Answer
-0.44 V
Work Step by Step
Half-reactions are
$Fe\rightarrow Fe^{2+}+2e^{-}$ (anode)
$Cu^{2+}+2e^{-}\rightarrow Cu$ (cathode)
$E^{\circ}_{cell}=E^{\circ}_{cathode}-E^{\circ}_{anode}$
$\implies E^{\circ}_{anode}= E^{\circ}_{cathode}-E^{\circ}_{cell}$
$=0.34\,V-0.78\,V=-0.44\,V$
$E^{\circ}$ for the $Fe(s)|Fe^{2+}(aq)$ half-cell is $-0.44\,V$
