Silver cyanide is used for silver electroplating because of the concentration of the free silver ions in solution which is so low that it is not far from zero. Due to the high values of its complex equilibrium formation constant, the complex $\ce{[Ag(CN)2]^-}$ is practically not dissociated in $\ce{Ag^+}$ ions. If such a complex solution is electrolyzed, only the rare free silver ions are discharged at the cathode. So at the beginning of the electrolysis, these free $\ce{Ag^+}$ ions produce a discontinuous deposit of metallic atoms on the cathode.

Let's consider the mechanism of electrolysis in the microscopic level. As the first metallic atoms produced on the cathode by the beginning of the electrolysis look like small points or bumps, these tiny bumps will attract electric field lines. As a consequence, the next positive ions will get discharged most probably on this bump, making it bigger, if the solution is concentrated enough. More and more positive ions will be attracted and discharged on this point. The metallic deposit will be localized around specific points. The surface of the cathode becomes rough and course.

But if the free ion concentration is extremely low, as with $\ce{Ag^+}$ ions in $\ce{[Ag(NH3)2]^-}$ solutions, the first $\ce{Ag}$ atom is too far from the next $\ce{Ag+}$ ion. This metallic ion is not attracted by the $\ce{Ag}$ "bump". It is attracted by the whole cathode. It touches the cathode and get discharged equally on its surface, independently from the position of the previously deposited Ag atom. The Ag deposit makes a smooth layer which looks like a mirror.

To summarize, the formation of a mirror can only occur in a solution of a silver complex where the free (non-complexed) ion concentration is extremely weak, like in $\ce{[Ag(CN)2]^-}$ solutions. Solutions of other silver compounds, like silver nitrate, will only produce rough silver deposits by electrolysis. No mirror !

Silver cyanide is used for silver electroplating because of the concentration of the free silver ions in solution which is so low that it is not far from zero. Due to the high values of its complex equilibrium formation constant, the complex $\ce{[Ag(CN)2]^-}$ is practically not dissociated in $\ce{Ag^+}$ ions. If such a complex solution is electrolyzed, only the rare free silver ions are discharged at the cathode. So at the beginning of the electrolysis, these free $\ce{Ag^+}$ ions produce a discontinuous deposit of metallic atoms on the cathode.

Let's consider the mechanism of electrolysis in the microscopic level. As the first metallic atoms produced on the cathode by the beginning of the electrolysis look like small points or bumps, these tiny bumps will attract electric field lines. As a consequence, the next positive ions will get discharged most probably on this bump, making it bigger, if the solution is concentrated enough. More and more positive ions will be attracted and discharged on this point. The metallic deposit will be localized around specific points. The surface of the cathode becomes rough and course.

But if the free ion concentration is extremely low, as with $\ce{Ag^+}$ ions in $\ce{[Ag(CN)2]^-}$ solutions, the first $\ce{Ag}$ atom is too far from the next $\ce{Ag+}$ ion. This metallic ion is not attracted by the $\ce{Ag}$ "bump". It is attracted by the whole cathode. It touches the cathode and get discharged equally on its surface, independently from the position of the previously deposited Ag atom. The Ag deposit makes a smooth layer which looks like a mirror.

To summarize, the formation of a mirror can only occur in a solution of a silver complex where the free (non-complexed) ion concentration is extremely weak, like in $\ce{[Ag(CN)2]^-}$ solutions. Solutions of other silver compounds, like silver nitrate, will only produce rough silver deposits by electrolysis. No mirror !