已知:将SO2通入FeCl3溶液中,溶液颜色会变为浅绿色,其原理可表示为:(1)请配平上述离子方程式;(2)

2025-03-16 09:18:34
推荐回答(1个)
回答1:

(1)Fe3+→Fe2+,化合价从+3→+2,降低1价,SO2→SO42-,化合价从+4→+6,升高2价,化合价升高和降低总数相等,
故铁元素前面系数为2,硫元素前面系数为1,则2Fe3++1SO2+H2O═2Fe2++1SO42-+H+,再根据氧原子守恒可知H2O前面系数为2,H+前系数为4,
故答案为:2;1;2;2;1;4;
(2)由反应可知,1molSO2反应转移2mol电子,则转移电子的物质的量为0.1mol,则参加反应的SO2在标准状况下的体积为0.1mol×

1
2
×22.4L/mol=1.12L,
故答案为:1.12L;
(3)浅绿色为亚铁离子颜色,黄色为三价铁离子颜色,溶液由浅绿色变为黄色,说明亚铁离子被氧化为铁离子,离子反应为2Fe2++Cl2=2Fe3++2Cl-
故答案为:2Fe2++Cl2=2Fe3++2Cl-
(4)①假设1:Fe2+还原性强于I-;假设2:I-还原性强于Fe2+,故答案为:I-还原性强于Fe2+
②步骤1:向FeSO4溶液中滴加1~2滴氯水,溶液由浅绿色变成黄色,向KI溶液中滴加1~2滴氯水,溶液由无色变成黄色,因此取2mL FeSO4溶液和2mL KI溶液混合于试管中,再滴加1~2滴氯水的现象为溶液变为黄色;
步骤2:可以用淀粉溶液检验碘单质的存在,溶液变蓝,证明I-的还原性强于Fe2+,也可以用20%KSCN溶液检验铁离子的存在,向试管中继续滴加几滴KSCN溶液,若溶液不变红,也说明I-的还原性强于Fe2+
故答案为:
实验步骤 预期现象和结论
步骤1:取2mLFeSO4溶液和2mLKI溶液混合于试管中,再滴加1~2滴氯水 溶液变成黄色.结论:可能是生成了Fe3+、也可能是生成了I2
步骤2:向试管中继续滴加几滴20%KSCN溶液或 溶液不变红,说明I-的还原性强于Fe2+
溶液变蓝,说明I-的还原性强于Fe2+

③若实验结果证明I-还原性强于Fe2+,则向含有1mol FeI2的水溶液中通入1.5mol Cl2充分反应,由电子守恒可知碘离子、亚铁离子均全被氧化,离子反应为2Fe2++4I-+3Cl2=6Cl-+2Fe3++2I2,故答案为:2Fe2++4I-+3Cl2=6Cl-+2Fe3++2I2

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