思考题和习题解答 △2Gm=△2G=△,H-T△,S =(5.54×103-29815×334)Jmol 4.42×103 V=0.0458V 1×96485 OE° =0.346×10-3V.K 3.写出下列电池的电极反应和电池反应: (1)PL, H2(. MPa) HCI(O. I mol-kg")Cl2(.1MPa), Pt (2)Pt, H2 (0.1 MPa)KOH (O. I mol-kg-b) o2(0.1MPa),Pt (3)Zn ZnCl, (0. 1 mol-kg-)Hg,Cl,, Hg (4)Zn, ZnO(s) KOH (0.5 mol.")HgO(s),Hg 解:(1)负极:H2(O.MPa)->2H.1 mol-kg-)+2 正极:Cl2O.MPa)+2e-→2C-(0. I mol kg-) 电池:H2O.1MPa)+Cl2(0.1 2H(0. 1 molkg )+2CI"(0. 1 molkg) (2)负极:2H2(.MPa)+4OH(01 mol kg)-4H2O+4e 正极:O2(0.MPa)+2H2O+4e-4OH(0.1 mol-kg-2) 电池:2H2(O.1MPa)+O2(0MPa)-→2H2O (3)负极 Zn(. I molkg 正极:Hg2Cl2+2e→2Hg+2Cl-(02 mol- kg-) 电池:Zn+Hg2Cl2 Hg+Zn(0. 1 molkg ")+2CI(0.2 molkg") (4)负极:Zn+2OH(0.5 molkg-)-ZnO+H2O+2 正极:HgO+HO+2 →Hg+2OH(0.5 mol- kg-) 电池:Zn+H Hg+ zno 4.计算电池 Ag,AgBr(s)Br(a=0.34)Fe"(a=0.1), Fe*(a=0.02)Pt·248· 思考题和习题解答 ( ) 1 3 1 o r m o r m o r m r m 4.42 kJ mol 5.54 10 298.15 33.4 J mol − − = − ⋅ = × − × ⋅ Δ G = Δ G = Δ H −TΔ S V 0.0458 V 1 96485 4.42 103 r m = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × − × = − Δ = − zF G E 3 1 1 o r m o 0.346 10 V K V K 1 96485 33.4 − − − = × ⋅ ⎟ ⋅ ⎠ ⎞ ⎜ ⎝ ⎛ × = Δ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ∂ ∂ ⎟ = ⎠ ⎞ ⎜ ⎝ ⎛ ∂ ∂ zF S T E T E p p 3. 写出下列电池的电极反应和电池反应： (1) Pt, H (0.1MPa) HCl(0.1mol kg ) Cl (0.1MPa), Pt 2 1 2 − ⋅ (2) Pt, H (0.1MPa) KOH (0.1mol kg ) O (0.1MPa), Pt 2 1 2 − ⋅ (3) Zn ZnCl (0.1mol kg ) Hg Cl , Hg 2 2 1 2 − ⋅ (4) Zn, ZnO(s) KOH (0.5 mol kg ) HgO(s), Hg −1 ⋅ 解：(1) 负极： + − − H (0.1MPa) ⎯⎯→2H (0.1mol⋅kg ) + 2e 1 2 正极：Cl (0.1MPa) 2e 2Cl (0.1mol kg ) 1 2 − − − + ⎯⎯→ ⋅ 电池： 2H (0.1mol kg ) 2Cl (0.1mol kg ) H (0.1MPa) Cl (0.1MPa) 1 1 2 2 + − − − ⋅ + ⋅ + ⎯⎯→ (2) 负极： − − − 2H (0.1MPa) + 4OH (0.1mol⋅kg ) ⎯⎯→4H O + 4e 2 1 2 正极：O (0.1MPa) 2H O 4e 4OH (0.1mol kg ) 1 2 2 − − − + + ⎯⎯→ ⋅ 电池：2H2 (0.1MPa) + O2 (0.1MPa) ⎯⎯→2H2O (3) 负极： + − − Zn ⎯⎯→ Zn (0.1mol⋅kg ) + 2e 2 1 正极：Hg Cl 2e 2Hg 2Cl (0.2 mol kg ) 1 2 2 − − − + ⎯⎯→ + ⋅ 电池： 2Hg Zn (0.1mol kg ) 2Cl (0.2 mol kg ) Zn Hg Cl 2 1 1 2 2 + − − − + ⋅ + ⋅ + ⎯⎯→ (4) 负极： − − − Zn + 2OH (0.5 mol⋅ kg ) ⎯⎯→ ZnO + H O + 2e 2 1 正极：HgO H O 2e Hg 2OH (0.5 mol kg ) 1 2 − − − + + ⎯⎯→ + ⋅ 电池： Zn + HgO ⎯⎯→Hg + ZnO 4. 计算电池 Ag, AgBr(s) Br ( 0.34) Fe ( 0.1), Fe ( 0.02) Pt 3 2 = = = − + + a a a