Disproportionation constant from a Latimer diagram

Worked example

Using E°(Cu²⁺/Cu⁺) = 0.157 V and E°(Cu⁺/Cu) = 0.525 V, find K at 298 K for the disproportionation of aqueous copper(I).

1. 1. Cell potential from half-cells: solve for E°cell = E°(Cu⁺/Cu) − E°(Cu²⁺/Cu⁺)

   $E^\circ_\mathrm{cell} = E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode}$

   = 0.3680 V

2. 2. Gibbs energy and cell potential: solve for ΔG°

   $\Delta G^\circ = -nFE^\circ$

   = −35.51 kJ mol⁻¹

3. 3. Standard Gibbs energy and K: solve for K

   $K = \exp\!\left(-\dfrac{\Delta G^\circ}{RT}\right)$

   = 1.673 × 10⁶

Answer: K = 1.67 × 10⁶

Method

Cu⁺ is both reduced (to Cu) and oxidised (to Cu²⁺). E°cell = E°(Cu⁺/Cu) − E°(Cu²⁺/Cu⁺); one electron is transferred. K = exp(nFE°/RT) is about 10⁶, which is why Cu⁺ does not survive in water unless stabilised by complexation or precipitation.

Common mistakes

• In 2Cu⁺ → Cu²⁺ + Cu, Cu⁺ is reduced to Cu (cathode) and oxidised to Cu²⁺ (anode): E°cell = E°(Cu⁺/Cu) − E°(Cu²⁺/Cu⁺).

Formulas used

• Cell potential from half-cells

  $E^\circ_\mathrm{cell} = E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode}$

• Gibbs energy and cell potential

  $\Delta G^\circ = -nFE^\circ$

• Standard Gibbs energy and K

  $\Delta G^\circ = -RT\ln K$

More electrochemistry questions

• ΔG° from E°
• Nernst equation
• Mass deposited by electrolysis
• E°cell from reduction potentials
• Equilibrium constant from E°cell
• Electrode potential away from standard conditions
• Latimer diagram: skipping an oxidation state
• Daniell cell away from standard conditions