Redox Reactions and Electrochemistry

• Redox reactions

• Galvanic cells

• Standard reduction potential

• Cell potential

– Free energy

– Effect of concentration

• Battery

• Corrosion

• Electrolysis

I. Oxidization–Reduction (Redox) Reaction

2Fe 2Fe2+ + 4e-

O2 + 4e- 2O2-

Oxidation half-reaction (lose e-):Fe is oxidized and Fe is the reducing agent

Reduction half-reaction (gain e-):

O2 is reduced and O2 is the oxidizing agent

2Fe (s) + O2 (g) 2FeO (s)

Electrochemical processes are redox reactions in which:• energy released by a spontaneous reaction is converted to electricity• electrical energy is used to cause a nonspontaneous reaction

0 0 2+ 2-

A. Oxidation number/state

• Free elements have an oxidation number (ON) =0• Monatomic ions, the ON is equal to the charge on the ion.

• The ON of oxygen is usually –2. In H2O2 and O22-:–1, in O2

- : -1/2

• The ON of hydrogen is +1 except when it is bonded to metals in

binary compounds. In these cases, its oxidation number is –1.• Fluorine is always –1, Group IA metals are +1, IIA metals are +2

in compounds.• The sum of the ON of all the atoms in a molecule or ion is equal to

the charge on the molecule or ion.

Ex. Oxidation numbers of all the atoms in HSO3–

O =-2 H= +1 S=? 1 +3x (-2)+S= -1

S=4

B. Balance redox reaction

1. By inspection

2. Ion-electron method (half-reaction method)

By balancing the number of electrons lost and gainedduring the reaction

Cu+ H+ Cu2+ + H22

Sn2+ (aq) + Fe3+ (aq) Sn4+(aq) + Fe2+(aq)

Oxidation half reaction

Reduction half reaction

Sn2+ (aq) Sn4+(aq) + 2 e–

Fe3+ (aq) + e– Fe2+(aq)

2 2

( ) x 2

2 Fe3+ (aq) + 2e– 2Fe2+(aq)

Ion-electron method

Fe2+ + Cr2O72- Fe3+ + Cr3+

2. Separate the equation into two half-reactions.

Oxidation:

Cr2O72- Cr3+

+6 +3

Reduction:

Fe2+ Fe3++2 +3

3. Balance the atoms other than O and H in each half-reaction.

Cr2O72- 2Cr3+

1. Write the unbalanced equation for the reaction ion ionic form.

The oxidation of Fe2+ to Fe3+ by Cr2O72- in acid solution?

Ion-electron method

4. For reactions in acid, add H2O to balance O atoms and H+ to balance H atoms.

Cr2O72- 2Cr3+ + 7H2O

14H+ + Cr2O72- 2Cr3+ + 7H2O

5. Add electrons to one side of each half-reaction to balance the charges on the half-reaction.

Fe2+ Fe3+ + 1e-

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

6. If necessary, equalize the number of electrons in the two half-reactions by multiplying the half-reactions by appropriate coefficients.

6Fe2+ 6Fe3+ + 6e-

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

Ion-electron method

7. Add the two half-reactions together and balance the final equation by inspection. The number of electrons on both sides must cancel.

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

6Fe2+ 6Fe3+ + 6e-Oxidation:

Reduction:

14H+ + Cr2O72- + 6Fe2+ 6Fe3+ + 2Cr3+ + 7H2O

8. Verify that the number of atoms and the charges are balanced.

14x1 – 2 + 6x2 = 24 = 6x3 + 2x3

9. For reactions in basic solutions, add OH- to both sides of the equation for every H+ that appears in the final equation.

