用惰性电极电解一定浓度的硫酸铜溶液,通电一段时间后,向所得的

用惰性电极电解一定浓度的硫酸铜溶液,通电一段时间后,向所得的溶液中加入0.1 mol CuO后恰好恢复到电解前的浓度。则电解过程中转移的电子数为

A.0.1 mol               B.0.2 mol                  C.0.3 mol                 D.0.4 mol

答案

解析:电解生成的Cu和O2从溶液中沉积和挥发出来,且两者物质的量之比为2∶1的关系,相当于溶液中损失CuO,故存在以下关系:2CuSO4+2H2O2Cu+O2↑+2H2SO4— 2CuO—4e-,据此计算,可知溶液损失0.1 mol CuO时转移0.2 mol 电子。

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