electrochemistry

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Electrochemistry = the area of chemistry that deals with the interconversion of electrical energy and chemical energy.

Electrochemical processes are redox reactions in which the energy released by a spontaneous reaction is converted to electricity or in which electricity is used to drive a non-spontaneous chemical reaction.

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Lithium-ion battery.

ELECTROCHEMISTRY

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Galvanic or Voltaic cell = an electrochemical cell that produces electricity as a result of spontaneous chemical change

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Why Does a Voltaic Cell Work?

The spontaneous reaction occurs as a result of the different abilities of materials (such as metals) to give up their electrons and the ability of the electrons to flow through the circuit.

Ecell > 0 for a spontaneous reaction

1 Volt (V) = 1 Joule (J)/ Coulomb (C)

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Salt bridge

Zn2+ Cu2+

K+Zn CuCl–

Voltmeter

(–) (+)

Example of Voltaic Cell

Zn and Cu cell

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Salt bridge

Zn2+ Cu2+

K+Zn CuCl–

Voltmeter

(–) (+)

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Oxidation half-reactionZn(s)

Salt bridge

Zn2+ Cu2+

K+Zn CuCl–

Zn2+(aq) + 2e–

Voltmetere–

Anode(–) (+)

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Zn2+Zn

Oxidation half-reactionZn(s)

Salt bridge

Zn2+ Cu2+

K+Zn CuCl–

Zn2+(aq) + 2e–

Voltmetere–

2e– lostper Zn atomoxidized

Anode(–) (+)

e–

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Zn2+Zn

Oxidation half-reaction

Reduction half-reaction Cu2+(aq) + 2e–

Zn(s)

Salt bridge

Zn2+ Cu2+

K+Zn CuCl–

Zn2+(aq) + 2e–

Cu(s)

Voltmetere– e–


Anode(–)

Cathode (+)

e–

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Cu2+ e–Cu

2e– gainedper Cu2+ ionreduced

Zn2+Zn


Reduction half-reaction Cu2+(aq) + 2e–

Zn(s)

Salt bridge Anode(–)

Cathode (+)

Zn2+ Cu2+

K+Zn CuCl–

Zn2+(aq) + 2e–

Cu(s)

Voltmetere– e–


e–

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Cu2+ e–Cu

2e– gainedper Cu2+ ionreduced

Zn2+Zn


Reduction half-reaction

Overall (cell) reactionZn(s) + Cu2+(aq)

Cu2+(aq) + 2e–

Zn(s)

Salt bridge

Zn2+ Cu2+

K+Zn CuCl–

Zn2+(aq) + 2e–

Cu(s)

Zn2+(aq) + Cu(s)

　 1.10 Ve– e–

Anode(–)

Cathode (+)


e–

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Figure 21.5 A voltaic cell based on the zinc-copper reaction.

Oxidation half-reactionZn(s) Zn2+(aq) + 2e-

Reduction half-reactionCu2+(aq) + 2e- Cu(s)

Overall (cell) reactionZn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

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Symbolic representation of cell diagram

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Notation for a Voltaic Cell

components of anode compartment

(oxidation half-cell)

components of cathode compartment

(reduction half-cell)

phase of lower oxidation state

phase of higher oxidation state

phase of higher oxidation state

phase of lower oxidation state

phase boundary between half-cells

Examples: Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu (s)

Zn(s) Zn2+(aq) + 2e- Cu2+(aq) + 2e- Cu(s)

| = represents boundary between an electrode and another phase (solution or gas)

|| = signifiess that solutions are joined by a salt bridge

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* Another way of representing the cell diagram:

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

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Salt bridge

Cu(NO3)2 AgNO3

K+Cu AgCl–

Voltmeter

(–) (+)

Exercise: Write the cell diagram of the following electrochemical cell and identify the oxidation and reduction couple:

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Salt bridge

Cu(NO3)2 AgNO3

K+Cu AgCl–

Voltmeter

(–) (+)

Exercise: Write the cell diagram of the following electrochemical cell and identify the oxidation and reduction couple:

Cell Diagram: Cu (s) | Cu2+(aq) || Ag+(aq) | Ag (s)

Oxidation Cu → Cu2+ + 2 e-

Reduction (Ag+ + e- → Ag)2

2 Ag+ + Cu → Cu2+ + 2 Ag

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Eo = Standard Electrode Potential = is based on the tendency for a reduction process to occur at the electrode

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•By convention, electrode potentials are written as reductions. The Eo values apply to the half-cell reactions as read in the forward direction (left to right).

•The more positive Eo, the greater the tendency to be reduced.

•The half-cell reactions are reversible, (sign changes). The reduction half-cell potential and the oxidation half-cell potential are added to obtain the E0

cell.

•Eo values are unaffected by multiplying half-equations by constant coefficients.

