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= (－0.5873－0.05916)V= －0.6465 V

4. 解： (1)

(-) H2→2H++2e

(+)1/2O2+2H++2e→H2O(l) 电池反应：H2(g)+1/2O2(g)→H2O(l)

(2) ΔrGm=-nFE=-2×96500×1.228=-2.37×105 (J·mol-1)

根据 ΔrHm==-Nfe+nFT(?E/?T)p

-2.861×105=-2.37×105+2×96500×298×(?E/?T)p (?E/?T)p =-8.537×10-4 (V·K-1)

(3) 根据 ΔrHm=nF[E-T(?E/?T)p]； 得 E = 1.25(V) 5. 解：

负极：Ag + Cl- - e- → AgCl(s) 正极：Ag+ + e- → Ag

电池反应：Ag+ + Cl -→ AgCl(s) E=E?-RT/Fln[a(AgCl)/a(Ag+)a(Cl-)] ∵a(AgCl)=1；

∴E?=E-RT/Fln[a(Ag+)a(Cl-)] = E-RT/Fln(γ±m/m?)

=0.4321-(8.314×291/96500)ln(0.84×0.05)=0.5766 V lnK?=nFE?/RT=22.9985；故K?=9.73×109 AgCl的溶度积 Ksp=1/K?=1.03×10-10 6. 解: 通过1F电量时，z=1

电极反应: (-)Hg(l) + Br-(aq)→1/2Hg2Br2(s) + e- (+)AgBr(s) + e-→Ag(s) + Br-(aq) 电池反应: Hg(l)+ AgBr(s)→1/2 Hg2Br2(s)+ Ag(s) 25℃,100kPa时, E??68.04mV?6.804?10?2V

??rGm??zFE???1?96484.6?6.804?10?2J?mol?1??6.565kJ?mol?1

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??E??2?1则 ???0.312?10V?k，??T?p??E???rSm?zF???1?96484.6?0.312?10?3J?mol?1?K?1?30.103J?mol?1?K?1

??T?p????rHm??rGm?T?rSm?（?6565?298.15?30.103）J?mol?1?2410.21J?mol?1

若通电量为2F，则电池所做电功为：

W`?zFE?2?96484.6?6.084?10?2J?mol?1?13.13kJ?mol?1

7. 解：

??Kcell13.7?1?1?m???2.092?10?2S?m?1 R6552.092?10?2?m??S?m2?mol?1?1.32?10?3S?m2?mol?1

c15.81???m(HAc)???m(H?)???m(Ac?)?(349.82?40.9)?10?4S?m2?mol?1

?3.91?10?2S?m2?mol?1

?m1.32?10?3?2?????3.38?10 ?2?m3.91?10c215.81?2?2??(3.38?10)??1 KC?c??2?51??1?3.38?10?1.87?10

8、（Pt）H2（g）│HBr (a = 0.1) │AgBr-Ag ( 1 ) Eo = Ψo AgBr –Ψo H2/H+ = 0.0711 – 0 = 0.0711V

E = Eo – RT/nFln a2HBr = 0.0711 – 8.314×298/2×96500 ln0.14 = 0.1894V ( 2 ) ΔG = – nFE = – 2 × 96500 × 0.1894 = – 36.554 kJ (2分) ( 3 ) ΔGo = – nFEo = – RTlnKa Ka = 254 ( 4 ) ΔG < 0 能自发进行 ( 5 ) E1 = E2 = 0.1894V

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