E（Cl2/Cl-）=1.36EA（HClO/Cl2）=1.63 Ka(HClO)=2.8*10-8.

题目

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!

答案

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)