The complete ionization of KBr into its constituents
is:
KBr (s) --->
K+ (aq) + Br- (aq)
During electrolysis, oxidation takes place at the anode electrode. This means
that an ion is stripped off its electron hence becoming more positive:
2 Br- (aq) --->
Br2 (g) + 2e-
We can see that Bromine gas Br2 is evolved at the anode.
Meanwhile at the cathode, the reduction reaction occurs.
Which means that the electron from the anode electrode is used to make an ion
more negative:
2K+ (aq) + 2e- ---> 2K (s)
Hence, through reduction, solid potassium is deposited on the
plate.
Half reactions:
Anode: 2 Br- (aq) ---> Br2 (g) + 2e-
Cathode: 2K+ (aq) + 2e- ---> 2K (s)
