復(fù)旦 物理化學(xué) 第五章 習(xí)題答案

1、第五章習(xí)題解答1(1) =QLQ=KR=0.1413112.3=15.87 m1(2) Sm1(3) Sm2mol12 Sm1 Sm2mol1=(349.82+40.9)104=390.72104 Sm2mol1=0.042313=(1.351+4.295-4.211)1022=2.87102 Sm2mol1molm3若以為單位，c=0.0214 molm34=0.032HA溶解濃度= c= 0.010.032=3.2104 molkg1 molkg1HAKClNa+SO425=0.015 molkg1+=0.5631+=0.8662或 =0.7504(1) 2Ag+ (a1) +H2(p)

2、= 2Ag + 2H+ (a2)負(fù)：H2(p) 2e = 2H+ (a2)正：2Ag+ (a1) +2e = 2Ag電池：Pt, H2(p) H+ (a2) Ag+ (a1)Ag(2) Sn + Pb2+(a1) = Sn2+ (a2) + Pb負(fù)：Sn2e = Sn2+ (a2)正：Pb2+(a1) +2e = Pb電池：SnSn2+ (a2) Pb2+ (a1)Pb(3) 負(fù)：正：AgCl +e = Ag + Cl(a)電池：Pt, H2(p) HCl (a) AgCl, Ag(4) Fe2+(a1) + Ag+(a3) = Fe3+(a2) + Ag負(fù)：Fe2+(a1)e = Fe3+(

3、a2)正：Ag+(a3) +e = Ag 電池：Pt, Fe2+(a1) , Fe3+(a2) Ag+(a3), Ag7 (1)負(fù)：Cd2e = Cd2+(a=0.01)正：Cl2(p) + 2e = 2Cl(a=0.5)電池反應(yīng)：Cd + Cl2(p) = Cd2+(a=0.01) + 2Cl(a=0.5)(2) VVE=+=1.8378 V8負(fù)：Pb2e = Pb2+(a=0.10)正：Cu2+(a=0.50) + 2e = Cu電池反應(yīng)：Pb + Cu2+(a=0.50) = Pb2+(a=0.10) + Cu=VG=nFE=2F0.4837=93.34 kJmol1Cu2+Cu為正極。

4、G=nFE=2F1.015 = 195.86 kJmol1=2F(4.92104)=94.94 JK1Hm=Gm+TSm=224.27 kJmol1Qr=TSm=28.31 kJ短路放電是熱效應(yīng)相當(dāng)于H。（T，p不變，W=0，Qp=H）H=40Qr即G+TS=40TSnFE=41TS=412981.4104=1.711 V11Sm=(96.2+77.4)(42.7+0.5195.6) = 33.1 JK1 VK1G=HTS=nFEV12絡(luò)合平衡反應(yīng)：Ag+ + 2NH3 = Ag(NH3)2+設(shè)計(jì)電池反應(yīng)：負(fù) Ag + 2NH3e = Ag(NH3)2+=0.373 V正 Ag+ + e =

5、Ag+=0.799 VE=0.426 V=1.5910713電池反應(yīng)負(fù)H2(p)2e = 2H+(aq)正HgO + H2O + 2e = Hg +2OHHgO + H2(p) = Hg + H2O(1)G1水生成反應(yīng)(2)G2HgO分解反應(yīng)=(1)-(2)：(3)G3G1=nFE=2F0.9265=178.8 kJG2=HTS =258.81298.15(70.80.52058.1130.67)103=237.38 kJG3=G1G2=98.58 kJ分解平衡常數(shù)=5.391011=2.941016 Pa14負(fù)：H2(p)2e = 2H+ (m=1.0)正：電池反應(yīng)=0.15715負(fù)極電位：Sb2O3 + 6H+ + 6e = 2Sb + 3H2O電動(dòng)勢(shì)E=甘= 甘(0.05915pH) = A + 0.05915pH(A=甘)ES = A + 0.05915pHS0.228=A+0.059153.98Ex = A + 0.05915pHx0.3459=A+0.5915pHxpHx=5.9616 PtH+ (a=1) Pt陰：2H+ + 2e = H2(p)陽(yáng)：E分解=E可逆+陰+陽(yáng)=1.229 + 0 + 0.487 = 1.716 V17電解時(shí)Pb陰極發(fā)生H+還原反應(yīng)電極反應(yīng)（H+濃度由一級(jí)電離決定H2SO4 H+ + HSO4）可逆電位不可逆電位不可逆