Consider the following half-reactions: $$\mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}(a q)+5 e^{-} \overrightarrow{\mathrm{Mn}^{2+}}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l)$$ $$\mathrm{NO}_{3}^{-}(a q)+4 \mathrm{H}^{+}(a q)+3 e^{-} \longrightarrow_{\mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)}$$

The first half-reaction is a reduction because electrons are gained: \( \mathrm{MnO}_{4}^{-}(aq) + 8 \mathrm{H}^{+}(aq) + 5e^{-} \rightarrow \mathrm{Mn}^{2+}(aq) + 4 \mathrm{H}_{2} \mathrm{O}(l)\). The second half-reaction is an oxidation because electrons are lost: \( \mathrm{NO}_{3}^{-}(aq) + 4 \mathrm{H}^{+}(aq) + 3e^{-} \rightarrow \mathrm{NO}(g) + 2 \mathrm{H}_{2} \mathrm{O}(l)\).

In order to couple the two half-reactions, the number of electrons transferred must be the same in both. This can be achieved by finding the least common multiple of the number of electrons in the reduction and oxidation half-reactions. The least common multiple of 5 (reduction) and 3 (oxidation) is 15. Therefore, the reduction half-reaction has to be multiplied by 3 and the oxidation half-reaction by 5.

By adding up the two balanced half-reactions we obtain the balanced redox reaction. \[ 3(\mathrm{MnO}_{4}^{-}(aq) + 8 \mathrm{H}^{+}(aq) + 5e^{-} \rightarrow \mathrm{Mn}^{2+}(aq) + 4 \mathrm{H}_{2} \mathrm{O}(l)) + 5(\mathrm{NO}_{3}^{-}(aq) + 4 \mathrm{H}^{+}(aq) + 3e^{-} \rightarrow \mathrm{NO}(g) + 2 \mathrm{H}_{2} \mathrm{O}(l)) \] gives: \[ 3 \mathrm{MnO}_{4}^{-} + 24 \mathrm{H}^{+} + 15 \mathrm{e}^{-} + 5 \mathrm{NO}_{3}^{-} + 20 \mathrm{H}^{+} + 15 \mathrm{e}^{-} \rightarrow 3 \mathrm{Mn}^{2+} + 12 \mathrm{H}_{2}O + 5 \mathrm{NO} + 10 \mathrm{H}_{2}O \] The terms with electrons cancel each other out and the reaction simplifies to: \[ 3 \mathrm{MnO}_{4}^{-} + 5 \mathrm{NO}_{3}^{-} + 44 \mathrm{H}^{+} \rightarrow 3 \mathrm{Mn}^{2+} + 5 \mathrm{NO} + 22 \mathrm{H}_{2}O \].