posted by Lana .
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
8H+ + 5e- + MnO4- --> Mn2+ + 4H2O
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.