Rules:

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Table 21.2 Selected Standard Electrode Potentials (298K)

Half-Reaction E0(V)

2H+(aq) + 2e- H2(g)

F2(g) + 2e- 2F-(aq)

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

MnO2(g) + 4H+(aq) + 2e- Mn2+(aq) + 2H2O(l)

NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l)

Ag+(aq) + e- Ag(s)

Fe3+(g) + e- Fe2+(aq)

O2(g) + 2H2O(l) + 4e- 4OH-(aq)

Cu2+(aq) + 2e- Cu(s)

N2(g) + 5H+(aq) + 4e- N2H5+(aq)

Fe2+(aq) + 2e- Fe(s)

2H2O(l) + 2e- H2(g) + 2OH-(aq)

Na+(aq) + e- Na(s)

Li+(aq) + e- Li(s)

+2.87

-3.05

+1.36

+1.23

+0.96

+0.80

+0.77

+0.40

+0.34

0.00

-0.23

-0.44

-0.83

-2.71

strength

of reducin

g agent

stre

ngt

h o

f ox

idiz

ing

agen

t

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Example:

Cu (s) | Cu2+(aq) || Ag+(aq) | Ag (s)

Oxidation Cu → Cu2+ + 2 e- - 0.34

Reduction (Ag+ + e- → Ag)2 0.80

2 Ag+ + Cu → Cu2+ + 2 Ag 0.46 V

*reaction is spontaneous

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Answers:

a) Ox: Zn → Zn2+ + 2e- 0.76 V

Red: Sn2+ + 2e- → Sn -0.14 V

Zn + Sn2+ → Zn2+ + Sn Eo = 0.62 V, spontaneous

b) Ox: 2 (Fe2+ → Fe3+ + 1e-) -0.77 V

Red: Sn4+ + 2e- → Sn2+ 0.15 V

2 Fe2+ + Sn4+ → 2 Fe3+ + Sn2+ Eo = -0.62 V, non-spo

c) Ox: 2 (Al → Al3+ + 3e-) 1.66 V

Red: 3 (Cu2+ + 2e- → Cu) 0.34 V

2 Al + 3 Cu2+ → 3 Cu + 2 Al3+ Eo = 2.00 V,spontaneous

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Problem:2) Give the cell diagram and determine the Eo cell for the reaction: Zn(s) + Cl2(g) ZnCl2(aq).

Answers:

Cell diagram:

Zn(s) | Zn2+(aq) | | Cl2(g), Cl-(aq) | Pt (s)

Ox: Zn → Zn2+ + 2e- 0.76 V

Red: Cl2 + 2e- → 2 Cl- 1.36 V

Zn + Cl2 → Zn2+ + 2 Cl- 2.12 V = Eo

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Free Energy = ▲G

▲G = ▲H - T▲S

▲G = 0 , the system is at equilibrium

▲G < 0 , spontaneous reaction

▲G > 0 , non-spontaneous reaction

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Free Energy = ▲G

▲G = ▲H - T▲S

▲G = ( - ), always a spontaneous process

in ELECTROCHEMISTRY

▲G = Wmax (max. amt. of work that can be done)

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Voltages of Some Voltaic or Galvanic Cells

Voltaic Cell Voltage (V)

Common alkaline battery

Lead-acid car battery (6 cells = 12V)

Calculator battery

Electric eel

in a voltaic cell, chemical energy is converted into electrical energy

2.0

1.5

1.3

0.15

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Electrical energy = volts X coulombs = Joules

The total charge is determined by the n of electrons that pass through the circuit.

By definition:

total charge = n F (1 F = 96,500 C/mol)

since 1 J = 1 C X 1 VWe can also express the units of Faraday as

1 F = 96,500 J/V.mol

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The total work done is the product of 3 terms:

1. The emf (voltage) of the cell

2. The n of e-s transferred between the electrodes

3. The electric charge per mole of e-s.

Welec = (n) (F) (Eocell)

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Welec = (n) (F) (Eocell)

Since for a spontaneous reaction ▲G < 0

▲G = - (n) (F) (Eocell)

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Problem: Determine ▲G for the reaction: Alo(s) | Al3+(aq) | | Cu2+(aq) | Cuo (s)

oxi: Al → Al3+ + 3 e- X 2 1.66 V

red: Cu2+ + 2 e- → Cu X 3 0.34 V

2 Al + 3 Cu2+ → 2 Al3++ 3 Cu 2.00 V = Eo

▲G = - (n F Eocell)

= - ( 6 mol e- X 96500 C/mol e- X 2.00 V )

= - 1,158,000 J or

= - 1,158 kJ

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Problem:2) Give the cell diagram and determine ▲G for the reaction: Zn(s) + Cl2(g) ZnCl2(aq).

