Ag(NH3)2++e=Ag+2NH3,计算[Ag(NH3)2+]=0.01mol .L-1,[NH3]=0.1mol .L-1时_其他解答

Ag(NH3)2++e=Ag+2NH3,计算[Ag(NH3)2+]=0.01mol .L-1,[NH3]=0.1mol .L-1时的电极电位

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解答

Ag+ + 2NH3 = Ag(NH3)+K稳 = [Ag(NH3)+] / [Ag+]·[NH3]2[Ag+] = [Ag(NH3)+] / K稳·[NH3]2 查表得 K稳 = 1.1×10^7,代入上式得：[Ag+] = 0.01 / 1.1×10^7 × (0.1)2 = 9.1×10^-8φ[Ag(NH3)2+/Ag] = φθ[Ag+/Ag] ...