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第五章习题解答1.1=QLQ=KR=
0.1413
112.3=
15.87m–12Sm–13Sm2mol–12.Sm–1Sm2mol–1=
349.82+
40.910–4=
390.7210–4Sm2mol–1=
0.042313.=
1.351+
4.295-
4.21110–22=
2.8710–2Sm2mol–1molm–3若以为单位,c=
0.0214molm–34.=
0.032HA溶解浓度=c=
0.01
0.032=
3.210–4molkg–1molkg–1HAKClNa+SO42–5.=
0.015molkg–1+=
0.5631+=
0.8662或=
0.75046.12Ag+a1+H2p=2Ag+2H+a2负H2p–2e=2H+a2正2Ag+a1+2e=2Ag电池PtH2pH+a2Ag+a1Ag2Sn+Pb2+a1=Sn2+a2+Pb负Sn–2e=Sn2+a2正Pb2+a1+2e=Pb电池SnSn2+a2Pb2+a1Pb3负正AgCl+e=Ag+Cl–a电池PtH2pHClaAgClAg4Fe2+a1+Ag+a3=Fe3+a2+Ag负Fe2+a1–e=Fe3+a2正Ag+a3+e=Ag电池PtFe2+a1Fe3+a2Ag+a3Ag7.1负Cd–2e=Cd2+a=
0.01正Cl2p+2e=2Cl–a=
0.5电池反应Cd+Cl2p=Cd2+a=
0.01+2Cl–a=
0.52VVE=+––=
1.8378V8.负Pb–2e=Pb2+a=
0.10正Cu2+a=
0.50+2e=Cu电池反应Pb+Cu2+a=
0.50=Pb2+a=
0.10+Cu=VG=–nFE=–2F
0.4837=–
93.34kJmol–1Cu2+Cu为正极9.G=–nFE=–2F
1.015=–
195.86kJmol–1=2F–
4.9210–4=–
94.94JK–1Hm=Gm+TSm=–
224.27kJmol–1Qr=TSm=–
28.31kJ10.短路放电是热效应相当于H(T,p不变,W’=0,Qp=H)H=–40Qr即G+TS=–40TS–nFE=–41TS=41298
1.410–4=
1.711V11.Sm=
96.2+
77.4–
42.7+
0.5
195.6=
33.1JK–1VK–1G=H–TS=–nFEV12.络合平衡反应Ag++2NH3=AgNH32+设计电池反应负Ag+2NH3–e=AgNH32+–=
0.373V正Ag++e=Ag+=
0.799VE=
0.426V=
1.5910713.电池反应负H2p–2e=2H+aq正HgO+H2O+2e=Hg+2OH–HgO+H2p=Hg+H2O1G1水生成反应2G2HgO分解反应=1-23G3G1=–nFE=–2F
0.9265=–
178.8kJG2=H–TS=–
258.81–
298.15
70.8–
0.5
2058.1–
130.6710–3=–
237.38kJG3=G1–G2=
98.58kJ分解平衡常数=
5.3910–11=
2.9410–16Pa14.负H2p–2e=2H+m=
1.0正电池反应=
0.15715.负极电位Sb2O3+6H++6e=2Sb+3H2O电动势E=甘––=甘–––
0.05915pH=A+
0.05915pHA=甘––ES=A+
0.05915pHS
0.228=A+
0.05915
3.98Ex=A+
0.05915pHx
0.3459=A+
0.5915pHxpHx=
5.9616.PtH+a=1Pt阴2H++2e=H2p阳E分解=E可逆+阴+阳=
1.229+0+
0.487=
1.716V17.电解时Pb阴极发生H+还原反应电极反应(H+浓度由一级电离决定H2SO4H++HSO4–)可逆电位不可逆电位不可逆=可逆–测量时,甘汞电极电位较大,Pb阴极电位教低,组成原电池应是PbH2SO
40.10molkg–1=
0.265甘汞求算E=甘–不可逆=甘–可逆–=
0.6950V。