Identify the oxidation and reduction half-reactions. In this case, Fe^{2+} is oxidized to Fe^{3+}, and N_2 is reduced to N_2H_4.
Write the oxidation half-reaction: Fe^{2+} -> Fe^{3+} + e^-. Balance the charge by adding one electron to the right side.
Write the reduction half-reaction: N_2 + 4 e^- + 4 H_2O -> N_2H_4 + 4 OH^-. Balance the atoms and charges by adding electrons, water, and hydroxide ions.
Combine the balanced half-reactions. Ensure that the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction.
Add the half-reactions together and simplify to get the balanced redox equation in a basic solution: 2 Fe^{2+}(aq) + N_2(g) + 4 OH^-(aq) -> 2 Fe^{3+}(aq) + N_2H_4(aq) + 2 H_2O(l).