Question

如果能给这题的详细答案，感激不尽 大一，无机化学

Answer 1

[Ag(NH3)2]+ + I- = 2NH3 + AgI (1) K0=K0f/K0sp
Ag+ + 2NH3 = [Ag(NH3)2]+ (2) K0f={[Ag(NH3)2]+}/{[Ag+][NH3]²}
AgI = Ag+ + I- (3) K0sp=[Ag+][I-]
Ag+ + e = Ag (4) E0(Ag+/Ag)=0.7996V
[Ag(NH3)2]+ + e = Ag + 2NH3 (5) E0{[Ag(NH3)2]+/Ag}=0.3720V

①反应式(1)可由[(5)-(4)-(3)]而得，故
E01=0.059lgK0
=E0{[Ag(NH3)2]+/Ag}-E0(Ag+/Ag)-{0.059lgK0sp}
=0.3720-0.7996-{0.059lg(8.3*10^-17)}
=0.3720-0.7996+0.9218
=0.4942
=> lgK0=0.4942/0.059=8.38
=> K0=10^8.38=2.38*10^8
此即为反应式(1)的标准平衡常数
②反应式(2)可由[(4)-(5)]而得，故
E02=0.059lgK0f
=E0(Ag+/Ag)-E0{[Ag(NH3)2]+/Ag}
=0.7996-0.3720
=0.4276
=> lgK0f=0.4276/0.059=7.25
=> K0f=10^7.25=1.78*10^7
此即为反应式(2)的稳定常数

原电池反应的电池符号如下：
(-) Ag|[Ag(NH3)2]+(1mol/L),NH3(1mol/L) || I-(1mol/L)|Agl|Ag (+)

追问

谢谢

Answer 2

三年前问我还我还是打出来的现在问我真的，无能为力呀！

追答

建议去问一下同学吧！