Question

题:298.15k时,反应Ag^+ +Fe^2+ ======Ag+Fe^3+,已知δ^θ(Ag^+/Ag)=0.08v

Answer 1

1.E^θ=RTlnK/zF所以K=e^(zFE^θ/RT)=e^(1*96485*(0.08-0.77)/8.314/298.15)=2.17E-12

2.Ag(s)|Fe3+(aq)||Fe2+(aq),Ag+(aq)|Pt(s)

3.ΔrGm^θ=-zFE^θ=-1*96485*(0.08-0.77)=66.57KJ/mol