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: <span>2ClO−3(aq) + 12H+(aq) + 10e- = Cl2(g) + 6H2O(l) </span><span>E = +1.47 </span> <span>Br(l) + 2e- = 2Br- </span><span>E = +1.065 </span> 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
