The correct option is C I− will be oxidized to I2
The favorable reaction is the oxidation of iodide ion to iodine.
The couple which has higher reduction potential will be reduce and couple with lower reduction potential gets oxidised.
Here, Fe3+/Fe2+ couple has higher reduction potential so Fe3+ get reduced to Fe2+.
Therefore, I− get oxidised to I2.
e−+Fe+3→Fe2+;E=0.77 V2I−→I2+2e−;E=−0.536 V2Fe+3+2I−→2Fe+2+I2;E=Ered+EoxE=0.77−0.536=0.234 V
Since the value of the cell potential is positive, the reaction will take place,
Hence, option C is correct.