Given, Fe3+(aq)+e−──▸Fe2+(aq)E°=+0.771V ......(i) Fe2+(aq)+2e−──▸Fe(s)−0.447V .......(ii) (F=96485Cmol−1) To find ΔG° of Fe3+(aq)+3e−───▸Fe(s) ∵ΔG(f)°=ΔG1°+ΔG2° =−nFE1°+−nFE2°(∵ΔG°=−nFE°) for eq. no. (1),n=1 and for eq. no. (2),n=2 E1°=0.771V and E2°=(−)0.447V on adding Eq. (i) and (ii), we have ΔG(f)°=(−nFE1°)+(−nFE2°) =F[−0.771+(−)2×0.447] =F[−0.771(−)−(0.894)] =96485×0.123=11867.655kJ∕mol =11.87kJ∕mol Hence, option (d) is the correct answer.