Electrochemistry Chem 17 Full Handout New

8/19/2019 Electrochemistry Chem 17 Full Handout New

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ELECTROCHEMISTR


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R: B REDOX

• B :

1. MO4() + C

() ⇌ M2+() + C2()

2. H2O2() + CO2() ⇌ CO2() + O2()

3. NO2() + C2O72() ⇌ C3+() + NO3()

ELECTROCHEMISTR


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• OR .

– LEORA ( ) – GEROA ( )

C2+ + ⇄⇄⇄⇄ 2+ + C

R: C2+ + 2⇄ C

O: ⇄ 2+ + 2

(H )


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A

• T

– R C2+ C

– O 2+

• T :

R: C()2+ + 2⇄ C()O: ()⇄ ()

2+ + 2


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T (

)


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I E,

E C

( )

(+ )

V/G C

( )

(+ )

+


7/73

T ()C () V/G C

+1.10


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E

C2+() + 2 ⇌ C() F

3+() +

⇌ F2+()C() ACE P() E


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C C

• V/G

• A :M ( )

A .

A ,

, .


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T C C

• C C :

1. A C 1.0 (II) .

2. A 1.0 (II) .

3. A

• T 1.10


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T C C

1.0 M 1.0


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C N

• S .

– C ()

() ()2+ (1.0 ) C()

2+ (1.0 ) C()

R O

D :

S

S :

S :


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T C S C• C :

1. A C 1.0 (II) .

2. A A 1.0 (I) .

3. A .

• T 0.46 .


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T C S C

•

C C CA :

C()C()2+ (1.0 )A()+(1.0 )A()

W

.

()()2+ (1.0 )C()

2+(1.0 )C() C

C


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• T C2+ 2+.

– I C2+ 2+.

• S, A+ C2+.

– B A+ C C 2+.• I ,

:

R


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S E P• P

• T , .

• T S H

E (SHE). – T SHE

0.00000000 V

– T

SHE


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T SHE C

• F :

1. A 1.0 (II) .

2. T S H E.

3. A .

• T 0.763 .

• T S H E. – I H+ H2.

• T .

–

2+

.


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T SHE C


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T CSHE C• T :

1. A C 1.0 (II) .

2. T S H E.

3. A .• T 0.337 .

• I SHE – T C2+ H2 H

+.

• T C .

–

T C2+

C .


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T CSHE C


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S E P

• S ()

.

L (L0 )

( OX

SHE)

F (F20 )

( RED

SHE)

• E

.

• E

.

⇌ .

⇌

.


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U S E P

• S E

• S E

• N:

E = E E I

E = E

+ E

I ,


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S

O

A

S

R

A


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H S E P

1. C

2. W E0 , E0 .

3. W , ..

E

0

.

4. B . D E0 ! (E0 INTENSIVE )

5. A . T E0 , (). I E0 , ().


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• W : P , MO4, (II) (III) , (III) (II) ?

• I S

, ?


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• G E :

2A+() + () ⇌ 2+() + 2A()

MO4

() + 5F2+

() + 8H+

() ⇌ M2+

() + 4H2O() + 5F3+

()

2NO3−(aq) + 8H

+() + 3C() ⇌ 3C2+() + 2NO() + 8H2O()


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G F E, K, E

• F

ΔGΔGΔGΔGoooo nFEnFEnFEnFEoooocellcellcellcell #

F 96 485 J/V (F C)

E

ΔG

G F E

• R E0 E K

ΔGΔGΔGΔG0000

nFEnFEnFEnFEoooo

cellcellcellcell 2.303RTlogK2.303RTlogK2.303RTlogK2.303RTlogKeqeqeqeq∴

.

at 25, .


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kJ/mol302.23-orrxnJ/mol302230-G

V1.5662emoleV

J48596emole2-G

0

-

-0

=∆

+⋅⋅=∆

kJ/mol285.60-orrxnJ/mol285595-G

V0.74emoleV

J48596emole4-G

0

-

-0

=∆

+⋅⋅=∆

D ∆G :


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D K :

.

. .

. . .


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• C . W +E.

