基础化学李保山第二版8氧化还原反应习题答案

O42-)2MnO4-+5C2O42-+8H+=2Mn2++10CO2+4H2On剩余(C2O42-)=2.5n(MnO4-)=1.645×10-3moln反应(C2O42-)=0.7049134mol-1.645×10-3mol=3.615×10-3moln(MnO2)=3.615×10-3molm(MnO2)=0.3145gw(MnO2)=0.31450.5261×100%=59.79%31.MnO4-+8H++5e=Mn2++4H2OCr2O72-+14H++6e=2Cr3++7H2O所以V=n(MnO4-)c(Cr2O72-)×56=27.78mol32.lgKθ=Z·E0.0592Z=lgKθ×0.0592E=-233.解答:(1)[Cl-]=1.47mol·L-1的电极为(anode)负极(2)standardemf(标准电动势):Eθ=0.000V(3)Ecell=0.0592lgCl大-Cl小-=0.102V(4)The[Cl-]whichinthecathodeofthecellwillincrease.Theotherwilldecrease.34.解答:(1)O2+4H3O++4e=6H2OE1θ=1.299VH2O2+2H3O++2e=4H2OE2θ=1.77VO2+2H3O++2e=H2O2+2H2OE3θ=?4E1θ=2E2θ+2E3θE3θ=0.69V(2)酸性(3)H2O2不能稳定存在,因为E右θ>E左θ,会发生歧化反应.35.解答:Ecell=Eθ(Ag+/Ag)-E(PbSO4/Pb)E(PbSO4/Pb)=0.220-0.546=-0.326VE(PbSO4/Pb)=Eθ(Pb2+/Pb)+0.05922lgKspθC(SO42-)-0.326=-0.13+0.05922lgKspθ0.05,得Kspθ(PbSO4)=1.19×10-8