内蒙古大学物理化学期末答案

1、写出下列浓差电池的电池反应，计算在 298 K 时的电动势：                        

Zn(s)│Zn2+(a=0.004)‖Zn2+(a=0.02)│Zn(s)                              

Pb(s)│PbSO4(s)│SO (a=0.02)‖SO (a=0.001)│PbSO4(s)│Pb(s)      

答案： (1) 电池反应   Zn2+(a=0.02) ─→Zn2+(a=0.004)                           

     E = -RT/2F×ln(0.004/0.02) = 0.0207 V                                

(2) 电池反应   SO (a=0.01) ─→ SO (a=0.001)                           

     E = -RT/2F×ln(0.001/0.01) = 0.0296 V                                

2、298 K时,下列电池的电动势为 E/V＝0.160235＋1.0023×10-3T/K－2.541×10-6(T/K)2 

Ag(s)+AgCl(s)│HCl(1.0 mol·kg-1, ＝0.809)│H2(p)│Pt         

求  (AgCl,Ag,Cl-)的值。                                                     

答案：解: 电池反应  Ag(s)＋HCl(1.0 mol·kg-1)─→AgCl＋H2(p)                     

E = 0.2333 V                                                          

E ＝E－RT/F×ln[1/(a+a-)]          (a+a-＝1×2)            

E＝ (AgCl,Ag,Cl-)＝0.2224 V                                          

3、在 p压力、18℃下，白锡与灰锡处于平衡。 从白锡到灰锡的相变热为-2.01 kJ·mol-1，请计算以下电池在 0℃和 25℃时的电动势。                        

   Sn(s,白)│SnCl2(aq)│Sn(s,灰)                                        

答案：电池反应：  Sn(s,白) ─→ Sn(s,灰)                                         

      291 K 达平衡    G = G= 0                                   

      S= (H- G)/T = -6.91 J·K-1·mol-1                            

     在 273 – 298 K 间视 H 为常数                                        

  273 K：G= H- TS= -124 J·mol-1                                 

         E= -G/zF = 0.000 V                                       

  298 K：G= 49 J·mol-1                                                 

         E= -0.00025 V                                               

4、一个原电池是由固态铝电极和固态的 Al- Zn 合金电极以及熔融的 AlCl3- NaCl混合物作电解质形成，当铝在合金电极中的物质的量分数是 0.38，电池电动势在653 K时为 7.43 mV 试计算 Al(s) 在 Al - Zn(s) 合金中的活度。                           

答案：电池： Ag(s)│AlCl3(在NaCl中)│Al(s)(在Zn中,x(Al)=0.38)                  

电池反应：  Al(s) Al(s)  [x(Al)=0.38]                                     

E = -RT/3F × lna(合金)/a(Al)     a(Al) = 0.673                                  

5、在 298 K 时有下述电池:                                                       

           Pb(s)│Pb2+(a=0.1)‖Ag+(a=0.1)│Ag(s)                      

(a) 写出电极反应并计算其电极电势                                          

(b) 计算电池的电动势和电池反应的 rGm                                 

已知     (Ag+,Ag) = 0.7991 V，   (Pb2+,Pb) = -0.126 V                        

答案： (1)  (-)  Pb(s) - 2ePb2+(a=1)                                     

     (+)  2Ag+(a=1) + 2e-2Ag(s)                                   