Summary of balancing redox reaction in acidic solutions

1. Divide reaction into two incomplete half-reactions2. Balance each half-reaction by doing the following:

a. Balance all elements, except O and H

b. Balance O by adding H2O

c. Balance H by adding H+d. Balance charge by adding e– as needed

3. If the electrons in one half-reaction do not balance those in the other, then multiply each half-reaction to get a common multiple

4. The overall reaction is the sum of the half-reactions5. The oxidation half-reaction is the one that produces electrons as

products, and the reduction half-reaction is the one that uses electrons as reactants

Summary of balancing redox reaction in basic solutions

Steps 1-3 are the same as the reaction in acidic solution4. Since we are under basic conditions, we neutralize any H+ ions

in solution by adding an equal number of OH– to both sides5. The overall reaction is the sum of the half-reactions

Ex. Balance I– + OCl– I2 + Cl– in basic solution?

Oxidation I– I2

Reduction

OCl– Cl–

2 H2O+2 e– + OCl– Cl– + H2O + 2 OH–

2 e– + H2O + OCl– Cl– + 2 OH–

2 I– I2 + 2 e–

2 I– + H2O + OCl– I2 + Cl– + 2 OH–

Add up

+ H2O2 H+ +2 OH– + + 2 OH–2e– +

II. Galvanic Cells

• Consider reaction Zn(s) + Cu(NO3)2 --> Cu(s) + Zn(NO3)2

Spontaneous (activity series, chapter 4)half reaction: Zn(s) --> Zn2+ + 2e– Cu2+ +2 e– --> Cu(s)

• Galvanic cell: a device uses the spontaneous redox reaction to produce an electric current

OxidationZn --> Zn2+ + 2e–

Anode

ReductionCu2+ +2 e– --> Cu

Cathode

e– flowe–

Salt bridge

Flow of cation

Flow of anion

Wire

Electrode: anode (oxidation rxn) and cathode (reduction rxn)Salt bridge: contains conc. soln of strong electrolyte (KCl)

II. Galvanic Cells

anodeoxidation

cathodereduction

II. Galvanic cell

Cell diagram : a conventional notation to represent galvanic cell

Zn (s) + Cu2+ (aq) Cu (s) + Zn2+ (aq)

[Cu2+] = 1 M and [Zn2+] = 1 M

Zn (s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu (s)anode cathode

1. Anode is on the left, cathode on the right2. | : represent boundary between phases3. || : represent salt bridge

Ex. Draw cell diagram for the following reaction:

Al (s) + Ag+ (aq) Al3+ (s) + Ag (s)

Al (s) | Al3+ || Ag+ | Ag (s)

III. Standard reduction potential

A. Cell voltage, cell potential, electromotive force (emf)

1. The potential energy of the e– is higher at the anode than at the cathode

2. The difference in electrical potential between the anode and cathode is

called as emf

3. Unit: V (volt) = 1 J/C

C: coulomb, unit of charge

4. Standard cell potential, Eocell

Cell potential under standard condition

Zn(s)| Zn2+(1 M) || H+ (1 M)| H2 (1 atm)| Pt (s)

2e- + 2H+ (1 M) H2 (1 atm)

Zn (s) Zn2+ (1 M) + 2e-Anode (oxidation):

Cathode (reduction):

Zn (s) + 2H+ (1 M) Zn2+ + H2 (1 atm)

B. Standard reduction potentialStandard reduction potential (E0) is the voltage associated with a reduction reaction at an electrode when all solutes are 1 M and all gases are at 1 atm.

E0 = 0 V

Standard hydrogen electrode (SHE) : half-cell contain 1M H+ and

1 atm H2, used as a reference to determine Eo for other species

2e- + 2H+ (1 M) H2 (1 atm)

Reduction Reaction

B. Standard reduction potential

E0 = 0.76 Vcell

Standard emf (E0cell )

0.76 V = 0 - EZn /Zn 0

2+

EZn /Zn = -0.76 V02+

Zn2+ (1 M) + 2e- Zn E0 = -0.76 V

E0 = Ecathode - Eanodecell0 0

Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)

E0 = EH /H - EZn /Zn cell0 0

+ 2+2

B. Standard reduction potential

Pt (s) | H2 (1 atm) | H+ (1 M) || Cu2+ (1 M) | Cu (s)

