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桑树之泪

金虫 (著名写手)

[求助] 电化学 已有1人参与

2Hg+O2+2H2O→2Hg2++4OH-已知25℃φHg2+/Hg=0.851V,φH+,H2O/O2=1.229V,求在含氧量20%的空气下,中性溶液中述反应中达到平衡时的Hg2+的活度
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iiok2

银虫 (小有名气)

【答案】应助回帖

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感谢参与,应助指数 +1
桑树之泪: 金币+20, ★★★★★最佳答案, 谢谢 2013-12-31 23:07:08
先写出电池的正负极反应:负极 2Hg-4e——2Hg2+;正极O2+2H2O +4e——4OH-;总反应:2Hg+O2+2H2O→2Hg2++4OH-
电池的标准电动势=正极的标准电势-负极的标准电势=1.229-0.851=0.378v,由于中性溶液,故H+和OH-的浓度等于10-7,根据能斯特方程得出E=RT/ZFln([OH-]4*[Hg2+]*2/Po2/P0)

Po2=20%P0
代入数据可求得
2楼2013-12-31 00:56:43
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