问题补充:
单选题电解100mL含c(H+)=0.30mol/L的下列溶液。当电路中通过0.04mol电子时,理论上析出金属质量最大的是A.0.10 mol/L Ag+B.0.20 mol/L Zn2+C.0.20 mol/L Cu2+D.0.20 mol/L Pb2+
答案:
C解析由题可知各项中含有的离子的物质的量分别为:A项0.01 mol Ag+,0.03 mol H+;B项0.02 mol Zn2+,0.03 mol H+;C项0.02 mol Cu2+,0.03 mol H+;D项0.02 mol Pb2+,0.03 mol H+。根据金属活动性顺序,可知只有H后金属才能析出,故当电路中通过0.04 mol电子时,A项析出0.01 mol Ag(1.08 g),C项析出0.02 mol Cu(1.28 g),其他项无金属析出,故选C项。