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OXIDATION-REDUCTION REACTIONS(Redox Reactions) (SJ, p. 316)

1. Oxidation-reduction (Redox) reactions

a. Redox reaction consists of two half-reactions:

1) Oxidation reaction (LEOs) - in which a substance loses or donates electrons (e -)

2) Reduction reaction (GERs) - in which a substance gains or accepts electrons (e-)

LEO said GER

Oil Rig oxidation in lose, reduction in gain

c. An oxidation reaction and a reduction reaction must be always coupled.

- Two rules apply to redox reaction:

1) "free" electron cannot exist in solution.

2) electron must be conserved.

d. All redox reactions involve a change in Oxidation State(Oxidation Number).

2. Oxidation State (Oxidation Number)

- is the hypothetical electronic charge that each atom has in a reaction.

- is a very useful concept in balancing chemical reactions.

All redox reactions involve a change in oxidation state.

- In general,

H has oxidation state of 1+ (+1)

O has oxidation state of 2- (-2)Cl has oxidation state of 1- (-1)

H2 has oxidation state of 0O2 has oxidation state of 0

H2O2has oxidation state of 0

Cl2has oxidation state of 0N2 has oxidation state of 0

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b. Rules

1) The oxidation state of a monoatomic substance is equal to its own electronic charge (valence).

e.g., Cu2+ Cu(II)

Cd

2+

Cd(II)Ni2+

Ni(II)

2) In a covalent compound, the oxidation state is the charge remaining on the atom when each

shared pair of electrons is assigned completely to the more electronegative of the two atomssharing them (divide the electrons equally if the two atoms are the same)

3) Sum of charges must equal charge on ion.

Example: Find the oxidation number of

S H2S S8 SO32- SO42-

C HCO3- HCOOH C6H12O6 C6 H5

N NH4+ N2 NO2

- NO3

- HCN

(Solution)

2(1+)2- 0 4+3(2-) 6+4(2-)

S H2 S S8 [SO3]2-

[SO4]2-

sulfite sulfate

1+4+3(2-) 1+2+2- 2- 1+ 012(1+) 6(2-) 6 (-5/6) 5(1+)

C [HCO3]- HCOOH C6H12O6 C6 H5

3-4(1+) 0 3+ 2(2-) 5+ 3(2-) 1+ 2+ 3-

N [NH4]+ N2 [NO2]

- [NO3]

- H C N

nitrite nitrate

Example: Write the oxidation number of

7+ 2(7+) 7(2-)

Cl in Cl2O7 Cl in Cl2 O7

4+ 4+3(2-)

S in SO32- S in [SO3]

2-

3+ 1+ 3+2(2-)

N in HNO2 N in H N O2

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6+ 2(1+) 2(6+) 7(2-)

Cr in K2Cr2O7 Cr in K2 Cr2 O7

+1 2+ 2(+1)

H in CaH2 H in CaH2

Example

2+ 3 2(1+)(2) 4+ 3 2 2(1+)

[CN ] + H2 O C N O + 2H+ + 2e

Example Nitrogen

NH4+ + 1.5 O2 NO2

- + H2O + 2H

+

NO2- + 0.5 O2 NO3

-

-All oxidation reduction reactions involve a change in oxidation state.

0

N2

Anammox Denitrificationprocess (denitrifyer)

3- 4(1+) --------- 3+ 2(2-) --------------- 5+ 3(2-)

N H4+ --------- N O2

- ---------------- N O3

-

1st stage 2nd stagenitrification nitrification

Nitrosomonassp. Nitrospirasp.,Nitrobacter sp.

Slower rxn Faster rxn

ExampleSulfur

0 6+ 4(2) 2(1+) 2

S8 -------> S O42-

------> H2 S

REO GER

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3. Oxidizing and Reducing Agents

a. Oxidizing Agent - a substance which causes an oxidation to occur, while being reduced itself.

e.g.,

O2 OxygenO3 Ozone

H2O2 Hydrogen peroxide

Cl2 ChlorineOH Hydroxyl radical

MnO4 Permanganate (as KMnO4)

Cr2O72

Dichromate (as K2Cr2O7)

0 2(1+) 2 0 4+ 2(2) 2(1+) (2)

Formaldehyde C H2 O + O2 --- C O2 + H2 O

O: 02- O is reduced (GERs)

C: 04+ C is oxidized (LEO)

Organic Chemistry: Aldehydes

General Functional

Formula Group

R-C=O -C=O H-C=O CH3-C=O

\ \ \ \

H H H H

Common: Formaldehyde AcetaldehydeIUPAC: Methanal Ethanal

No of C IUPAC1 Methanal

2 Ethanal

3 Propanal4 Butanal

5 Pentanal6 Hexanal7 Heptanal

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b. Reducing Agent - a substance which causes a reduction to occur, while being oxidized itself.

e.g., SnCl2Nao

Na2SO4

4+ 2(2) 6+ 4(2)

Sn2+

Sn4+

+ 2e S O2 (S O4 )

2-

tin

Nao ---> Na

++ e

Balancing Redox Reaction- the half reaction method

Example7-1 (SJ, 317): Balance the reaction in which ferrous ion (Fe2+

) is oxidized to ferric ion (Fe3+

)by permanganate (MnO4

-) which itself is reduced to manganese dioxide (MnO2(s)). The reaction takes

place in alkaline solution.

Reaction: Fe2+ + MnO4- Fe

3+ + MnO2(s)

STEP

1. Identify principal Reactants Products

reactants and2+ 3+

products Fe2+ Fe3+ Oxidation

7+ 4(2-) 4+ 2(2-)

MnO4- MnO2(s) Reduction

2. Obtain balanced Fe2+

Fe3+

half-reactions(balance the charge MnO4

- MnO2(s)

w/ e-)

3. Balance the oxygen Fe2+ Fe3+

w/ H2O MnO4- MnO2(s) + 2H2O

4. Balance the hydrogen Fe2+ Fe3+

w/ H+ 4H

+ + MnO4

- MnO2(s) + 2H2O

5. Balance the charge Fe2+

Fe3+

+ e-

w/ e - 3e - + 4H+ + MnO4- MnO2(s) + 2H2O

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6. Multiply each half -rxn 3Fe2+ 3Fe3+ + 3e -

by an appropriate integer

so that both contain the 3e- + 4H

+ + MnO4

- MnO2(s) + 2H2O

same number of e - -------------------------------------------------------------------------

4H

+

+ 3Fe

2+

+ MnO4-

3Fe

3+

+ MnO2(s) + 2H2O

7. If the rxn takes place in alkaline 4H+ + 3Fe

2+ + MnO4

- 3Fe

3+ + MnO2(s) + 2H2O

solution,add the water dissociation 4H2O 4H+ + 4OH

-

equation to eliminate H+ -----------------------------------------------------------------------

as a reactant 2H2O + 3Fe2+

+ MnO4- 3Fe

3+ + MnO2(s) + 4OH

-

8. We know that in alkaline 2H2O + 3Fe2+ + MnO4

- 3Fe3+ + MnO2(s) + 4OH

-

solution, Fe3+

and OH- 3Fe

3+ + 9OH

- 3Fe(OH)3(s)

will combine to form -----------------------------------------------------------------------Fe(OH)3(s) 2H2O + 3Fe2+ + MnO4

-+ 5OH- 3Fe(OH)3(s)+ MnO2(s)

Net:

1. Ferrous iron (Fe 2+

) is oxidized, it loses e- (REO)

2. Permanganate (MnO4-) is reduced, it gains e

- (GER)

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Other Method(Sato)

Balance the following equation by the half reaction method:

Reaction: Fe2+

+ MnO4- Fe

3+ + MnO2(s)

1. Identify principal Reactants Productsreactants and2+ 3+

products Fe2+ Fe3+ LEO - Oxidation

7+ 4(2-) 4+ 4-

MnO4- MnO2(s) GER - Reduction

2. Obtain balanced Fe2+

Fe3+

+ e-

half-reactions

(balance the charge Mn7+

+ 3e- Mn

4+

w/ e-)

3. Multiply each half rxn (x3) 3Fe2+

3Fe3+

+ 3e-

by an appropriate integerso that both contain the Mn7+ + 3e - Mn

4+

same number of e - ----------------------------------------------------------------

3Fe2+

+ Mn7+

3Fe3+

+ Mn4+

4. Replace y compound

5. Write the equation 3Fe2+

+ MnO4- 3Fe

3+ + MnO2(s)

w/ compounds

6. Balance Ow/ H2O 3Fe2+

+ MnO4- 3Fe

3+ + MnO2(s) + 2H2O

7. Balance H+ 4H+ + 3Fe2+ + MnO4- 3Fe

3+ + MnO2(s) + 2H2O

8. In alkaline solution 4H+ + 3Fe

2+ + MnO4

- 3Fe

3+ + MnO2(s) + 2H2O

- balance H+with H2O

4H2O 4H+ + 4OH

-

-------------------------------------------------------------------------2H2O + 3Fe

2+ + MnO4- 3Fe

3+ + MnO2(s) + 4OH-

9. We know that in alkaline 2H2O + 3Fe2+ + MnO4

- 3Fe3+ + MnO2(s) + 4OH

-

solution, Fe3+

and OH- 3Fe

3+ + 9OH

- 3Fe(OH)3(s)

will combine to form -----------------------------------------------------------------------

Fe(OH)3(s) 2H2O + 3Fe2+

+ MnO4-+ 5OH

- 3Fe(OH)3(s)+ MnO2(s)

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Example A B + C

Reactants Products

S2O42

S2O32

HSO3

(aq)1) Identify principal reactants and products

Reactants Products

2(3+) 4(2) 2(2+) 3(2)

S2 O42

S2 O32

1+ 4+ 3(2-)

H SO3

Note:2S3+

+ 2S2+

S3+

S4+

2) Balance e-

S3+

+ e- S2+

S3+

S4+

+ e------------------------------------------------------------------------------

3) Balance S

(x 2) 2S3+ + 2e- 2S2+

(x 2) 2S3+

2S4+

+ 2e-

------------------------------------------------------------------------------4S3+ 2S

2+ + 2S4+

4) Replace by compound

2S2 O42

(aq) S2 O32

+ 2HSO3

5) Balance O with H2O

2 S2 O42 + H2O S2 O3 2 + 2HSO3

6) Balance H+with H2O - balanced

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ExampleReactants Products

CN + Cl2 CO2 + N2 + Cl

2+ 3 0 4+2(2) 0 1

C N

+ Cl2 CO2 + N2 + Cl

1) Identify principal reactants and products

Reactants Products

2+ 4+2(2)

C CO23 0

N N2

0 1

Cl2 Cl

---------------------------------------------------------------------------------------------------------2)

Ox: C2+

C4+

+ 2e-

0

Ox: (x2) 2N3 N2 + 6e-

0

Re: Cl2 + 2e- 2Cl

----------------------------------------------------------------------------------------------------------

3)

Ox: (x2) 2C2+

2C4+

+ 4e-

0

Ox: 2N3 N2 + 6e-

Re: (x5) 5Cl2 + 10e- 10Cl

-------------------------------------------------------------------------------------------------------------

2CN

+ 5 Cl2 2CO2 + N2 + 10 Cl

4) Balance O with H2O

2CN + 5 Cl2 + 4H2O 2 CO2 + N2 + 10Cl

5) Balance H+with H2O

2CN + 5 Cl2 + 4H2O 2 CO2 + N2 + 10Cl + 8H+

Note:

NaCN + H2O HCN + NaOH

KCN + H2O HCN + KOH

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Organic Chemistry - Nitriles

Cyanide any chemical compound that contains the cyano group, -CN

General Functional Example:

Formula Group

R-CN -CN CH3-CN

Common: Acetonitrile or Methyl cyanideIUPAC: Ethanenitrile

- Nitrile do not release cyanide ions.

Direction of Reaction(SJ, 322)

Example: Chemical Oxygen Demand (COD) test

- Redox reaction between potassium dichromate K2Cr2O7and ferrous iron, Fe2+.

1) Balance the reaction2) Determine the equilibrium constant, K

3) Determine if the reaction proceeds spontaneously if [Fe2+

] = 10-4

, [Fe3+

] = 10-3

,

[Cr3+

] = 10-2.4

, [Cr2O72-

] = 10-2.7

, [H+] = 10

+1.3(Ignore ionic strength).

(Solution)

1. Balance the reactionReactants Products

step 1:

2(6+) 7(2-) (3+)

Cr2 O72-

W Cr3+

Fe2+ W Fe3+step 2:

Cr6+ + 3e - W Cr3+

Fe2+

W Fe3+

+ e-

step 3:

(x 2) 2Cr6+

+ 6e- W 2Cr3+

(x 6) 6Fe2+ W 6Fe3+ + 6e -------------------------------------------------------

Cr2 + 6Fe2+ W 2Cr3+ + 6Fe3+

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step 4: Replace with compounds

Cr2O72- + 6Fe 2+ W 2Cr3+ + 6Fe3+

step 5: Balance O with H2O

Cr2O72-

+ 6Fe

2+

W 2Cr3+

+ 6Fe

3+

+ 7 H2O

step 6: Balance H with H+

14H+ + Cr2O7

2- + 6Fe

2+W 2Cr3+ + 6Fe3+ + 7 H2O

Note:

1) dichromate is reduced to chromic ion - gained e-

2) ferrous ion is oxidized to ferric ion - lost e-

or

Dichromate is reduced to chromic ion (GER):

Cr2O72- + 14H+ + 6e -W 2Cr3+ + 7 H2O Reduction

Ferrous ion is oxidized to ferric ion (LEO):

6Fe 2+W 6Fe 3+ + 6e - Oxidation---------------------------------------------------------------------------------------------------

14H+ + Cr2O72- + 6Fe 2+W 2Cr3+ + 6Fe 3+ + 7 H2O

fG

(kcal/mol) 0 -315.4 -20.30 -51.5 -2.52 -56.69

GE= (3

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or

ln Keq= 131.9

2.302log10Keq= 131.9

log Keq= 57.27

Keq= 1057.27 = 1.92 x 1057

or

G E = RT ln Keq= (2.303) RT logKeq

3.57)298)(/1099.1(303.2

)/8.77(

303.2log

310 =

=

=

KKmolkcalx

molkcal

RT

GKeq

Keq= 1057.3

3) Determine if the reaction proceeds spontaneously if [Fe2+

] = 10-4

, [Fe3+

] = 10-3

,

[Cr3+

] = 10-2.4

, [Cr2O72-

] = 10-2.7

, [H+] = 10

+1.3.

G = G E+ RT ln Q

= -77.8 kcal/mol + (1.99 x 10-3kcal/mol K) (298 K) ln 10-14.3

= - 97.2 kcal/mol < 0 This reaction is spontaneous as written.

or

This reaction is spontaneous as written.

This reaction is utilized in the titrimetric finish of the COD test.

3 6 3 2 3 6 2.4 214.3

2 6 2 14 4 6 2.7 1.3 14

2 7

[ ] [ ] (10 ) (10 )10

[ ] [ ][ ] (10 ) (10 )(10 )

Fe Cr Q

Fe Cr O H

+ +

+ + += = =

14.3

57.3

101

10eq

Q

K

=

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Free Energy (G) and Electrode Potential (E) (SJ, p. 324)

G = G + RT ln Q (1)

G = - RT ln Keq (2)

G = - n FE (3)

G = - n FE (4)

where G = free energy change

G = standard free energy change

E = electrode potential, voltE = standard electrode potential, volt (Read Chap 7.3.2, SJ 324-330)

n = number of electrons involved in the reaction

F= Faradays constant

= 23.06 kcal/(volt e- equivalent) = 96,500 coulombs/ e- equivalent

- the charge per mole of electron or the charge per equivalent

Note: 1 coulomb is the amount of electric charge transported by a currentof 1 amperein 1 second.1C = 1A 1s

Unit check: G = n F E

kcal e- kcal volt-------- = ------- -------------------- --------mole mole volt e- equiv

Note: From (2) G = - RT ln Keq = - (2.303) RT log10Keq

at 25CG = - (2.303)(1.99 x 10-3)(298) = -1.364 log10Keq

G = -1.364 log10Keq

orlog10Keq = - 0.733 G

Keq = 10

- 0.733 G

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7.3.2 Free Energy and Potential of Half-Reaction (SJ, 324)

Example. Winkler Method for DO

The titration of iodine, I2(aq), to iodide, I-, using thiosulfate, S2O3

2-(Na2S2O3)

blue clear

(SJ 328-329; 450-451)

Reactants Products

1) (0) (1-)I2 I

-

2(2+) 3(2) 4(2.5+) 6(2)

S2 O32-

S4 O62

thiosulfate tetrathionate

2) balance e-

(0) 2(1-)I2 + 2e

2 I-

(2+) (2.5+)

S S + 0.5 e

(0) 2(1-)

I2 + 2e 2 I

-

4(2+) 4(2.5+)

(x4) 4S 4S + 2 e

---------------------------------------------------

4(2+) 4(2.5+)

I2 + 4S 4S + 2I

I2 + 2 S2O32- S4O6

2- + 2I

Note:

I2 + 2 S2O32-

S4O62-

+ 2I-

fG +3.93 -127.2 -246.3 -12.35

(kcal/mole)