Answers:

Cell diagram:

Zn(s) | Zn2+(aq) | | Cl2(g) , Cl-(aq) | Pt (s)

Ox: Zn → Zn2+ + 2e- 0.76 V

Red: Cl2 + 2e- → 2 Cl- 1.36 V

Zn + Cl2 → Zn2+ + 2 Cl- 2.12 V = Eo

▲G = - (n F Eocell)

= - ( 2 mol e- X 96500 C/mol e- X 2.12 V )

= - 409,160 J or - 409.2 kJ

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EXERCISES:

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I. Give the following:a) cell diagram, c) Eo for the net reaction andb) anode and cathode reactions, d) ▲G for the following reactions

1. Ag+(aq) + Mgo(s) → Mg2+(aq) + Ago(s)

2. I2(s) + Cl-(aq) → Cl2(g) + I-(aq)

3. Br2o(aq) + Fe2+(aq) → FeBr3(aq)

4. Pb2+(aq) is displaced from solution by Al(s)

5. MgBr2(aq) is produced from Mg(s) and Br2(l)

6. Cl2(g) is reduced to Cl-(aq) and Fe(s) is oxidized to Fe2+(aq)

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Sketch or draw and label the voltaic cell from the spontaneous reaction of Cu2+ and Sn+2

solutions. Indicate the following: the a) anode,

b) cathode,

c) oxidation reaction,

d) reduction reaction,

e) final equation, and

f) direction of electron flow.

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Arrange the following metals in the order of decreasing oxidizing strength:

Ni, Au, Ag, Al, and Cr.

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CORROSION OF IRON

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CORROSION OF IRON

No indication of corrosion Basic or alkaline solutions (NaOH, KOH, Mg(OH)2)

With indication of corrosion Neutral solutions (including water, Na2SO4, K3PO4) Acidic solutions (HCl, H2SO4, NH4Cl)

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Table 21.2 Selected Standard Electrode Potentials (298K)

Half-Reaction E0(V)

2H+(aq) + 2e- H2(g)

F2(g) + 2e- 2F-(aq)

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

MnO2(g) + 4H+(aq) + 2e- Mn2+(aq) + 2H2O(l)

NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l)

Ag+(aq) + e- Ag(s)

Fe3+(g) + e- Fe2+(aq)

O2(g) + 2H2O(l) + 4e- 4OH-(aq)

Cu2+(aq) + 2e- Cu(s)

N2(g) + 5H+(aq) + 4e- N2H5+(aq)

Fe2+(aq) + 2e- Fe(s)

Cr3+(aq) + 3e- Cr(s)

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

Mg2+(aq) + 2e- Mg(s)

+2.87

-2.37

+1.36

+1.23

+0.96

+0.80

+0.77

+0.40

+0.34

0.00

-0.23

-0.44

-0.74

-0.76

strength

of reducin

g agent

stre

ngt

h o

f ox

idiz

ing

agen

t

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Corrosion of the nail occurs at strained regions (more anodic). Contact with zinc protects the nail from corrosion. Zinc is oxidized instead of the iron (forming the faint white precipitate of zinc ferricyanide). Copper does not protect the nail from corrosion.

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The use of sacrificial anodes to prevent iron corrosion.

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Energy is absorbed to drive a nonspontaneous redox reaction

General characteristics of voltaic and electrolytic cells.

VOLTAIC CELL ELECTROLYTIC CELLEnergy is released from

spontaneous redox reaction

Reduction half-reactionY++ e- Y

Oxidation half-reactionX X+ + e-

System does work on its surroundings

Reduction half-reactionB++ e- B

Oxidation half-reactionA- A + e-

Surroundings(power supply)do work on system(cell)

Overall (cell) reactionX + Y+ X+ + Y; G < 0

Overall (cell) reactionA- + B+ A + B; G > 0

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Comparison of Voltaic and Electrolytic Cells

Cell Type G Ecell

Electrode

Name Process Sign

Voltaic

Voltaic

Electrolytic

Electrolytic

< 0

< 0

> 0

> 0

> 0

> 0

< 0

< 0

Anode

Anode

Cathode

Cathode

Oxidation

Oxidation

Reduction

Reduction

-

-

+

+

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Electrolysis of PbBr2(s)

Oxi: 2 Br- → Br2(g) + 2 e-

Red: Pb2+ + 2 e- → Pb(s)

Final: Pb2+ + 2 Br- → Pb(s) + Br2(g)

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Electrolysis of Dilute H2SO4

Electrolysis of water H2SO4 serves as a catalyst

As the electrolysis proceeds, the H2SO4 will become more and more concentrated as the water is used up.

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The electrolysis of water.

