Electrochemistry

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.

1

ELECTROCHEMISTRY
Lithium-ion battery.

2

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

3

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)

4

Example of Voltaic Cell

Voltmeter

Salt bridge
Zn (–) K +
(+) Cu
Cl–

Zn2+ Cu2+

Zn and Cu cell
5

Voltmeter

Salt bridge
Zn (–) K +
(+) Cu
Cl–

Zn2+ Cu2+

6

Voltmeter
e–
Anode Salt bridge
Zn (–) K (+) Cu
+
Cl–

Zn2+ Cu2+

Oxidation half-reaction
Zn(s) Zn2+(aq) + 2e–

7

e–

2e– lost
per Zn atom
oxidized
Zn
Zn2+
Voltmeter
e–
Anode Salt bridge
Zn (–) K (+) Cu
+
Cl–

Zn2+ Cu2+


8

e–

2e– lost
per Zn atom
oxidized
Zn
Zn2+
Voltmeter
e– e–
Anode Salt bridge Cathode
Zn (–) K (+) Cu
+
Cl–

Zn2+ Cu2+


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

9

e–
2e– gained
per Cu2+ ion
2e– lost
reduced
per Zn atom
oxidized
Zn Cu2+ Cu e
–
Zn2+

Voltmeter
e– e–
Zn (–) K (+) Cu
+
Cl–

Zn2+ Cu2+



10

e–
2e– gained
per Cu2+ ion
2e– lost
reduced
per Zn atom
oxidized
Zn Cu2+ Cu e
–
Zn2+

　 1.10 V
e– e–
Zn (–) K (+) Cu
+
Cl–

Zn2+ Cu2+


Overall (cell) reaction 11
Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

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

Oxidation half-reaction Reduction half-reaction
Zn(s) Zn2+(aq) + 2e- Cu2+(aq) + 2e- Cu(s)

Overall (cell) reaction
Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
12

Symbolic representation of cell diagram

14

Notation for a Voltaic Cell

components of components of
anode compartment cathode compartment
(oxidation half-cell) (reduction half-cell)

phase of lower phase of higher phase of higher phase of lower
oxidation state oxidation state oxidation state 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
15

* Another way of representing the cell
diagram:

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

16

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

Salt bridge
Cu (–) K +
(+) Ag
Cl–

Cu(NO3)2 AgNO3

17

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

Salt bridge
Cu (–) K +
(+) Ag
Cl–

Cu(NO3)2 AgNO3

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 18

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

19

Rules:

•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
E0cell.

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

20

Table 21.2 Selected Standard Electrode Potentials (298K)

Half-Reaction E0(V)
F2(g) + 2e- 2F-(aq) +2.87
Cl2(g) + 2e- 2Cl-(aq) +1.36
MnO2(g) + 4H+(aq) + 2e- Mn2+(aq) + 2H2O(l) +1.23
NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l) +0.96

strength of reducing agent
Ag+(aq) + e- Ag(s) +0.80
strength of oxidizing agent

Fe3+(g) + e- Fe2+(aq) +0.77
O2(g) + 2H2O(l) + 4e- 4OH-(aq) +0.40
Cu2+(aq) + 2e- Cu(s) +0.34
2H+(aq) + 2e- H2(g) 0.00
N2(g) + 5H+(aq) + 4e- N2H5+(aq) -0.23
Fe2+(aq) + 2e- Fe(s) -0.44
2H2O(l) + 2e- H2(g) + 2OH-(aq) -0.83
Na+(aq) + e- Na(s) -2.71
Li+(aq) + e- Li(s) -3.05
21

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

22

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
24

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

25

Free Energy = ▲G
▲G = ▲H - T▲S

▲G = 0 , the system is at equilibrium

▲G < 0 , spontaneous reaction

▲G > 0 , non-spontaneous reaction
26

Free Energy = ▲G
▲G = ▲H - T▲S

▲G = ( - ), always a spontaneous process

in ELECTROCHEMISTRY

▲G = Wmax (max. amt. of work that can be done)
27

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

Voltages of Some Voltaic or Galvanic Cells
Voltaic Cell Voltage (V)

Common alkaline battery 1.5

Lead-acid car battery (6 cells = 12V) 2.0

Calculator battery 1.3

Electric eel 0.15

28

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 V
We can also express the units of Faraday as

1 F = 96,500 J/V.mol
29

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)

30

Welec = (n) (F) (Eocell)

Since for a spontaneous reaction ▲G < 0

▲G = - (n) (F) (Eocell)

31

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

oxi: Al → Al3+ + 3 e- X2 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
32

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 )
33
= - 409,160 J or - 409.2 kJ

I. Give the following:
a) cell diagram, c) Eo for the net reaction and
b) 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)

35

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.

