] + 2Ag$
And Hence, after going through the step 1 and 2, we can conclude that cyanide solution acts as a Leaching agent to bring the gold and silver into the solution as cyanide complexes and thus separate these metals from the ores.

So, clearly we can conclude that the correct answer is Option (B).

Note: The leaching process is only carried out when the ore is soluble in a particular solution and not the substances that make it impure. Many times some other substance is also able to dissolve in the leaching solution along with the ore but this is later separated.