Couple Half-reaction E / V
Zn2+/Zn Zn2+ + 2e¯ Zn −0.76
Sn2+/Sn

Sn2+ + 2e¯ Sn

 −0.14
Cu2+/Cu Cu2+ + 2e¯ Cu 0.34
I2/I¯

I2 + 2e¯ 2I¯

 0.54
Fe3+/Fe2+ Fe3+ + e¯ Fe2+ 0.77
Ag+/Ag Ag+ + e¯ Ag 0.80
Br2/Br¯ Br2 + 2e¯ 2Br¯ 1.10
O2/H2O O2 + 4H+ + 4e¯ 2H2O 1.23
Cl2/Cl¯ Cl2 + 2e¯ 2Cl¯ 1.40


As the I2/I¯ couple is negative with respect to the Br2/Br¯ couple the cell will be:

Pt|I2,I¯||Br2,Br¯|Pt
and iodide ions will reduce Br2 in this reaction.

The half reactions and the cell reaction will be:
2I¯ I2 + 2e¯
Br2 + 2e¯2Br¯
2I¯ + Br2I2 + 2Br¯

and the cell potential, E = E(Br2/Br¯) − E(I2/I¯) = 1.10 V − 0.54 V = 0.56 V