在含有Cu(NO3)2、Zn(NO3)2、Fe(NO3)3、AgNO3各0.1 mol的混合溶液中加入0.1 m

在含有Cu(NO32、Zn(NO3)2、Fe(NO3)3、AgNO3各0.1 mol的混合溶液中加入0.1 mol铁粉,充分搅拌后,Fe溶解,溶液中不存在Fe3+离子,同时析出0.1 mol Ag。下列结论错误的是(    )

A.氧化性:Zn2+>Cu2+>Fe3+>Ag+                         B.Fe3+的氧化性大于Cu2+

C.溶液中Cu2+与Fe2+物质的量比为1∶2             D.1 mol Fe可还原2 mol Fe3+

答案

A


解析:

0.1 mol Fe全部溶解且无Fe3+,所以变化中共失去0.2 mol e-,又有0.1 mol Ag生成,其共得到0.1 mol e-,据得失电子守恒,则Fe3+→Fe2+,而发生了Fe+2Fe3+====3Fe2+,所以氧化性是Ag+>Fe3+>Cu2+>Zn2+,所以A错。B、C、D正确。

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