G = [(-246.3) +2 (-12.35)] - [ (3.93) + 2 (-127.2)] = -20.53 kcal/mole

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From Table 7-1, (SJ 328-329; 450-451)

Half Reaction (same as above example reaction) E (v)

Re: I2 + 2e- 2 I

+ 0.62

Ox: 2 (S2O3)2 (S4O6)

2 + 2e- - 0.18

--------------------------------------------------------------------------------------------

Reduction: I2 + 2 e 2 I

+ 0.62

Gf 3.93 0 -12.35

(kcal/mole)

Oxidation 2 (S2 O3)2-

(S4 O6)2-

+ 2 e-

- 0.18

Gf -127.2 - 246.3 0(kcal/mole)

Species Gf (kcal/mol)------------------------------------------e - 0

I2 +3.93

I- -12.35

S2O32-

-127.2S4O6

2- -246.3

------------------------------------------

Standard free energy change:

Reduction rxn: G = (Gf)products- (Gf)reactants

= 2 (-12.35) [(+3.93) + 2 (0)] = -28.63kcal/mole

Oxidation rxn: G = (Gf)products- (Gf)reactants

= [- 246.3 + 2(0)] [2 (-127.2)] = + 8.1kcal/mole

The overall reaction E G - n FE

(v) (kcal/mole) (kcal/mole)

2 e - + I2(aq) 2 I - + 0.62 -28.63 -2 (23.06)(+0.62) = -28.59

2 S2O3)2-

S4O62-

+ 2 e-

- 0.18 +8.1 -2 (23.06)(-0.18) = +8.3

-----------------------------------------------------------------------------------------------------------------I2(aq) + 2 S2O3

2- S4O62- + 2I- -20.53 -20.29

F = 23.06 kcal/ (volt e-equivalent)

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Combination of Half-Reactions

Example 7.5 (SJ, p. 331) see Table 7-1 (p. 329)

Determine EFe3+, Fe(s)for the half reaction: Fe3+

+ 3e Fe(s)

G = - n FE

E (V) - n FE

Fe2+ + 2e- Fe(s) -0.44 - 2F(-0.44)

Fe3+ + e- Fe2+ 0.77 -1F(+0.77)

Fe3+ + 3e- Fe(s) -3F (EFe3+, Fe(s))

-n F (EFe3+, Fe(s)) = (-n FEFe2+, Fe(s)) + (-n F EFe3+, Fe2+)

-3F (EFe3+, Fe(s)) = [-2F (-0.44)] + [-1F(0.77) ]

-3F (EFe3+, Fe(s)) = F (0.88 0.77)

(EFe3+, Fe(s)) = (F/-3F) (0.88 0.77) = - 0.037 volt

Example

E(volt)

Fe2+ Fe3+ + e- -0.771

O2 + H+ + e

- H2O +1.229

Determine free energy for each half reaction and equilibrium constant.

Given: Table 1 (SJ 64)

Species G (kcal/mole)

e-

0

Fe2+(aq) -20.30Fe3+(aq) -2.52

O2(g) 0H+(aq) 0

H2O (aq) -56.69

_________________________

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(Solution)Fe2+ Fe

3+ + e -

G -20.30 -2.52 0

(kcal/mol)

G (kcal/mol) = (-2.52) + 0 (-20.30) = +17.78

O2 + H+ + e- H2O

G 0 0 0 -56.69

(kcal/mol)

G (kcal/mol) = (-56.69) 0 = -28.35

Combine the half reactions (balanced) G (kcal/mole) E(volt)__- nFE____

Fe2+ Fe3+ + e - + 17.78 -0.771 - (1)F(-0.771)

O2 + H+ + e - H2O -28.35 +1.229 - (1)F(+1.229)

_________________________________________________________________________Fe 2+ + O2 + H+ Fe

3+ + H2O -10.57 (-0.458) -(1)F Ecell

[- (1)F(-0.771) ] + [ - (1)F(+1.229)] = -(1)FEcell

-F (-0.771+1.229)

Ecell= --------------------- = +0.458volt- F

G = - n FE = - (1) (23.06 kcal/volt. e-equiv)(0.458 volt)

= -10.56 kcal/ e-equiv (-10.56 kcal/mole)

At equilibrium,

G = - RT ln Keq = - (2.303) RT log10Keq= -(2.303)(1.9872 x 10-3kcal/mol K)(298K) log10Keq

