236.已知Eø( )=0.799V,Ag2C2O4的溶度积常数为3.5×10-11。求Ag2C2O4+2e-=2Ag+C2O42-Eø(Ag2C2O4/Ag)为()

解析:根据标准电极电势的定义,Eø(Ag2C2O4/Ag)=Eø(Ag2C2O4/Ag+)-Eø(Ag/Ag+)。将已知数据代入可得:Eø(Ag2C2O4/Ag)=Eø(Ag2C2O4/Ag+)-Eø(Ag/Ag+)=Eø(Ag2C2O4/Ag+)-0.799V。由于Ag2C2O4+2e-=2Ag+C2O42-,根据标准电极电势的定义,Eø(Ag2C2O4/Ag+)=Eø(Ag/Ag+)+E(Ag2C2O4+2e-=2Ag+C2O42-)。将已知数据代入可得:Eø(Ag2C2O4/Ag+)=Eø(Ag/Ag+)+E(Ag2C2O4+2e-=2Ag+C2O42-)=Eø(Ag/Ag+)+0.49V。将上述两个式子联立,可得:Eø(Ag2C2O4/Ag)=Eø(Ag/Ag+)+0.49V-0.799V=Eø(Ag/Ag+)-0.309V。又因为Ag++e-=Ag,所以Eø(Ag/Ag+)=-Eø(Ag+/Ag)=-0.799V。代入可得:Eø(Ag2C2O4/Ag)=-0.799V-0.309V=-1.108V。根据电化学反应的方向性,Eø(Ag2C2O4/Ag)应该为正值,因此需要将其取反,即Eø(Ag2C2O4/Ag)=1.108V。因此,选项A为正确答案。