The reducing agent in the given equations : \(4\ Ag_{(s)}+8\ CN_{(aq)}^-+2H_2O_{(aq)}+O_{2_{(g)}}\rightarrow4[Ag(CN)_2]_{(aq)}^-+4\ OH_{(aq)}^-\) \(2[Ag(CN)_2]_{(aq)}^-+Zn_{(s)}\rightarrow[Zn(CN)_4]_{(aq)}^{2-}+2Ag_{(s)}\)
\[ 4 \, \text{Ag}^+ (s) + 8 \, \text{CN}^- (aq) + 2 \, \text{H}_2\text{O} (aq) + \text{O}_2 (g) \rightarrow 4 \, [\text{Ag(CN)}_2]^- (aq) + 4 \, \text{OH}^- (aq) \] In this reaction, silver ion (\(\text{Ag}^+\)) is reduced to \([\text{Ag(CN)}_2]^-\), and oxygen (\(\text{O}_2\)) is involved as the oxidizing agent. Since silver ions are reduced, **cyanide ions (CN⁻)** are oxidized, making them the reducing agent.
Reaction 2:
\[ 2 \, [\text{Ag(CN)}_2]^- (aq) + \text{Zn} (s) \rightarrow [\text{Zn(CN)}_4]^{2-} (aq) + 2 \, \text{Ag} (s) \] Here, zinc metal (\(\text{Zn}\)) is oxidized to \(\text{Zn}^{2+}\), while the silver ions (\(\text{Ag}^+\)) are reduced to metallic silver. Since zinc is losing electrons, it is the reducing agent in this reaction.
Conclusion:
In both reactions, the **reducing agent is CN⁻ (cyanide ion)**, as it is responsible for reducing the metal ions (\(\text{Ag}^+\) and \(\text{Zn}^{2+}\)).
