Identify the oxidation and reduction half-reactions from the overall reaction. Here, MnO_4^- is reduced to Mn^{2+}, and Fe^{2+} is oxidized to Fe^{3+}.
Write the half-reactions separately with their respective electrons to balance the changes in oxidation states:
Reduction: \(\mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O}\)
Oxidation: \(\mathrm{Fe^{2+} \rightarrow Fe^{3+} + e^-}\)
Balance the electrons lost in oxidation with the electrons gained in reduction by multiplying the oxidation half-reaction by 5 so that 5 electrons are lost, matching the 5 electrons gained in the reduction half-reaction.
Add the two half-reactions together, canceling out the electrons, to get the overall balanced redox reaction:
\(\mathrm{MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 4H_2O + 5Fe^{3+}}\)
Verify that the atoms and charges are balanced on both sides of the equation, confirming the correct coefficients for the balanced redox reaction in acidic solution.