The half-reaction occurring at the anode in the balanced reaction shown below is 3mno4- (aq) + 24h+ (aq) + 5fe (s) → 3mn2+ (aq) + 5fe3+ (aq) + 12h2o (l)

Reduction takes place in cathode while oxidation takes place at the anode.
Given that the reaction 3MnO4- +24H + +5Fe→3Mn+2+5Fe+3+12H2O
Now that oxidation is termed as an increase in oxidation number or loss of electrons while reduction is a decrease in oxidation number and gain of electrons,
∴oxidation will be
Fe→Fe +3+3e-
reduction will
MnO4- +8H+ →Mn+2+4H2O
If oxidation takes place at anode then Anode: an oxidation reaction
Fe→Fe+3+3e- . Then the answer is 
Fe(s)→Fe+3(aq)+3e-