d and f Block Elements: Tricky Questions Solved (1)

Question

Why does $\text{Cu}^+$ disproportionate in aqueous solution to give $\text{Cu}^{2+}$ and $\text{Cu}$, while $\text{Cu}^{2+}$ is stable?

Solution — Step by Step

Why This Works

Disproportionation is a redox reaction where the same species is both oxidised and reduced. The reaction proceeds spontaneously when the overall standard potential is positive. For $\text{Cu}^+$ in water, the energy released by hydrating $\text{Cu}^{2+}$ outweighs the energy needed to remove the second electron — so $\text{Cu}^+$ would rather become $\text{Cu}^{2+}$ (and a free Cu atom).

In non-aqueous environments, the picture flips. $\text{CuCl}$ (with $\text{Cu}^+$) is stable as a solid, and $\text{Cu}^+$ complexes like $[\text{Cu(NH}_3)_2]^+$ exist comfortably. Water is the destabiliser.

Alternative Method

Use the thermodynamic cycle. Hydration enthalpy of $\text{Cu}^{2+}$ ≈ $-2100\text{ kJ/mol}$. For $\text{Cu}^+$ ≈ $-580\text{ kJ/mol}$. The second ionisation energy of Cu is about $1958\text{ kJ/mol}$. Net for $\text{Cu}^+(\text{aq}) \to \text{Cu}^{2+}(\text{aq}) + e^-$:

Difference in hydration: $-2100 - (-580) = -1520\text{ kJ/mol}$ (more stable for +2). This more than offsets the IE2 of $1958\text{ kJ/mol}$ once we account for the energetics of the other half (reduction of $\text{Cu}^+$ to $\text{Cu}$).

Common Mistake

Confusing disproportionation with comproportionation (the reverse). Also, using $E°_{\text{cell}}$ with the wrong sign — always: $E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}}$ where cathode is the species being reduced.