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-

Since there is the same amount of agn03 in both silver chloride and silver nitrate they produce the same amount of moles. therefore silver chloride produces the same amount of silver nitrate, so your answer is 15.0 moles of silver nitrate. i hope this !

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