Q2(c). Equilibrium constant for disproportionation of Cu⁺

LINKln K=nFE°/RTn1 electronRESULTK=1.23×10⁶

1. Split into half-reactions. Reduction (cathode): Cu⁺ + e⁻ → Cu, E°=+0.52 V. Oxidation (anode): Cu⁺ → Cu²+ + e⁻, the reverse of the couple with E°=+0.16 V.
2. Sum: 2Cu+(aq) → Cu(s) + Cu²+(aq), one electron transferred, n=1.
3. E°_(cell)=E°_(cathode)-E°_(anode)=0.52-0.16=+0.36 V
4. Gibbs energy: Δ G°=-nFE°=-(1)(96500)(0.36)=-3.474×10⁴ J mol⁻¹=-34.7 kJ mol⁻¹
5. Equilibrium constant: ln K=(nFE°)/(RT)=((1)(96500)(0.36))/((8.314)(298))=14.02; log K=6.09 K=[Cu²+][Cu⁺]²=1.23×10⁶
6. Meaning. K≫1: at equilibrium almost all Cu⁺ has become Cu and Cu²+. In 0.1 M Cu²+, [Cu⁺]=0.1/1.23×10⁶=2.9×10⁻⁴ M only. On the Frost diagram Cu⁺ lies above the line from Cu to Cu²+.
7. Check: the two couples give E°(Cu²+/Cu)=(0.52+0.16)/2=+0.34 V, the familiar value.
8. Why Cu⁺ is unstable in water although it has a filled 3d¹⁰ shell: the small, doubly charged Cu²+ has a much larger hydration enthalpy, which outweighs the second ionisation energy.

Application. Cu(I) survives only where it is stabilised by precipitation or complexation: CuI, CuCN and Cu₂O are insoluble, and [CuCl₂]⁻ and [Cu(NH₃)₂]⁺ are stable complexes; this is used in iodometric estimation of copper (2Cu²+ + 4I⁻ → 2CuI + I₂).