Question

For the disproportionation reaction
\(\mathrm{2Cu^+(aq) \rightleftharpoons Cu(s) + Cu^{2+}(aq)}\) at 298 K. \(\ln K\) (where K is the equilibrium constant) is ___________ \(\times 10^{-1}\).
Given : ( \(E^0_{\mathrm{Cu^{2+}/Cu^+}} = 0.16\) V, \(E^0_{\mathrm{Cu^+/Cu}} = 0.52\) V, \(\dfrac{RT}{F} = 0.025\) )

Hint:

Cathode: Cu⁺/Cu (0.52 V); anode: Cu²⁺/Cu⁺ (0.16 V). E⁰_cell = 0.36 V, n = 1, and ln K = nE⁰F/RT.

Answer 1

Correct Answer: 144

Reduction half: Cu⁺ + e^- → Cu, E⁰ = 0.52 V.
Oxidation half: Cu⁺ → Cu²⁺ + e^-, with E⁰(Cu²⁺/Cu⁺) = 0.16 V.
E⁰_cell = 0.52 - 0.16 = 0.36 V, with n = 1 electron for the reaction as written.
At equilibrium, E⁰_cell = (RT/nF) ln K, so ln K = nE⁰_cell/(RT/F) = 0.36/0.025 = 14.4 = 144 × 10^-1.
Answer: 144