Given standard E⊖:
Fe3++3e−→Fe; E⊖=−0.036V
Fe2++2e−→Fe; E⊖=−0.440V
The E⊖ of Fe3++e−→Fe2+ is:
Fe3++3e−→Fe(E1) .... (i)
Fe2++2e−→Fe(E2) .... (ii)
Net equation Fe3++e−→Fe2+(E3) .... (iii)
[Obtained by equation (i) - (ii)]
E3=n1E1−n2E2n3=3(−0.036)−2(−0.440)1
=0.772V
Option D is correct.