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在高三物理课本第84节中,举了两个电解反应的例子,一个是盐酸溶液,另一个是硫酸铜溶液。这两种电解质的溶液都是属于酸性的。关于盐酸溶液的电解反应的分析: 在阴极 2H~++2e→H_2↑在阳极 2Cl~--2e→Cl_2↑是对的,但对硫酸铜溶液的电解反应的分析就有提出再讨论之点:原书中的反应如下: CuSO_4→Cu~(++)+SO_4~= 在阴极 Cu~(++)+2e→Cu 在阳极 SO_4~=-2e→SO_4 2SO_4+2H_2O→2H_2SO_4+O_2↑关于阴极Cu的析出,这种解释是对的;但是关于氧的放出,此种解释就与实际不符合。因为在CuSO溶液中,还有H_2O→H~++OH~-的反应存在,因此在阳极放出氧,可能是两种反应的结果:即除了SO_4~=按以下反应放电的初反应:
In Section 84 of the Year 3 physics textbook, two examples of electrolysis reactions are given, one for hydrochloric acid and the other for copper sulfate. Both electrolyte solutions are acidic. Analysis of the electrolysis reaction of hydrochloric acid solution: In the cathode 2H ~ + 2e → H_2 ↑ in the anode 2Cl ~ - 2e → Cl_2 ↑ is right, but the electrolysis reaction of copper sulfate solution has been proposed to discuss the point : The reaction in the original book is as follows: CuSO_4 → Cu ~ (++) + SO_4 ~ = At the cathode Cu ~ (++) + 2e → Cu at the anode SO_4 ~ = -2e → SO_4 2SO_4 + 2H_2O → 2H_2SO_4 + O_2 ↑ About Cathodic Cu precipitation, this explanation is correct; but on the release of oxygen, this explanation and the actual does not meet. Because of the presence of H 2 O → H ~ + OH ~ - in the CuSO solution, oxygen evolution at the anode may be the result of two reactions: the initial reaction with the exception of SO 4 - =