Firstly, we need to write a balanced equation for the oxidation of Fe2+ to Fe3+ . We know that when something is oxidised, it is losing electrons, making the half equation: Fe2+ ---> Fe3+ + e- (we can check this by ensuring the charges on each side of the equation are equal). If iron is being oxidised, the manganate ion must be being reduced and so we write a half equation for the reduction of Mn(VII) to Mn(II): MnO4- ---> Mn2+. We must now balance the equation, first for number of atoms of each element by adding water and hydrogen ions: MnO4- + 8H+ ---> Mn2+ + 4H2O. We then balance for charge by adding electrons: MnO4- + 8H+ + 5e- ---> Mn2+ + 4H2O. Finally, we must combine the two half equations by multiplying everything in the equation for oxidation of iron by 5 so that, once combined, the number of electrons on each side of the equation cancel each other out. This leave us with the final equation: 5Fe2+ + MnO4- + 8H+ ---> 5Fe3+ + Mn2+ + 4H2O.