Cu+2Ag+ → Cu2++2Ag;[Ag+]=0.001 (M),[Cu2+]=0.1 (M)
E=(E陰極-E陽極)-(RT/nF)*ln(K) =(0.8-0.34)-[(8.314*298)/(2*96500)]*ln(0.1/0.0012) =0.46-0.0128*ln(100000) =0.3122
25.一電池的反應為: Cu(s)+ 2Ag+ ..-阿摩線上測驗