 Solution

Read the potential labels rather than the alphabetical order: b, the less negative event, is $\mathrm{Q}+e^{-}\rightleftharpoons Q^{\bullet -}$; a, the more negative event, is ${\mathrm{Q}}^{\bullet -}+e^{-}\rightleftharpoons Q^{2-}$. These are the first and second reductions respectively, with the reverse branches giving the corresponding oxidations. The second electron is harder to add to an already negatively charged species.