2e- + Cu2+ (1 M) Cu (s)

H2 (1 atm) 2H+ (1 M) + 2e-Anode (oxidation):

Cathode (reduction):

H2 (1 atm) + Cu2+ (1 M) Cu (s) + 2H+ (1 M)

E0 = Ecathode - Eanodecell0 0

E0 = 0.34 Vcell

Ecell = ECu /Cu – EH /H 2+ +2

0 0 0

0.34 = ECu /Cu - 00 2+

ECu /Cu = 0.34 V2+0

• E0 is for the reaction as written

• The more positive E0 the greater the tendency for the substance to be reduced

• The half-cell reactions are reversible

• The sign of E0 changes when the reaction is reversed

• Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E0

What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Cr electrode in a 1.0 M Cr(NO3)3 solution?

Cd2+ (aq) + 2e- Cd (s) E0 = -0.40 V

Cr3+ (aq) + 3e- Cr (s) E0 = -0.74 V

Cd is the stronger oxidizer

Cd will oxidize Cr

2e- + Cd2+ (1 M) Cd (s)

Cr (s) Cr3+ (1 M) + 3e-Anode (oxidation):

Cathode (reduction):

2Cr (s) + 3Cd2+ (1 M) 3Cd (s) + 2Cr3+ (1 M)

x 2

x 3

E0 = Ecathode - Eanodecell0 0

E0 = -0.40 – (-0.74) cell

E0 = 0.34 V cell

Ex.

Calculate Eocell

for the following reactions:

(a) 2 Al (s) + 6 H+ -> 2 Al3+ + 3 H2 (g)

(b) H2O2 + 2 H+ + Cu (s) -> Cu2+ + 2 H2O

6 H+(aq) + 6e- 3 H2(g)

2 Al(s) 2 Al3+ + 6e-Anode (oxidation):

Cathode (reduction):

Ex.

(a) 2 Al (s) + 6 H+ -> 2 Al3+ + 3 H2 (g)

E0 = EH /H - EAl /Al cell0 0

+ 3+2 = 1.66 V

E0 = Ecathode - Eanodecell0 0

H2O2+2 H+ +2e- 2 H2O

Cu(s) Cu2+ + 2e-Anode (oxidation):

Cathode (reduction):

(b) H2O2 + 2 H+ + Cu (s) -> Cu2+ + 2 H2O

= 1.77 V- 0.34 V = 1.43 VE0 = E - ECu /Cu cell0

2+H2O2/H2 O

IV. Cell Potential

• Free energy : spontaneity of redox reactions

G = -nFEcell

G0 = -nFEcell0

n = number of moles of electrons in reaction

F = 96,500J

V • mol = 96,500 C/mol

G0 = -RT ln K = -nFEcell0

Ecell0 =

RTnF

ln K(8.314 J/K•mol)(298 K)

n (96,500 J/V•mol)ln K=

=0.0257 V

nln KEcell

0

=0.0592 V

nlog KEcell

0

A. Free energy and spontaneity of redox reaction

G = -nFEcell

G0 = -nFEcell0

=0.0257 V

nln KEcell

0

=0.0592 V

nlog KEcell

0

Ex.

2e- + Fe2+ Fe

2Ag 2Ag+ + 2e-Oxidation:

Reduction:

=0.0257 V

nln KEcell

0

E0 = -0.44 – (0.80) E0 = -1.24 V

0.0257 Vx nE0 cellexpK =

n = 2

0.0257 Vx 2-1.24 V

= exp

K = 1.23 x 10-42

E0 = EFe /Fe – EAg /Ag0 0

2+ +

What is the equilibrium constant for the following reaction at 250C? Fe2+ (aq) + 2Ag (s) Fe (s) + 2Ag+ (aq)

B. Concentration effect on cell potential

G = G0 + RT ln Q G = -nFE G0 = -nFE 0

-nFE = -nFE0 + RT ln Q

E = E0 - ln QRTnF

Nernst equation

At 298 K

-0.0257 V

nln QE0E = -

0.0592 Vn

log QE0E =

Ex.