36

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

Ni, Au, Ag, Al, and Cr.

37

CORROSION OF IRON

39

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)
40

Table 21.2 Selected Standard Electrode Potentials (298K)

Half-Reaction E0(V)
F2(g) + 2e- 2F-(aq) +2.87
Cl2(g) + 2e- 2Cl-(aq) +1.36
MnO2(g) + 4H+(aq) + 2e- Mn2+(aq) + 2H2O(l) +1.23
NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l) +0.96

strength of reducing agent
Ag+(aq) + e- Ag(s) +0.80
strength of oxidizing agent

Fe3+(g) + e- Fe2+(aq) +0.77
O2(g) + 2H2O(l) + 4e- 4OH-(aq) +0.40
Cu2+(aq) + 2e- Cu(s) +0.34
2H+(aq) + 2e- H2(g) 0.00
N2(g) + 5H+(aq) + 4e- N2H5+(aq) -0.23
Fe2+(aq) + 2e- Fe(s) -0.44
Cr3+(aq) + 3e- Cr(s) -0.74
Zn2+(aq) + 2e- Zn(s) -0.76
Mg2+(aq) + 2e- Mg(s) -2.37
41

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.

42

The use of sacrificial anodes to prevent iron corrosion.

43

General characteristics of voltaic and electrolytic cells.
VOLTAIC CELL ELECTROLYTIC CELL
System does work on its
Energy is released from Surroundings(power supply)
Energy is absorbed to drive a
spontaneous redox reaction
surroundings nonspontaneous redox reaction
do work on system(cell)

Oxidation half-reaction Oxidation half-reaction
X X+ + e - A- A + e-
Reduction half-reaction Reduction half-reaction
Y++ e- Y B+ + e - B
Overall (cell) reaction Overall (cell) reaction
X + Y+ X+ + Y; ∆G < 0 A- + B+ A + B; ∆G > 0 44

Comparison of Voltaic and Electrolytic Cells
Electrode

Cell Type ∆G Ecell Name Process Sign

Voltaic <0 >0 Anode Oxidation -

Voltaic <0 >0 Cathode Reduction +

Electrolytic >0 <0 Anode Oxidation +

Electrolytic >0 <0 Cathode Reduction -

45

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)

46

Electrolysis of Dilute H2SO4

 Electrolysis
of water
 H SO serves as a catalyst
2 4

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

47

The electrolysis of water.

2H2O(l) 4H+(aq) + O2(g) + 4e- 2H2O(l) + 4e- 2H2(g) + 2OH-(aq)
48
2H2O(l) 2H2(g) + O2(g)

Electrolysis of KI(aq)

49

Electrolysis of KI(aq)

Oxi: 2 I - → I2 + 2 e -
Red: 2 H + + 2 e - → H2
Final: 2 H + + 2 I - → H 2 + I2

50

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

Cu(s) Cu2+(aq, 0.1M) + 2e- Cu2+(aq, 1.0M) + 2e- Cu(s)

Cu2+(aq,1.0M) Cu2+(aq, 0.1M)
51

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.

52

E.g.
1) cathode reaction for CuSO4 solution
Cu2+ + 2 e- → Cuo

2) cathode reaction for the electrolysis of molten NaCl
Na+ + 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. 53

 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

54

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

1C=1AX1s

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.
55

A summary diagram for the stoichiometry of electrolysis.

MASS (g)
MASS (g)
of substance
of substance
oxidized or M(g/mol)
oxidized or
reduced
reduced
AMOUNT (MOL)
AMOUNT (MOL) AMOUNT (MOL)
AMOUNT (MOL)
of substance
of substance Faraday
of electrons
of electrons
oxidized or
oxidized or constant
transferred
transferred
reduced
reduced (C/mol e-)

balanced CHARGE (C)
CHARGE (C)
half-reaction
time(s)
CURRENT (A)
CURRENT (A)

56

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

57

Answers:

?C = 0.452 A X 1.50 hr X 3600s X 1C
1 hr 1AXs
= 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 2F 1 mol Ca
= 0.507 g Ca

? g Cl2 = 2.44X103 C X 1F X 1 mol Cl2 X 70.90 g Cl2
96,500 C 2F 1 mol Cl2
= 0.896 g Cl2

58

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

59

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-

60

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
61

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

62

The corrosion of iron.

67

Enhanced corrosion at sea.

68

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

faster corrosion cathodic protection

69

Electrochemistry

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