V0.83- OH2He2O2H

V0.40 4OHe4O2HO

V1.23 O2H4e4HO

-(aq)2(g)

-(l)2

-

(aq)

-

(l)22(g)

(l)2

-

(aq)2(g)

+→+

→++

→++ +

V2.06 O2H2HO

V0.83 )e2OH2OH22(H

V1.23 O2H4e4HO

(l)2(g)22(g)

-

(l)2

-

(aq)2(g)

(l)2

-

(aq)2(g)

→+

+→+

→++ +

V1.23 O2H2HO

V0.83 )e2OH2OH22(H

V0.40 4OH4eO2HO

(l)2(g)22(g)

-

(l)2

-

(aq)2(g)

-(aq)

-(l)22(g)

→+

+→+

→++

M

,

H+ (

H2 )

L ,

H2O


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V2.06 O2H2HO

V0.83 )e2OH2OH22(H

V1.23 O2H4e4HO

(l)2(g)22(g)

-

(l)2

-

(aq)2(g)

(l)2

-

(aq)2(g)

→+

+→+

→++ +


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L D

• A

I F() , F(III) F(II).

ΔG H L, E0

ΔG0

H L


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T

(L

,

)

T ( )


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I ,

. T

ΔG:

(1) C

(2) T

(3) O

(4) H ( (3))


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L

F E

0 +1 +2F P ( E)

( ).

T E X+/X0

(X+ + → X0). E

(0 ).

X: E (N: E = ∆G/F)

:


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C M:

1 M 2+:

W M0 :

M2+ + 2 → M0 E = 1.18 V

C E:

= 2, ∴ E = −2.36

2 M 3+:

H: M3+ + 3 → M0

D E E

M2+ + 2 → M E = 1.18 V ∆G = 2×F×(−1.18 V) = +2.36F

M3+ + → M2+ E = +1.50 V ∆G = 1×F×(+1.50 V) = −1.50 F

M3+ + 3 → M ∆G= 3×F×(E) = +0.86 F∴ E = −0.287 V

C E:

= 3, ∴ E = −0.86


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A)

T . T

.

,

A. ,

A.

B)

R RA OA

.

, .


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) (

)

S , . T

()

)

I

,

. T

.

)

S

D

D


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W N2 ?

W N2 ?

I 2

I 4


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E C E P

E

T N

.


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T N E• S

N 25

C :


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• F :

T N :

• I C2+ C+ 1.0 , .. , E = E0

.


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• C C2+/C+

250C C+ 1/3 C2+

.

C


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• C

F

3+

/F

2+

F

3+

= 1.0 10

2

M F2+ = 0.10 M S4+/S2+

S4+ = 1.00 M S2+ = 0.100

M. A .

C E

S Q


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• S Q

E.


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CCEA CE


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M E E

AgBr s ⇌ Ag

a q B r

aq E 0.729 V

E 0.0592

1

log Ag Br 0.0592

1

logK

= ?

C:

AgBr s e ⇌ Ag s Br aq 0.071 V

A:

C :

Ag

a q e

⇌ Ags

0.800 V

AgBr s e ⇌ Ag s Br aq E 0.071 V

Ag s ⇌ Ag a q e E 0.800 V

AgBr s ⇌ Ag aq Braq

M E E


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M E E

C :

Au

aq 3e

⇌ Aus

1.520 V

AuCl aq 3e ⇌ Au s 4Cl aq 1.002 V

Au aq 4Claq ⇌ AuCl aq ?

C:

A:

Au aq 3e ⇌ Aus 1.520 V

Au s 4Cl aq ⇌ AuCl 3e aq 1.002 V

Auaq 4Cl aq ⇌ AuClaq 0.518 V

E 0.0592

3

log AuCl

Au

Cl

0.0592

3

logK

• M K


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Ag

0.100 M e

⇌ AgsAg saturated e ⇌ Ags

E E 0.0592

1 log

Agsaturated

Ag 0.100

Ag 0.100 M e ⇌ AgsC:

A: Ag s ⇌ Ag saturated e

Ag 0.100 M ⇌ Ag saturated

E C


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E C• N

• E0

•

ΔG0

ΔG0

= F(E0

)= +

E

( ) S, 212.3

J/

N, 212.3 J/

E C


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E C

• E

• I , ,

• P

o :

o :

• E 1 2

• T :

– E .

– E .

E C 1


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E C: 1

A

.

T

.

A .

T = .E

o P ().

o N ().

