In order to solve this, we need to know the standard cell potentials of the half reaction from the given overall reaction.
The half reactions with their standard cell potentials are:
2ClO−3(aq) + 12H+(aq) + 10e- = Cl2(g) + 6H2O(l)
E = +1.47
Br(l) + 2e- = 2Br-
E = +1.065
We solve for the standard emf by subtracting the standard emf of the oxidation from the reducation, so:
1.47 - 1.065 = 0.405 V
