INFORME DEL DIAGRAMA POURBAIX.docx
Transcript of INFORME DEL DIAGRAMA POURBAIX.docx
NOMBRE: GUTIERREZ PATIÑO, EMILIO
CODIGO 11160061
INFORME DEL DIAGRAMA POURBAIX
Como resultado del balanceo de ecuaciones:
CALCULANDO ∆G 0 :
Una vez que tengamos todas los valores se procede a reemplazar en la formula:
1.- CU+ + 1ē CU
∆G0=1× µ0(CU) - 1× µ0(CU+)
∆G0= 0 - 12100
∆G0= -12100
E0 = ∆G 0−nF =
−12100−1×23060=+0.52
2.- CU2+ + 2ē CU
∆G0=1× µ0(CU) - 1× µ0(CU2+ )
∆G0= 0 - 15530
∆G0= - 15530
E0 = ∆G 0−nF =
−15530−2×23060=+0.34
3.- CU2O + 2H+ +2ē H2O + 2 CU
∆G0=2×µ0(CU)+1×µ0(H2O)-1×µ0
(CU2O)-
2×µ0(2H+)
∆G0= 0 +(-56687) – (-30350) – 2×(0)
∆G0= -26337
E0 = −26337
−2×23060 = +0.57
4.- CU2O2- +2H+ H2O + 2CU
∆G0=2×µ0(CU)+1×µ0(H2O)-1×µ0
(CU2O2-)- 2×µ0(H+)
∆G0= 0 + (-56687) – (-43900)- 0
∆G0= - 12987
5.- CUO2H- +3H++ 2ē 2H2O+ CU
∆G0=1×µ0(CU)+2×µ0(H2O)-1×µ0
(CUO2H- )- 3×µ0(H+)
∆G0= 0+2×(-56687) - 1×(-61800) -
3×(0)
∆G0= -51574
E0 = −51574
−2×23060 = +1.12
6.- CUO + 2H+ + 2ē H2O + CU
∆G0=1×µ0(CU)+1×µ0(H2O)-1×µ0
(CUO)- 2×µ0(H+)
∆G0= 0+(-56687) –(-30350) - (0)
∆G0= -26337
E0 = −26337
−2×23060 = +0.57
7.- CU2+ + 1ē CU+
∆G0=1×µ0(CU+) - 1×µ0(CU2+)
∆G0= 12100 – 15530
∆G0= -3430
E0= −3430
−1×23060 = +0.15
8.- CU2O +2H+ H2O+ 2CU+
∆G0=2×µ0(CU+)+1×µ0(H2O)-1×µ0
(CU2O)- 2×µ0(H+)
∆G0=2×12100 + (-56687) –(30350)-0
∆G0= -2137
9.- H2O+ 2CU+ +2ē CU2O2-+2H+
∆G0=1×µ0(CU2O2-)+2×µ0(H+)-2×µ0
(CU+)- 2×µ0(H2O)
∆G0= -43900 + 2×(0) -2×1200 – (-
56687)
∆G0= -11413
E0 = −11413
−2×23060 = +0.25
10.- CUO2H-+3H++1ē 2H2O+ CU+
∆G0=1×µ0(CU+)+2×µ0(H2O)-1×µ0
(CUO2H-)- 3×µ0(H+)
∆G0= 12100 + 2×(-56687)-(-61800)-
(0)
∆G0= -39474
E0 = −39474
−1×23060 = +1.71
11.-CUO +2H++ 1ē CU+ + H2O
∆G0=1×µ0(CU+)+1×µ0(H2O)-1×µ0
(CUO)- 2×µ0(H+)
∆G0= 12100 + 1×(-56687)-(-30350)-
(0)
∆G0= -14237
E0 = −14237
−1×23060 = +0.62
12.-H2O + 2CU2++ 2ē CU2O+ 2H+
∆G0=2×µ0(H+)+1×µ0(CU2O)-2×µ0
(CU2+)- 1×µ0(H2O)
∆G0= 0 +(30350) -2(15530)-(-56687)
∆G0= -4723
E0 = −4723
−2×23060 = +0.10
13.- H2O+2CU2++4ē CU2O2-+2H+
∆G0=1×µ0(CU2O2-)+2×µ0(H+)-2×µ0
(CU2+)- 1×µ0(H2O)
∆G0= -43900 +0 -2(15530)-(-56687)
∆G0= -18273
E0 = −18273
−4×23060 = +0.20
14.- CUO2H-+3H+ 2H2O + CU2+
∆G0=1×µ0(CU2+)+2×µ0(H2O)-1×µ0
(CUO2H-)- 3×µ0(H+)
∆G0= 15530 + 2(-56687)-(-61800)-0
∆G0= 63956
15.- CUO + 2H+ H2O + CU2+
∆G0=1×µ0(CU2+)+1×µ0(H2O)-1×µ0
(CUO)- 2×µ0(H+)
∆G0= 15530 + (-56687)-(-30350)-2(0)
∆G0= -10807
16.- CU2O +2ē CU2O2-
∆G0=1×µ0(CU2O2-) - 1×µ0(CU2O)
∆G0= -30350 –(- 43900)
∆G0= 13550
E0= 13550
−2×23060 = -0.30
17.2CUO2H-+4H++2ē 3H2O+CU2O
∆G0=1×µ0(CU2O)+3×µ0(H2O)-2×µ0
(CUO2H-)- 4×µ0(H+)
∆G0= (-30350)+3(-56687)-2(-61800)-
4(0)
∆G0= -76811
E0 = −76811
−2×23060 = +1.67
18.-2CUO +2H++2ē H2O+ CU2O
∆G0=1×µ0(CU2O)+1×µ0(H2O)-2×µ0
(CUO)- 2×µ0(H+)
∆G0= -30350 + (-56687) -2(-
30350)+2(0)
∆G0= -26337
E0 = −26337
−2×23060 = +0.57
19.- 2CUO2H- + 4 H++4 ē
3H2O+CU2O2-
∆G0=1×µ0(CU2O2-)+3×µ0(H2O)-2×µ0
(CUO2H-)- 4×µ0(H+)
∆G0= -43900 +3 (-56687) -2(-61800)-
4(0)
∆G0= -90361
E0 = −90361
−4×23060 = +0.98
20.- 2CUO+2H++4ē H2O+CU2O2-
∆G0=1×µ0(CU2O2-)+1×µ0(H2O)-2×µ0
(CUO)- 2×µ0(H+)
∆G0= -43900 +1 (-56687) -2(-30350)-
2(0)
∆G0= -39887
E0 = −39887
−4×23060 = +0.43
21.- H+ + CUO2H- CUO + H2O
∆G0=1×µ0(CUO)+1×µ0(H2O)-1×µ0
(CUO2H-)- 1×µ0(H+)
∆G0= -30350 +1 (-56687) -1(-61800)-
1(0)
∆G0= -25237
Diagramando con las formulas adecuadas: