The standard reduction potential of Br2/Br- is higher than that of I2/I-, which means Br2 has a higher tendency of gaining an electron and forming Br- than I2 does. Therefore when Br2 and I- are present in the same solution, Br2 will gain electrons from I- and form Br-, meanwhile I- loses electrons and forms I2.
Br2 + 2I- = 2Br- + I2