A) \[NaCN\]
B) \[NaCN\] in presence of\[{{O}_{2}}\]
C) \[NaCl\]
D) \[AgN{{O}_{3}}\]
Correct Answer: B
Solution :
The reaction is based on the fact that silver and its compounds form soluble complex with\[NaCN\]. This reaction is reversible and accumulation of \[N{{a}_{2}}S\] must be prevented. A free excess of air \[({{O}_{2}})\] is continuously passed through the solution which oxidises \[N{{a}_{2}}S\] to\[N{{a}_{2}}S{{O}_{4}}\]and\[N{{a}_{2}}{{S}_{2}}{{O}_{3}}\]. \[A{{g}_{2}}S+4NaCN2Na[Ag{{(CN)}_{2}}]+N{{a}_{2}}S\] \[2N{{a}_{2}}S+2{{O}_{2}}+{{H}_{2}}O\to N{{a}_{2}}{{S}_{2}}{{O}_{3}}+2NaOH\] Silver is recovered from solution by precipitating with zinc. \[2Na[Ag{{(CN)}_{2}}]+4NaOH+Zn\xrightarrow{{}}N{{a}_{2}}Zn{{O}_{2}}\] \[4NaCN+2{{H}_{2}}O+2Ag\]
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