 (Ag+,Ag) =   (Ag+/Ag) - RT/2F × ln[1/a2(Ag+)] = 0.7991 V           

 (Pb2+,Pb) =   (Pb2+,Pb) - RT/2F × ln[1/a(Pb2+)] = -0.126 V                        

(2)  E =  (右) -  (左) = 0.7991 V - (-0.126 V) = 0.925 V                

rGm = -zEF = -178.5 kJ·mol-1                                               

* 如果电池反应的电子得失数为 1，E不变，而 rGm= -.26 kJ·mol-1           

6、298 K时，有如下两个反应：

A、  2Hg(l)+O2(g)+2H2O(l)=2Hg2++4OH –

B、    2Hg(l)+O2(水中溶解态)+2H2O(l)=2Hg2++4OH –

将反应A、设计成电池，其E = - 0.453 V。工业排放的废汞，可与水中溶解氧发生

如B、所示的反应，设废液呈中性，在液面上O2(g)分压为0.21×p ，活度系数均为1。

⑴ 写出反应A、的电池表示式。

⑵ 当废液中的[Hg2+]为1×10-5  mol·kg -1时，反应B、能否自发进行？

⑶ [Hg2+]在废液中浓度为多少时，不再与溶解O2反应？

答案：⑴ Hg(l)|Hg2+||OH -|O2(g),Pt                                                    

⑵，能自发进行。                    

⑶

    [Hg2+]=2.09×10-2 mol·kg-1         Hg2+浓度大于此值时不再反应。                                                                                                         

7、用电化学的方法计算2MnO4 -+10I -+16H+→5I2+2Mn2++8H2O反应的平衡常数。已知下列两电极的标准电极电势为：

    2MnO4 -+16H++10e -→2Mn2++8H2O        1.515 V

    5I2+10e -→10I -                        0.536 V

答案：设计电池为：Pt,I2(s)|I -||MnO4 -,Mn2+,H+|Pt                                    

E =1.515 V – 0.536 V=0.979 V                                                    

lnK==381.3；K=101                                                    

8、298 K 时，下述电池的电动势为 4.55×10-2 V , Ag + AgCl(s)│HCl(aq)│Hg2Cl2(s) + Hg(l)其温度系数 (E/Ｔ)p= 3.38×10-4 V·K-1。当有 1 mol电子电量产生时，求电池反应的rGm、rHm、rSm 值。                                     

答案： rGm= -zEF = -4.391 kJ·mol-1                                              

rSm= zF(E/T)p= 32.6 J·K-1·mol-1                                        

rHm= rGm+ TSm= 5.324 kJ·mol-1                                         

9、对半径为 R' 的球形液滴，在温度 T 时，其表面张力为  ，表面积为 A，试求 (/p)T, A  的表示式。                                                    

答案：ps= p - p0= 2 /R'                                                                 

(/p)T, A= R'/2                                                                 

10、已知某电池的 E＝0.058 V (298 K), 其电池反应为：                              

Ag(s)＋Cu2+(a=0.48)＋Br-(a=0.40)─→AgBr(s)＋Cu+(a=0.32)                

(1) 写出两电极上发生的反应                                                 

(2) 写出电池的书面表达式                                                   

(3) 计算电池的电动势                                                       

答案：解:(1)  (-) Ag(s)＋Br-(0.40)─→AgBr(s)＋e-                                   

    (+) Cu2+(0.48)＋e-─→Cu+(0.32)                                     

(2) Ag(s)│AgBr(s)│Br-(a=0.40)‖Cu+(a=0.32),Cu2+(a=0.48)│Pt              

(3) E＝E－(RT/F)ln[a(Cu+)/(a(Cu2+)×a(Br-))]＝0.045 V                      

11、下列电池在 298 K 时, 当溶液 pH= 3.98 时, E1=0.228 V； 当溶液为 pHx时,          

E2=0.3451V, 求 pHx为多少?                                                     

Sb│Sb2O3(s)┃H+(pH)‖KCl(饱和)│Hg2Cl2(s)│Hg(l)                           

答案：阳极反应为    2Sb＋3H2O─→Sb2O3(s)＋6H+＋6e-                            

-＝＋RT/(6F) ln a6(H+)＝－(0.05916pH)V                       

E1＝+－＋(0.05916pH) V＝0.228 V                                        

E2＝+－＋(0.05916pHx)V＝0.3451V                                     

E2－E1＝0.05916(pHx-pH) V                                                 

pHx＝(E2 - E1)/0.05916 V＋pH＝5.96                                          

12、电池  Pt│H2(p)│HBr(0.100 mol·kg-1)│AgBr(s)│Ag(s)   在298 K时, 测得E= 0.200 V。已知  (Ag│AgBr│Br-)＝0.071 V。写出电极反应和电池反应, 并求所指浓度下的(HBr) 。                                                             

答案：解:  H2(g, p) H+＋e-                                                  

AgBr(s)＋e- Ag(s)＋Br-                                                

───────────────                                                

H2(g，p)＋AgBr(s) Ag(s)＋H+＋Br-                                     

E＝E－RT/F×ln[aa/(p(H2)/p)1/2]                                

p(H2)＝p,         E＝0.071 V－0＝0.071V                                  

ln(a·a)＝(E－E)/(RT/F)＝－5.0244                                 

a·a＝(c/c·c/c)2(HBr)     ＝0.811                           

13、电池:  Pt│H2(p)│HCl(0.01 mol·kg-1)‖NaOH(0.01 mol·kg-1)│H2(p)│Pt在 298 K 时的电动势为 -0.587 V，求水的离子积。                                    

（0.01 mol·kg-1 水溶液中 ,HCl 和 NaOH 的平均活度系数为 0.904）                  

答案：电池反应   H+(a1) ─→ H+(a2)                                        

HCl 溶液中 H+ 的活度  a1= m+/m = m/m                                

E = -RT/F×ln(a1/a2)    a2 = 1.06×10-12                                      

已知 0.01 mol·kg-1 NaOH 溶液中  a(OH-) = m/m= 9.04×10-3                 

Kw = a(OH-) a2 = 0.96×10