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: