J.R. S. answered 07/30/22
Using a table of standard reduction potentials, I find the following:
I2 + 2e- ==> 2I- Eº = 0.53 V
Br2 + 2e- == 2Br- Eº = 1.07 V
According to these values, Br2 will be reduced more readily than I2, and so it becomes the cathode. The anode will thus be I2.
Eºcell = Eºcathode - Eºanode
Eºcell = 1.07 - 0.53
Eªcell = 0.54 V