Question

已知电池反应：Pb（s） + 2 Ag + （c1=0.1mol/L）= Pb 2+ （c2=0.01mol/L）+ 2

Answer 1

E(Ag+/Ag)=0.7996+0.05916log0.1

E(Pb2+/Pb)=-0.1263+0.02958log0.01

电池的电动势E=E(Ag+/Ag)-E(Pb2+/Pb)

Pb/Pb2+(0.01mol/L)//Ag+(0.1mol/L)/Ag