250· 思考题和习题解答 (4)Ag'(a=0.1)+Br(a=02)—>AgBr(s) 解:(1)负极:2Cr2(a=02)-→2Cr(a=0.1)+2 正极:I2(S)+2 电池:P(Cr2(a=0)Cr2(a=02)|r(a=01)|2(sP 0.353+0407)-005961g01×01 =l.019V (2)负极:Pb+SO;(ao →PbSO4+ 正极:2H(a1)+2e H2(0.1MPa) 电池:Pb,PbSO4(s)HSO4(a2=0.)H2OMPa)Pt RI RT PH/p =|0+03590 0.05916.1 =0.2703V (3)负极:H2(O.MPa)+2OH(a)-2H2O+2e 正极:Ag2O()+H2O+2e—2Ag+2OH(a) 电池:PH2OMPa)OH(a)4gOs)Ag In =(0.342+0.8277)V =1.170V (4)负极:Ag+Br(a=0.2)—→AgBr+e 正极:Ag(a=0.1)+e-)Ag 电池: Ag, AgBr(s)|Br(a=02)|2g(a=01Ag·250· 思考题和习题解答 (4) Ag ( = 0.1) + Br ( = 0.2) ⎯⎯→AgBr(s) + − a a 解：(1) 负极： + + − 2Cr ( = 0.2) ⎯⎯→2Cr ( = 0.1) + 2e 2 3 a a 正极：I (s) 2e 2I ( 0.1) 2 + ⎯⎯→ = − − a 电池：Pt Cr ( 0.1), Cr ( 0.2) I ( 0.1) I (s), Pt 2 3 2 = = = + + − a a a 2 Cr 2 I 2 o Cr 2 3 ln + + − = − a a a zF RT E E ( ) 1.019 V V 0.2 0.1 0.1 0.5353 0.407 0.05916 lg = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × = + − (2) 负极： − − Pb +SO ( − ) ⎯⎯→PbSO + 2e SO 4 2 4 2 4 a 正极：2H ( ) 2e H (0.1MPa) H + ⎯⎯→ 2 + − a + 电池：Pb, PbSO (s) H SO ( 0.1) H (0.1MPa), Pt 4 2 4 ± = 2 a ( ) 0.2703 V V 0.1 1 lg 2 0.05916 0 0.3590 / ln / ln 3 3 o o H SO 2 H o o H 2 2 4 2 = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = + − = − ⋅ = − + − ± a p p zF RT E a a p p zF RT E E (3) 负极： − − H (0.1MPa) + 2OH ( ) ⎯⎯→2H O + 2e 2 2 a 正极：Ag O(s) H O 2e 2Ag 2OH ( ) 2 2 a − − + + ⎯⎯→ + 电池：Pt, H (0.1MPa) OH ( ) Ag O(s), Ag 2 2 a − ( ) 1.170 V 0.342 0.8277 V 1 ln o H o 2 = = − = + zF p p RT E E (4) 负极： − − Ag + Br (a = 0.2) ⎯⎯→AgBr + e 正极：Ag ( = 0.1) + e ⎯⎯→Ag + − a 电池：Ag, AgBr(s) Br ( = 0.2) Ag ( = 0.1) Ag − + a a