Oxidation half-reaction2H2O(l) 4H+(aq) + O2(g) + 4e-

Reduction half-reaction2H2O(l) + 4e- 2H2(g) + 2OH-(aq)

Overall (cell) reaction2H2O(l) 2H2(g) + O2(g)

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Electrolysis of KI(aq)

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Electrolysis of KI(aq)

Oxi: 2 I- → I2 + 2 e-

Red: 2 H+ + 2 e- → H2

Final: 2 H+ + 2 I- → H2 + I2

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Figure 21.11 A concentration cell based on the Cu/Cu2+ half-reaction.

Overall (cell) reactionCu2+(aq,1.0M) Cu2+(aq, 0.1M)

Oxidation half-reactionCu(s) Cu2+(aq, 0.1M) + 2e-

Reduction half-reactionCu2+(aq, 1.0M) + 2e- Cu(s)

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Quantitative Aspects of Electrolysis

Faraday’s Law: the mass of product formed (or reactant consumed) at an electrode is proportional to both the amount of electricity transferred at the electrode and the molar mass of the substance in question.

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E.g. 1) cathode reaction for CuSO4 solution

Cu2+ + 2 e- → Cuo

2) cathode reaction for the electrolysis of molten NaClNa+ + 1 e- → Nao

According to Faraday’s law:

a) The mass of Na produced is proportional to its atomic mass divided by one, and the mass of Cu produced is proportional to its atomic mass divided by two.

b) 1 mole of Cu ion reacts with 2 F of electricity to form 1 mole of Cu atom, and 1 mole of Na ion reacts with 1 F of electricity to form 1 mole of Na atom.

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Cu+2 + 2 e- → Cuo 2 F = 1 mole Cu2+

Na+ + 1 e- → Na 1 F = 1 mole Na+

Al3+ + 3 e- → Alo 3 F = 1 mole Al+3

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In an electrolysis experiment, we generally measure the current (in amperes) that passes through an electrolytic cell in a given period of time.

1 C = 1 A X 1 s

that is, a coulomb is the quantity of electrical charge passing any point in the circuit in 1 second when the current is 1 ampere.

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A summary diagram for the stoichiometry of electrolysis.

MASS (g)of substance oxidized or

reduced

MASS (g)of substance oxidized or

reduced

AMOUNT (MOL)of electrons transferred

AMOUNT (MOL)of electrons transferred

AMOUNT (MOL)of substance oxidized or

reduced

AMOUNT (MOL)of substance oxidized or

reduced

CHARGE (C)CHARGE (C)

CURRENT (A)CURRENT (A)

balanced half-reaction

Faraday constant (C/mol e-)

M(g/mol)

time(s)

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Problems:

1. A current of 0.452 A is passed through an electrolytic cell containing molten CaCl2 for 1.50 hours. Write the electrode reactions and calculate the quantity of products (in grams) formed at the electrodes.

Anode: 2 Cl- → Cl2 + 2 e-

Cathode: Ca2+ + 2 e- → Ca

Final: Ca2+ + 2 Cl- → Ca + Cl2

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Answers:

?C = 0.452 A X 1.50 hr X 3600s X 1C

1 hr 1 A X s

= 2.44 X 103 C

? g Ca = 2.44 X 103 C X 1 F X 1 mol Ca X 40.08 g Ca

96,500 C 2 F 1 mol Ca

= 0.507 g Ca

? g Cl2 = 2.44X103 C X 1 F X 1 mol Cl2 X 70.90 g Cl2

96,500 C 2 F 1 mol Cl2

= 0.896 g Cl2

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Problems:

2. If you wish to silver plate a spoon will you make it the cathode or the anode? How much Ag will be deposited by 0.3 F of electric charge?

Ans. The object to be plated is always made the cathode.

Ag+ + 1 e- → Ago

x = (0.3 F) (1 mol Ag+ / 1 F) (107.87 g Ag / 1 mol Ag+)

x = 32.4 g of Ag

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Problems:

3. How many moles of electrons are needed to deposit 0.45 g of aluminum from fused AlCl3?

Ans. Al3+ + 3 e- → Alo

x = (0.45 g) (1 mol) (3 mol e-)

27 g Al 1 mol

x = 0.05 mol e-

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Problems:

4. Calculate the time required for a current of 1.8 Amperes to deposit 12.5 grams of copper from CuSO4 solution.

Ans. Cu+2 + 2 e- → Cuo

time = (g) (n) (F)

(A) (M)

= (12.5 g) (2 e-) (96,500 C)

(1.8 A) (64 g/mole)

= 20,942 secs. Or 5.82 hours

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Problems:

5. A constant current is passed through an electrolytic cell containing molten MgCl2 for 18 hours. If 4.8 X 105 grams of Cl2 are obtained, what is the current in Amperes?

Ans. = 2.0 X 104 A

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The corrosion of iron.

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Enhanced corrosion at sea.

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The effect of metal-metal contact on the corrosion of iron.

faster corrosion cathodic protection

electrochemistry

Education

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