-10.57 kcal/mole = -(1.3638 kcal/mol) log10Keq

-10.57

log10Keq = --------- = 7.79

-1.36

Keq = 107.79= 6.15 x 107

( )

3

7.79

1/ 42

2

10eqo

FeK

Fe H P

+

+ +

= =

= 6.15 x 10

7

Po2can be computed knowing Fe2+

, Fe3+

, and pH

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The Nernst Equation(SJ 331)

G = G + RT ln Q (1)

G = - RT ln Keq (2)

G = - n FE (3)G = - nFE (4)

From (3) EG

nF=

(3)

From (4) EG

nF =

(4)

Divide each term of Eq (1) by nF

G

nF

G

nF

RT

nFQ

=

+

ln (1() (1)

Substitute (3) and (4) into (1)

E ERT

nFQ= ln or or 102.303 log

RTE E Q

nF= (5)

where

R = 8.32 J/mole K = 8.32 volt. coulomb /mole K = 1.9872 x 10-3

kcal/mol K

T = 273 +25 =298 K at 25C

F = 23.06 kcal/ (volt e-equiv) = 96,500 coulomb/e-equiv

This is the Nernst equation.

Note: 1 Joule = 1 volt. coulomb = 0.2389 calorie

1 calorie = 4.185 Joule

At 25C (298 K)

( )3

10

1.9872 102.303 298

log23.06

x kcalK

mole KE E Q

kcalnvolt e equiv

=

i

or

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( )

10

8.32 .2.303 298

log96,500

volt coulombK

mole KE E Q

coulombn

e equiv

=

E En

Q= 0059 10. log

At equilibrium, G = 0, E = 0, and Q = Keq

En

Keq =0 059

10

.log

10log 16.9eqK nE= at 25C

Example Find the electrode potential, E (V), for the half reaction of sulfate, SO42-

to sulfite, SO32-

at

25C. The reaction takes place at [SO42-

] = 1 x10-5

and [SO32-

] = 1 x10-5

at pH = 5.0

E (V)

SO42-

+ 2H+ + 2e

- SO3

2- + H2O -0.04

10

23

10 22

4

0.059log

0.059log

E E Qn

SOE

n SO H

+

=

=

( ) ( )5

10 25 5

100.059 0.059log 0.04 10 0.335

10 10E E V

n n

= = =

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2

Example7-6: Daniel cell (SJ 332)

A zinc electrode in a zinc chloride solution connected to a copper electrode in a cupric chloride

solution. What is the equilibrium constant of the cell reaction at 25C?

e-

e-

Zn K+Cl

-Cu

Salt Bridge

Zn2+

Cl-

Cu2+

Cl-

Zn Zn2+ + 2e- Cu2++ 2e- CuAnode Cathode

(negative) (positive)

Voltim

eter

Note: Anode is where oxidation occurs An Ox

Cathode is where reduction occurs Red Cat

From Table 7-1 (SJ 329)

Half reaction E(V) G= -nFE

Ox: Zn(s) Zn2+ + 2e- +0.76 -2F(0.76)

Re: 2e-

+ Cu2+

Cu(s) +0.34 -2F(0.34)

---------------------------------------------------------------------------------------Redox: Zn(s)+ Cu

2+ Zn2+ + Cu(s) +1.10 -2FEZn(s),Cu2+

Note:-2FE Zn(s),Cu2+= -2F(0.76) -2F(0.34) = -2F(0.76 - 0.34)

E Zn(s),Cu2+ = +1.10

The Nernst Equation at equilibrium: ln eqRT

E KnF

=

At 25C, 100.059

logeq

E Kn

= 100.059

1.10 log2

eqK+ = Keq = 10

37.2

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or 10log 16.9 16.9(2)(1.10) 37.18K nE= = = Keq = 1037.2

Free Energy Eq: G G RT Q= + ln (1)

The Nernst Eq: E E

RT

nF Q= ln (2 (2 (2)

Dividing Eq (2) by RT/nFand changing E to EHyields

E

RT nF

E

RT nFQH

o

/ /ln=

where EH= redox potential (see SJ 335)

nFRT

E nFRT

E QHo= ln

nF

RTE

nF

RTE QH

o= 2 303. log

If n = 1

F

RTE

F

RTE QH

o

2 303 2 303. .log=

At equilibrium, G = 0 thus EH= 0

log2.303

eq

FE K

RT =

2.303log eq

RTE K

F =

log 2.303eq

FK E

RT=

exp2.303

eq

FK E

RT

=

or based on natural logarithm

If n 1

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2

ERT

nFK

KnF

RTE

KnF

RTE

eq

eq

eq

=

=

=

ln

ln

exp

Electrochemistry (FE manual)

An electrolyteis a substance that dissociate in solution to produce positive and negative ions.- It can be an aqueous solution of a soluble salt, or it can be an ionic substance in molten

form.

