chemistry
posted by Lana on .
Balance the following redox reaction.
MnO4 (aq) + Fe (s) > Mn2+ (aq) +Fe2+ (aq)

first step: break down the equation into half equations for oxidation and reduction
reduction reaction;
8H+ + 5e + MnO4 > Mn2+ + 4H2O
oxidation reaction;
Fe > Fe2+ + 2e
second step: manipulate the equations in order to get the same number of electrons; i.e. multiply reduction equation with 2 to get;
16H+ + 10e + 2MnO4 > 2Mn2+ + 8H2O
multiply oxidation reaction with 5 to get;
5Fe > 5Fe2+ + 10e
third step: add the two equations together (cancel variables on different side of the equation)i.e. the electrons are cancelled out.
16H+ + 5Fe + 2MnO4 > 2Mn2+ + 5Fe2+ + 8H2O 
I expect this is just an exercise in balancing redox equations; from a practical standpoint I doubt there is much reaction between MnO4^ and solid Fe.