this is the way i answered it;
at the anode, the highest one in the reduction potential list is oxidised
so thats Br- --> 1/2Br2(l) + e-
at the cathode the lowest one in the reduction potential list is reduced
which is H2O + 2e- --> 1/2H2(g) + OH-
now i dunno why in successone it says that the oxidised Br- goes to Br2(aq), the list suggests it should be liquid form coz it is higher in the list
can soemone clear that up 4 me plz
thx