Electrolysisis the passage of an electric current through an electrolyte driven by an externalvoltage source.

- Electrolysis occurs when the positive terminal (the anode) and negative terminal (the

cathode) of a voltage source are placed in an electrolyte.- Negative ions (anions) will be attracted to the anode, where they are oxidized.

- Positive ions (cations) will be attracted to the cathode, where they will be reduced.

- The passage of ions constitute the current.

Electrolytic (electrochemical) reactions

- Some reactions that do not proceed spontaneously can be forced to proceed by supplyingelectrical energy. Such reactions are called electrolytic (electrochemical) reactions.

Faradays Laws of Electrolysis- Faradays Laws of Electrolysiscan be used to predict the duration and magnitude of a

direct current needed to complete an electrolytic reaction.

Law 1: The mass of a substance generated by electrolysis is proportional to the amount of

electricity used. Law 2: For any constant amount of electricity, the mass of substance generated is

proportional to its equivalent weight.

Law 3: Onefaradayof electricity (96,485 C or 96,485 As) will produce one gram

equivalent weight.

The number of gram of a substance produced at an electrode in an electrolytic reaction can befound from the equation:

( ) ( )no. of

96,485 change in oxidation state

I t MWm faradays GEW

= =

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The number of gram-moles produced (n) is

( )( )

no. of

96,485 change in oxidation state change in oxidation state

m I t faradaysn

MW

= = =

Example (FE manual): A current of 0.075 A passes through a solution of silver nitrate for 10 minutes.

How much silver is deposited?

(Solution)

The number of gram of a substance produced at an electrode in an electrolytic reaction can be foundfrom the equation (Faradays law):

( ) ( )no. of

96,485 change in oxidation state

I t MWm faradays GEW

= =

where m = mass of a substance produced, gI = current, A

t = time, s

MW = molecular weight or AW = atomic weight

Given:

I = 0.075 A

t = (10 min)(60 s/min) = 600 schange in oxidation state = 1/mol

AW of Ag = 107.87 g/mol.

( ) ( )( )

( )( )

0.075 600 107.87 / 0.050

96,485 1 /

A s g molm g

A s mol= =

Note: AgNO3Ag+ + NO3

AgNO3 is soluble.

Ag+ + e

- Ag Ag

+is readily reduced to Ag (metallic)

- Change in oxidation state, z = 1

- Silver exists as single atoms, so the molecular weight is the atomic weight, 107.87 g/mol.

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Example(FE manual): How many grams of copper will be deposited at an electrode if a current of 1.5

A is supplied for 2 hours to a CuSO4solution?

Given:

The dissociation reaction for copper sulfate is: CuSO4 Cu2+

+ SO42-

The electrolytic reaction equation is: Cu

2+

+ 2e

-

Cu

The change in oxidation number is 2 per atom of copper deposited; i.e.,

The change in oxidation state, z = 2/atom

I = 1.5 A

t = (2 hrs)(3600 s/hr) = 7200 s

AW of Cu = 63.5 g/mol.

Use Faradays law

( ) ( )no. of

96,485 change in oxidation stateI t MWm faradays GEW

= =

where m = mass of a substance produced, g

I = current, A

t = time, s

MW = molecular weight or AW = atomic weight

( )( )( )

( )( )

1.5 7200 63.5 / 3.55

96,485 2 /

A s g molm g

A s mol= =

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Electron activity and the negative logarithm of electron activity (SJ 338)

Electron activity, {e-}

Negative logarithm of electron activity, log {e-} = p

- a measure of the availability of electrons in solution (even though no free electrons exist insolution).

