What is the cell voltage of the following reaction and is the reaction spontaneous? Ba + Pb2+ --> Ba2+ + Pb?
回答 (1)
2020-05-15 8:28 pm
Refer to: http://www2.ucdsb.on.ca/tiss/stretton/database/Standard_Reduction_Potentials.htm
Ba²⁺(aq) + 2e⁻ → Ba(s) E° = -2.90 V
Pb²⁺(aq) + 2e⁻ → Pb(s) E° = -0.13 V
Ba(s) + Pb²⁺(aq) → Ba²⁺(aq) + Pb(s) E°cell
E°cell = E°(red) - E°(oxid) = [(-0.13) - (2.90)] V = 2.77 V
As E°cell = 2.77 V > 0, the reaction is spontaneous.
Ba²⁺(aq) + 2e⁻ → Ba(s) E° = -2.90 V
Pb²⁺(aq) + 2e⁻ → Pb(s) E° = -0.13 V
Ba(s) + Pb²⁺(aq) → Ba²⁺(aq) + Pb(s) E°cell
E°cell = E°(red) - E°(oxid) = [(-0.13) - (2.90)] V = 2.77 V
As E°cell = 2.77 V > 0, the reaction is spontaneous.
收錄日期: 2021-04-30 18:01:52
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200515081502AAHQVEC