• / • E


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• / o C (+): M P

E0o A (): M N

E0o T MOST E0

• E o C (): L N

E0o A (+): L P

E0o T LEAST E0

, EDC CADE, DA ADE

(EDCA, A)

V0.83- OH2He2O2H

V0.40 4OHe4O2HO

V1.23 O2H4e4HO

-

(aq)2(g)

-

(l)2

-(aq)

-(l)22(g)

(l)2

-

(aq)2(g)

+→+

→++

→++ +

V2.06 O2H2HO

V0.83 )e2OH2OH22(H

V1.23 O2H4e4HO

(i)2(g)22(g)

-

(l)2

-

(aq)2(g)

(l)2-(aq)2(g)

→+

+→+→++

+

V1.23- 2HOO2H

V0.83- )2OHH2eO2(2H

V0.40- 4eO2HO4OH

(g)22(g)(i)2

-

(aq)2(g)

-

(l)2

-(l)22(g)-(aq)

+→

+→+++→

E V/G

E C: E A KI


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+ () + () 2.924

22 + 2 22 () + 0.828

2() + 2 2 () +0.535

2 () + 4+ +4 22 +1.229

C :

E1 = E(K+/K) E(I2/I) = (2.924 V) (+0.535 V)= 3.459V

E2 = E(H2O/H2,OH) E(I/I2) = (0.828V) (0.535 V)

= 1.363V

E3 = E(H2O/H2,OH) E(H2O/O2, H+)=(0.828V) (1.229)

= 2.057

2 ,

.

E C: E A NC


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E C: E A NC

N+ () + N() 2.714 V

2H2O + 2 2H2 () + OH 0.828 V

C2() + 2 C () +1.360 V

O2 () + 4H+ +4 2H2O +1.229 V

C :

E1 = E(N+/N) E(C/C2)= (2.714 V) (+1.360 V) = 4.074 V

E2 = E(H2O/H2,OH) E(C2/C)

= (0.828V) (1.360 V) = 2.188 VE3 = E(H2O/H2,OH) E(H2O/O2, H+)

=(0.828V) (1.229) = 2.057

E = E E


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• T V

O2() C2().

• H, (O2) .

•

T .

E2 = E(H2O/H2,OH) E(C/C2)= (0.828V) (1.360 V) = 2.188 V

E3 = E(H2O/H2,OH) E(H2O/O2, H+)=(0.828V) (1.229) = 2.057

OVERPOTENTIAL

C E: F E


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C F

• F L , (96,487 ) , ,

.

• A

.

M+ + M0 A0 A + • T M A

1 = 6.022 ×××× 1023

1 = 96485

C E F L E


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C E: F L E

Mass of substance oxidized or reduced

I ,

.

•

A (A) .

• Q () =

• 1 = 1 /

• I () =

• 1 = 1 /

• C (II) 3 20


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(II) 3.20

(II) 30.0.

: 2+ + 2 0

∴∴∴∴ = 2

A = 106.42

: 30

Mass of Pd 3.20

30 min

60 106.42

2

96485 C

mol e

= 3.18 P

• C ( STP)


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2.0 5.0

:

2 2 2() + 4+ + 4

∴∴∴∴ = 4

1 2 = 22.4

: 5

Vol 2.0

5 min

60 22.4

4

96485

C

mol e

= 0.0348 L 34.8 L

C A E C


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E E

• I ≈ 100% C. – T

: A, A, P.

• T CSO4 H2SO4

• T C C2+.

•

T C2+

C .

E C:

E NC


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E NC

• R () NC – C N+ C .

• I ( 3.852 V) ( )

NC, : – P (C2)

.

– M, N . T N

NC.

A

C

D C I

N()

C


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• M

CO2, O2, H2O.

( )

points.exposedatrapidlyoccursreactionThe

reactionoverallOFe2O3+Fe4 320

2

0 →

O2 + 4 + 2 H2O → 4 OH

F → F2+ + 2 −

4 F2+ + O2 → 4 F3+ + 2 O2−

2 + O2 + 2H2O → 4OH

F3+ + 3 H2O ⇌ F(OH)3 + 3 H+

F(OH)3 ⇌ FO(OH) + H2O

2FO(OH) ⇌ F2O3 + H2O


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C P• S .

1. P (

) .

2. C ,

.

C P


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3. A .

4 G, , .

C P


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C P

5. P

.

ceramic.withcoatedare bathtubsSteel

B


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B

• P ,

( )

• S ,

(

)

• F

,

V G


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C• E

• H ( )

.

• E : – A

– F

– B

– L+

P C


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L

S

:

A

:

L (


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)


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A B

L I


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Electrochemistry Chem 17 Full Handout New

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