The half-cell reactions are as follows:

At anode : $$F{e^{2 + }}(aq) \to F{e^{3 + }}(aq) + e$$ ; $${E^0} = - 0.68\,V$$

At cathode : $$C{e^{4 + }}(aq) + e \to C{e^{3 + }}(aq)$$ ; $${E^0} = 1.44\,V$$

Cell reaction : $$F{e^{2 + }}(aq) + C{e^{4 + }}(aq)$$ $$\rightleftharpoons$$ $$F{e^{3 + }}(aq) + C{e^{3 + }}(aq)$$ ; $$E_{cell}^0 = + 0.76\,V$$

We know, $$E_{cell}^0 = {{0.059} \over n}\log {K_{eq}}$$ or, $$ + 0.76 = {{0.059} \over 1}\log {K_{eq}}$$

or, $$\log {K_{eq}} = {{0.76} \over {0.059}} = 12.859$$ $$\therefore$$ $${K_{eq}} = 7.227 \times {10^{12}}$$