For the half rxn: M3+

+ 2 e- M

+

{M+}

K = -------------------

{M3+

} {e-}

2

1 {M+}

{e-}

2= ----- --------

K {M3+

}

{M+}

2 log {e-}

= log K - log -------

{M3+

}

Divide by 2:

1 1 {M+}

log {e-}

= --- log K --- log -------

2 2 {M3+

}

1 {M+}

p= p

--- log -------

2 {M3+}

where p= log {e

-}

p = 1/n log K

---------------------------

General half reaction (reduction)

For the half rxn, Ox + n e- Red

General eqn

1 {Red}

p= p

--- log -------

n {Ox}

where p= log { e

-} and p

= 1/n log K

Combining half reactions:

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Reduction M2+

+ e- M

+ K1

Oxidation H2(g) H+ + e

- K2

----------------------------------------------------------------------------------------------------------------

Redox M2+

+ H2(g) M+ + H

+ K3= K1 K2

Note: The reduction of H+to H2(g)at standard condition is assigned G = 0 and Keq= 1

(i.e., K2= 1 ). Since G = nFE , E = 0

{M+} {e

-}

K3= K1 because K2 = ------------ = 1 by convention

{H2}1/2

{M+}

K1= ---------------

{M2+

} {e-}

1 {M+}

{e-}

= ----- ---------

K1 {M2+

}

{M+}

- log {e-}

= log K1 - log ---------

{M2+

}

{M+}

p= p

log -------

{M2+

}

where p= log {e

-}

p

= 1/n log K1

Example (SJ 340): Reduction of ferric ion (Fe3+

) to ferrous ion (Fe2+

).

From Table 7-1 (p =16.9E see below)

Half Reaction ___ E (v) p___K = exp (nFE/RT)_______

(1) Fe3+

+ e Fe

2+ 0.77 13 K1= 10

13

(2) H2 (g) H+ + e

0 0 K2= 1 (by convention)

_______________________________________________________________________

(3) Fe3+

+ H2 (g) Fe2+

+ H+ 0.77 K3= K1K2 = 10

13

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From Rxn (1)

{ }

{ }{ }

{ } { }

{ }

{ } { }

{ }

KFe

Fe e

eK

Fe

Fe

e KFe

Fe

1

2

3

1

2

3

1

2

3

1

=

=

=

+

+

+

+

+

+log log log

{ }

{ }p p

Fe

Fe =

+

+log

2

3 (7-11)

where

p= - log {e-}

p = log K1 (p = 1/n log K1 )

when {Fe2+

} = {Fe3+

}, p= p = log K1

For the reaction (3)

{ }{ }

{ }{ }K K K

Fe H

Fe H g3 1 2

2

3

2

1 2= =

+ +

+

( )

/

Note {H+} / {H2(g)}

1/2 = 1 by convention

{ }

{ }{ }

{ }{ }

( )

{ }

{ }2( )

2 2

1/ 23 3

gH

Fe H e Fe

Fe e Fep

+ + +

+ +=

From Free energy equation:

{ }

{ } G G RT Fe

Fe= +

+

+ln

2

3 (7-12)

From Nernst Equation:

{ }

{ }E E

RT

nF

Fe

FeH =

+

+ln

2

3 (7-13)

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Potential E EH redox potential (see p. 335)

Divide (7-13) by RT/nF

{ }{ }+

+

=3

2

ln// Fe

Fe

nFRT

E

nFRT

EH

{ }

{ }

nF

RTE

nF

RTE

Fe

FeH

=

+

+ln

2

3

converting lnlog yields

{ }{ }

nF

RTE

nF

RTE

Fe

FeH

=

+

+2 303 10

2

3. log

Since n = 1

{ }

{ }

F

RTE

F

RTE

Fe

FeH2 303 2 303 10

2

3. .log

=

+

+ (7)

Comparing (7-11) and (7), we find

{ }

{ }

2

10 3log

Fep p

Fe

+

+= (7-11)

p F

RTE p

F

RTEH = =

2 303 2 303. .

Since

( )

pF

RTE

p

kcal

volt equiv

x kcal

mole KK

E E

H

H H

=

=

=

2 303

2306

2 3031987 10

298

16 93

.

.

.

..

.

(7-14)

or

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( )

pF

RTE

p

coulomb

equiv

Jmole K

K

E E

H

H H

=

=

=

2 303

96 500

2 303 8314 298

16 9

.

,

. .

.

16.9 16.92.303

H

Fp E E E

RT

= = = (7-15)

Since G = nFE, FE = G/nSince G = nFE , FE = G /n

pG

nR T =

2 303. (7-20)

and

pG

n RT =

2 303. (7-21)

p is proportional to the free energy change accompanying the transfer of 1 mole of electronfrom a reducing agent at unit activity to H

+at unit activity.