计算原电池电动势Hg│HgY2-(4.5010-5mol/L),Y4-(0.333mol/L)║SCE已知:lgKHgY2-=21.8,φθHg2+/Hg=0.854V,φSCE=0.241V求详细过程

Question

计算原电池电动势Hg│HgY2-(4.5010-5mol/L),Y4-(0.333mol/L)║SCE
已知:lgKHgY2-=21.8,φθHg2+/Hg=0.854V, φSCE=0.241V
求详细过程

Answer 1

K(HgY)=10^21.8=6.3x10^21
c(Hg2+)=4.5x10^-5/(0.333•K(HgY))
=2.15x10^-26 mol/L
据Nernst公式
E(HgY/Hg2+)=0.854 + (RT/2F)ln(c(Hg2+))
= 0.854 - 0.760 = 0.094v
电池电动势为
0.241-0.094=0.147v
说明：10^21表示10的21次方，10^-26表示10的-26次方