"However, the Ag^+ + I^- reacted to give AgI (but that is not a redox reaction) and if they reacted why didn't Br^- and Ag^+?"
Hi Dr.Bob, I am also given additional information for reaction #1
Cu(s)
I-(aq)
Ag(s)
Br- (aq)
To determine whether a reaction would occur between Br^- and Ag^+ in the presence of Cu(s), I-(aq), and Ag(s), we need to consider the reactivity of these species.
The reactivity of a metal is determined by its position in the reactivity series. Metals higher in the reactivity series can displace metals lower in the series from their compounds.
In this case, Cu (copper) is higher in the reactivity series than Ag (silver). Therefore, Cu can displace Ag from its compounds, but not vice versa. This means that Cu(s) can react with Ag+ to form Cu2+ and Ag(s):
Cu(s) + 2Ag+(aq) -> Cu2+(aq) + 2Ag(s)
Now let's consider the reaction between I^- and Cu(s). Since I^- is higher in the reactivity series than Cu, it can displace Cu from its compounds. Therefore, the reaction between I^- and Cu(s) would be:
I^-(aq) + Cu(s) -> CuI(aq)
However, the Ag+ and I^- reaction would not be a redox reaction. It would simply form a precipitate of AgI (silver iodide):
Ag+(aq) + I^-(aq) -> AgI(s)
In contrast, Br^- is lower in the reactivity series than Ag. Therefore, it cannot displace Ag from its compounds. As a result, no reaction would occur between Br^- and Ag+ in this scenario.