Zm(s)→Zn+2e AgC1(s)+e→Ag(s)+C1 电池反应：Zn(s)+2AgC1(s)→2Ag(s)+ZnCl,(1) (2)=exp(zFP/RD=exp[2×96500×(0.2220+0.7630)/8.315/298.2]= 1.983×10 (3)=®-(0.05916/2)1g[43(b/P)] ,而， 代入1.015=(0.2220+0.7630)-(0.05916/2)1g(4×3×0.555) 4=0.521 (4)可逆电池Q=zF7(aaD=2×96500×298.2×(-4.02×10) J·mol4 =-23136J·mo1 非电池反应： Q=△.H=△.G+△5=△，G+4=-zFAQ=[-2×96500×1.015+(-23136)] J·mol4 =-219031J·mo1 6、解：(1)Γ=-(c/RD)di/dc =-(c/RD[0.4(c/mol·dm)-0.5]×103N·m/(ol·dm =-[0.6/(8.315×298.2)](0.4×0.6-0.5)×103mo1·m2 =6.29×10-mol·m2 注意单位的处理！ (2)F2pgr=2×(72-0.5×0.6+0.2×0.6)×10/(986×9.8×10m) =0.0149m Zn(s) → Zn2+ + 2e— AgCl(s) + e—→ Ag(s) + Cl— 电池反应： Zn(s) + 2AgCl(s) →2 Ag(s) + ZnCl2(l) (2) K  = exp(zFE /RT) = exp[2×96500×(0.2220+0.7630)/8.315/298.2]= 1.983×1033 (3) ＝E －(0.05916/2)lg[4 3 (b/b ) 3 ] ，而 ， 代入 1.015=(0.2220+0.7630)－(0.05916/2)lg(4× 3×0.5553 ) ±=0.521 (4) 可逆电池 Qr= zFT( E/T)p= 2×96500×298.2×(－4.02×10-4 ) J·mol-1 = -23136 J·mol-1 非电池反应： Qp= rH = rG＋TrS = rG＋Qr = －zFE+ Qr= [-2×96500×1.015+(-23136)] J·mol-1 = -219031 J·mol-1 6、解：(1)  = －(c/RT)d/dc = －(c/RT)[0.4(c/ mol·dm-3 )－0.5]×10-3 N·m -1 / (mol·dm-3 ) =－[0.6/(8.315×298.2)](0.4×0.6－0.5) ×10-3 mol·m －2 = 6.29×10－8 mol·m －2 注意单位的处理！ (2) h=2/gr = 2×(72－0.5×0.6+0.2×0.62 ) ×10-3 /(986×9.8×10-3 m) =0.0149m