Using ion electron method, how would I balance Fe2+ and MnO4-, step by step?

Answer 1

The Iron will be oxidized from +2 to +3

Fe2+ --> Fe3+ + e-

The Mn(VII) in permanganate will be reduced to Mn+2

(MnO4)- + 8 H+ + 5 e- --> Mn2+ + 4 H2O

To balance the electrons, 5 times as much Iron will need to be oxidized, so:

5 Fe2+ + (MnO4)- + 8 H+ --> 5 Fe3+ + Mn2+ + 4 H2O