电池Ag(s)|Ag+(a1=0.090)|| Ag+(a2=0.072)|Ag(s)在25 ℃时的电动势E=

Question

如题所述

Answer 1

正极：φ2(Ag+/Ag)= φθ(Ag+/Ag) + 0.0592 lg C2(Ag+) = φθ(Ag+/Ag) + 0.0592 lg a2
负极：φ1(Ag+/Ag)= φθ(Ag+/Ag) + 0.0592 lg C1(Ag+) = φθ(Ag+/Ag) + 0.0592 lg a1
E = φ2(Ag+/Ag) - φ1(Ag+/Ag)= 0.0592 lg a2 - 0.0592 lg a1 = 0.0592 lg 0.072/0.090
= - 0.00574 (V)

Answer 2

ag