Chemistry-Redox reaction

since KBr is a stronger reducing agent ( oxidation ) than KCl
so Br- must lose electrons more readily than Cl-

in case 2, Br- will lose electrons ( oxidation )
and Cl will gain elections to form Cl-
hence redox reaction occur

but in case 1, Br cannot lose electrons
because : Br2 ---> 2Br- + 2e- is not correct
so Cl- cannot gain electrons
no redox reaction occur

This is because Br-(aq) loses electrons readily than Cl-(aq),
Br-(aq) is able to oxidize itself, give out electrons to Cl2.
Among the halide ions, I-(aq) loses electrons most readily
i.e. I-(aq) > Br-(aq) > Cl-(aq) > F-(aq)

You can also understand in this way.
Cl2 is higher than Br2 in the Group VII ( Cl2 is more reactive than Br2 )
Therefore, Cl2 is a stronger oxidizing agent than Br2 -- oxidize sb readily.
You will find that F2(g) > Cl2(g) > Br2(g) > I-(aq)
Power of oxidizing is related to the reactivity of halogens

Cl2 has stronger oxidizing power than Br2
so Cl2 will displace Br- out, but not the other way round.

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