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一、电子守恒法解题原理 氧化—还原反应的本质是电子得失或偏移,在同一个氧化—还原反应里,氧化剂得到电子的总数等于还原剂失去电子的总数,而电解或电镀实际上也是发生氧化—还原反应,因此,在同一时间内,阴极上物质得到电子的数目等于阳极上物质失去电子的数目,电子守恒法就是依据这种等量关系来解题的。 二、电子守恒法在解高考化学选择题中的应用 1.求元素的化合价 [例1](95年高考23题)24毫升浓度为0.05摩/升Na_2SO_3溶液恰好与20毫升浓度为0.02摩/升K_2Cr_2O_7溶液完全反应,则元素Cr在被还原的产物中的化合价是( )
First, the principle of solving the problem of the electronic conservation method The nature of the oxidation-reduction reaction is the gain or loss of electrons. In the same oxidation-reduction reaction, the total number of electrons obtained by the oxidant is equal to the total number of electrons lost by the reducing agent, and the electrolysis or plating is also actually An oxidation-reduction reaction occurs. Therefore, at the same time, the number of electrons on the cathode is equal to the number of electrons lost on the anode. The electron conservation method is based on this equal relationship. Second, the application of electronic conservation method in solving college entrance examination chemistry multiple choice questions 1. Find the valence of the element [Example 1] (95 college entrance examination in 23 questions) 24 ml concentration of 0.05 mol/l Na_2SO_3 solution happens to coincide with 20 ml concentration of 0.02 mol/ When the K 2 Cr 2 O 7 solution is completely reacted, the valence of element Cr in the reduced product is ()