第17章电化学 257 nK“(2985K)=FE°2×96485×0342-040=459 8.3145×298.15 K°(29815K)=00102 0.46 P 00 K(m)△,H K(29815K)R I5K T T=428K 18.电池 Pt, H2(P) HCI(0. 1 molkg")Hg2 C12(s), 在25℃测得p=0.1013MPa时,E=0.3990V;p=11.16MPa时, E=04596V。试求H2在25℃、11.16MPa时的逸度因子。设0.1013MPa 时H2可当作理想气体 解:H2(P)+Hg2Cl2(s)—)2Hg+2HCl(0. H2(P2)+Hg,, Cl2(s)->2Hg+ 2HCI(0. I mol-kg H2(P1)—H2(P2) △G(3)=△Gn()-△Gn(2)=-F(E1-E2) △Gn(3)=RTln P2中2 P 222×(0.3990-04596) =2.049 P,d, 0.05916 P2φ2 0.1013 19.计算下列浓差电池在18℃时的电池反应电势 (1)lzn(a4=0)|zn(a2=05)zn (p=0.1MPa)HCI(0. 1 mol.")H2(P2=0.01MPa),Pt第 17 章 电化学 ·257· ( ) ( ) 4.59 8.3145 298.15 2 96485 0.342 0.401 ln 298.15 K o o = − × × × − = = RT zFE K ( ) 298.15 K 0.0102 o K = ( ) 0.46 100 101.3 0.21 1/ 2 o o O2 = × = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = p p K T ( ) ( ) ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − × = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − Δ = / K 1 298.15 1 8.3145 31.05 10 1 298.15 K 1 298.15 K ln 3 o r m o o T R T H K K T T = 428 K 18. 电池 Pt, H ( ) HCl (0.1mol kg ) Hg Cl (s), Hg 2 2 1 2 − p ⋅ 在 25℃测得 p = 0.1013 MPa 时， E = 0.3990 V ； p = 11.16 MPa 时， E = 0.4596 V 。试求H2在 25℃、11.16 MPa 时的逸度因子。设0.1013 MPa 时H2可当作理想气体。 解： H ( ) Hg Cl (s) 2Hg 2HCl (0.1mol kg ) 1 2 1 2 2 − p + ⎯⎯→ + ⋅ ① H ( ) Hg Cl (s) 2Hg 2HCl(0.1mol kg ) 1 2 2 2 2 − p + ⎯⎯→ + ⋅ ② ①－② H ( ) H ( ) 2 1 2 2 p ⎯⎯→ p ③ ( ) () ( ) ( ) r m r m 1 2 o r m Δ G 3 = Δ G 1 − Δ G 2 = −zF E − E 又 ( ) 1 1 2 2 1 2 r m 3 ln ln φ φ p p RT f f Δ G = RT = ∗ ∗ ∴ ( ) 2.049 0.05916 2 0.3990 0.4596 lg 1 1 2 2 = × − = − φ φ p p 112 1 2 1 2 ⋅ = φ φ p p 1 1.02 11.16 0.1013 112 φ 2 = × × = 19. 计算下列浓差电池在 18℃时的电池反应电势： (1) Zn Zn ( 0.1) Zn ( 0.5) Zn 2 2 1 2 = = + + a a (2) Pt,H ( 0.1MPa) HCl (0.1mol kg ) H ( 0.01MPa), Pt 2 2 1 2 1 = ⋅ = − p p