During the electrolysis of brine, by using inert electrodes,
Consider the following 4 electrodes
$$\begin{aligned} & \mathrm{A}: \mathrm{Ag}^{+}(0.0001 \mathrm{M}) / \mathrm{Ag}(s) ; \\ & \mathrm{B}: \mathrm{Ag}^{+}(0.1 \mathrm{M}) / \mathrm{Ag}(s) ; \\ & \mathrm{C}: \mathrm{Ag}^{+}(0.01 \mathrm{M}) / \mathrm{Ag}(s) ; \\ & \mathrm{D}: \mathrm{Ag}^{+}(0.001 \mathrm{M}) / \mathrm{Ag}(s) ; E^{\circ}{ }_{\mathrm{Ag}^{+} / \mathrm{Ag}}=+0.80 \mathrm{~V} \end{aligned}$$
Then reduction potential in volts of the electrodes in the order.
When $$\mathrm{FeCl}_3$$ is added to excess of hot water gives a sol '$$X$$'. When $$\mathrm{FeCl}_3$$ is added to $$\mathrm{NaOH}(a q)$$ solution, gives sol '$$Y$$'
$$X$$ and $$Y$$ formed in the above processes respectively are
The reducing agent in the given equations
$$4 \mathrm{Ag}(s)+8 \mathrm{CN}^{-}(a q)+2 \mathrm{H}_2 \mathrm{O}(a q)+\mathrm{O}_2(g) \longrightarrow 4\left[\mathrm{Ag}(\mathrm{CN})_2\right]^{-}(a q)+4 \mathrm{OH}^{-}(a q)$$
$$2\left[\mathrm{Ag}(\mathrm{CN})_2\right]^{-}(a q)+\mathrm{Zn}(s) \longrightarrow {\left[\mathrm{Zn}(\mathrm{CN})_4\right]^{2-}(a q)+2 \mathrm{Ag}(s)}$$