In a cell, the following reactions take place Fe ^ 2 + → Fe ^ 3 + +...

The electrode which has more reduction potential will get reduced in spontaneous reaction in the cell. So, spontaneous reaction will be

$2{Fe}^{3 +}\left(aq\right) + 2{I}^{-}\left(aq\right) \to 2{Fe}^{2 +}\left(aq\right) + {I}_{2}$

Iron will act as cathode and Iodine will act as anode.

${E}^{0}cell = {\left({E}^{0}RP\right)}_{C} - {\left({E}^{0}RP\right)}_{A}$

$= 0 . 77 - 0 . 54$

$= 0 . 23V$

$= 23 \times {10}^{- 2}V$