Given Fe3++3e−→Fe EFe3+∕Fe∘=−0.036V ...(i) Fe2++2e−→Fe EFe2+∕Fe∘=−0.439V ...(ii) we have to calculate Fe3++e−→Fe2+,∆G∘=? To obtain this equationsubtract equ. (ii) from (i) we get Fe3++e−→Fe2+ ...(iii) As we know that ∆G∘=−nFE∘ Thus for reaction (iii) ∆G3=∆G1−∆G2 −nFE3∘=−nFE1∘−(−nFE2∘) −nFE3∘=nFE2∘−nFE1∘ −1FE3∘=2×0.439F−3×0.036F −1FE3∘=0.770F ∴E3∘=−0.770V