试题与答案

用惰性电极电解CuSO4溶液一段时间后，停止电解，向所得溶液中加入0.1 mol

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题型：选择题

题目：

用惰性电极电解CuSO4溶液一段时间后，停止电解，向所得溶液中加入0.1 mol Cu(OH)2，溶液浓度恢复至电解前。关于该电解过程的说法不正确的是

A．生成Cu的物质的量是0.1 mol

B．转移电子的物质的量是0.2 mol

C．随着电解的进行溶液的pH减小

D．阳极反应式是4OH-－4e-＝2H2O＋O2↑

答案：

答案：B

题目分析：加入0.1 mol Cu(OH)2，溶液浓度恢复至电解前，说明电解CuSO4溶液后，继续电解H2SO4溶液。A、根据Cu元素守恒，生成Cu的物质的量是0.1 mol，正确；B、加入0.1 mol Cu(OH)2，溶液浓度恢复至电解前，说明电解了0.1mol的CuSO4和0.1mol的H2O，转移电子的物质的量为0.4mol，错误；C、电解CuSO4溶液时生成H2SO4，电解H2SO4溶液，实质电解水，H2SO4浓度增大，pH减小，正确；D、电解CuSO4溶液和H2SO4溶液，阳极反应都是：4OH-－4e-＝2H2O＋O2↑，正确。

考点：电解池原理电解池电极反应式的书写
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