For the disproportionation reaction 2Cu⁺(aq) ⇌ Cu(s) + Cu²⁺ at 298 K, ln K (where K is the equilibrium constant) is × 10⁻¹.
Given: (E⁰(Cu²⁺/Cu⁺) = 0.16V, E⁰(Cu⁺/Cu) = 0.52V, RT/F = 0.025) [numerical value].
For the disproportionation reaction 2Cu⁺(aq) ⇌ Cu(s) + Cu²⁺ at 298 K, ln K (where K is the equilibrium constant) is × 10⁻¹.
Given: (E⁰(Cu²⁺/Cu⁺) = 0.16V, E⁰(Cu⁺/Cu) = 0.52V, RT/F = 0.025) [numerical value].
Asked by Shiksha User
-
1 Answer
-
E°_cell = E° (cathode)-E° (anode) = 0.16-0.52=-0.36V.
lnK = nFE°/RT = 1*0.36/0.025 = 14.4. Ans is x10? ¹, so 144.
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else
Sign Up on ShikshaOn Shiksha, get access to
- 65k Colleges
- 1.2k Exams
- 682k Reviews
- 1800k Answers