2e- + Fe2+ Fe

Cd Cd2+ + 2e-Oxidation:

Reduction:n = 2

= -0.44 – (-0.40) V = -0.04 V E0 = EFe /Fe – ECd /Cd0 0

2+ 2+

-0.0257 V

nln QE0E =

-0.0257 V

2ln -0.04 VE =

0.0100.60

E = 0.013 E > 0 Spontaneous

Will the following reaction occur spontaneously at 250C if [Fe2+] = 0.60 M and [Cd2+] = 0.010 M? Fe2+ (aq) + Cd (s) Fe (s) + Cd2+ (aq)

[Cd2+][Fe2+]

Q =

Ex.

2e- +2 Ag+ 2 Ag

Ni Ni2+ + 2e-Oxidation:

Reduction:n = 2

= 0.80 – (-0.25) V = 1.05V E0 = EAg /Ag – E Ni /Ni0 0

+ 2+

A redox reaction for the oxidation of nickel by silver ion is set up with an initial concentration of 5.0 M silver ion and 0.050 M nickel ion. What is the cell EMF at 298K?

-0.0257 V

nln QE0E =

-0.0257 V

2ln 1.05 VE =

0.0505.02 E = 1.13

[Ni2+][Ag+]2

Q =

Ni + 2 Ag+ Ni2+ + 2 Ag

V. Battery

Leclanché cell

Dry cell

Zn (s) Zn2+ (aq) + 2e-Anode:

Cathode: 2NH4 (aq) + 2MnO2 (s) + 2e- Mn2O3 (s) + 2NH3 (aq) + H2O (l)+

Zn (s) + 2NH4+ (aq) + 2MnO2 (s) Zn2+ (aq) + 2NH3 (aq) + H2O (l) + Mn2O3 (s)

About 1.5 V

V. Battery

Zn(Hg) + 2OH- (aq) ZnO (s) + H2O (l) + 2e-Anode:

Cathode: HgO (s) + H2O (l) + 2e- Hg (l) + 2OH- (aq)

Zn(Hg) + HgO (s) ZnO (s) + Hg (l)

Mercury Battery

1.35 V

V. Battery

Anode:

Cathode:

Lead storagebattery

PbO2 (s) + 4H+ (aq) + SO2- (aq) + 2e- PbSO4 (s) + 2H2O (l)4

Pb (s) + SO2- (aq) PbSO4 (s) + 2e-4

Pb (s) + PbO2 (s) + 4H+ (aq) + 2SO2- (aq) 2PbSO4 (s) + 2H2O (l)4

Total 12V

VI. Corrosion

• Deterioration of metal by an electrochemical process

VI. Corrosion

• Cathodic Protection of an Iron Storage Tank

VII. Electrolysis

Electrolysis is the process in which electrical energy is used to cause a nonspontaneous chemical reaction to occur.

A. Electrolysis of water

B. Quantitative aspect of electrolysis

charge (C) = current (A) x time (s)

1 mole e- = 96,500 C

Ex. It will take the greates amount electricity to produce 2 mol of

the ____ metal by electrolysis.

(1) potassium (2) calcium (3) aluminum (4) silver

Ex.

Anode:

Cathode: Ca2+ (l) + 2e- Ca (s)

2Cl- (l) Cl2 (g) + 2e-

Ca2+ (l) + 2Cl- (l) Ca (s) + Cl2 (g)

2 mole e- = 1 mole Ca

mol Ca = 0.452Cs

x 1.5 hr x 3600shr 96,500 C

1 mol e-

x2 mol e-

1 mol Cax

= 0.0126 mol Ca

= 0.50 g Ca

How much Ca will be produced in an electrolytic cell of molten CaCl2 if a current of 0.452 A is passed through the cell for 1.5 hours?