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E(Cl2/Cl-)=1.36EA(HClO/Cl2)=1.63 Ka(HClO)=2.8*10-8.
算EB(ClO/Cl2)和25度下氯气在碱液中歧化标准平衡常数(所给电势均为标准电势)拜托各位了,急用呀要考了!
人气:360 ℃ 时间:2020-07-01 07:42:02
解答
E = EB(HClO/Cl2)+(0.05916/2)log([HClO]^2/[Cl2])
=EB(HClO/Cl2)+(0.05916/2)log([H+]^2[ClO-]^2/Ka^2[Cl2])
therefore, EB(ClO-/Cl2)=EB(HClO/Cl2)-0.05916)logKa
In basic solution,
E=EB(ClO-/Cl2)+(0.05916/2)log(Kw^2[ClO-]^2/[OH-]^2[Cl2])
EB(ClO-/Cl2, in basic solution)=B(ClO-/Cl2)+(0.05916/2)logKw^2
=(RT/nF)lnK
Therefore K=exp{(nF/RT)(EB(ClO-/Cl2, in basic solution)
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