ELECTROCHEMISTRY
Introduction
• Electrochemistry is the study of electrical energy
and
chemical energy.
• Some chemical reactions produce electricity or
electricity causes the reactions take place.
• In 1771 Luigi Galvani, Italian anatomist, discovered a
new form of electricity could be produced by living tissue.
• In 1800’s Italian physicist Alessandro Volta built a
battery.
1. Oxidation-Reduction Reactions
• Electron transfer reactions are called oxidationreduction or redox reactions.
• Oxidation is loss of electrons.
• Reduction is gain of electrons.
2Na(s) + Cl2(g) → 2NaCl(s)
In the reaction above, Na undergoes oxidation process and
Cl2 undergoes reduction process. Na is called oxidized, and
Cl2 is called reduced substance.
• The substance being oxidized is called reducing agent.
• The substance being reduced is called oxidizing agent.
+1
Na → Na + 1e
oxidation
-1
Cl2 + 2e
→ 2Cl reduction
Here, Na is reducing agent, while Cl2 is oxidizing agent.
Example 1
Mg and O2 react to form MgO. What are the oxidizing and
reducing agents?
Solution
2Mg + O2 → 2MgO
Mg undergoes oxidation process, it is called reducing agent.
O2 undergoes reduction process, it is called oxidizing agent.
1.
Oxidation States
• Oxidation states of the elements must be known to
balance redox reactions.
• Oxidation states of elements in most stable form is
zero, like Fe, Cu, Ag, O2, H2, P, S, P4…etc
• Group IA have +1, and Group IIA have +2 and Group
IIIA
have +3, Halogens have -1 oxidation states.
• Hydrogen in metal hydrates has -1 oxidation state.
• Sum of the oxidation states in compounds is zero, in
ions is equal to charge of ion.
Example 2
Find the oxidation state of metals in the following species.
A. FeO B. KMnO4
C. Na2Cr2O7
D. HgOH
Solution
A. x + -2 = 0 → x = +2
B. +1 + x + 4(-2) = 0 → x = +7
C. 2(+1) + 2x + 7(-2) = 0 → x = +6
D. x + (-2) + 1 = 0 → x = + 1
Example 3
Find the oxidation state of indicated atoms in the following
species.
-3
-2
+1
A. PO4 B. CO3
C. K2CrO4
D. NH4
Solution
A. x + 4(-2) = -3 → x = +5
B. x + 3(-2) = -2 → x = +4
C. 2(+1) + x + 4(-2) = 0 → x = +6
D. x + 4(+1) = +1 → x = -3
2. Oxidation-Reduction Half Reactions
In redox reactions both oxidation and reduction processes
occur at the same time. Following redox reaction can be
divided into two half reactions;
+2
+2
Zn(s) + Cu (aq) → Zn (aq) + Cu(s)
+2
Oxidation:
Zn(s)
→ Zn (aq) + 2e
+2
Reduction:
Cu (aq) + 2e → Cu(s)
+2
+2
Net Reaction: Zn(s) + Cu (aq) → Zn (aq) + Cu(s)
Example 4
Find the half reactions of the reaction between potassium
metal with water to produce hydrogen gas and potassium
hydroxide.
Solution
K(s) + H2O(l) → KOH(aq) + H2(g)
+1
Oxidation:
K → K + 1e
+1
Reduction:
2H + 2e → H2
3. Balancing Redox Reactions
By Half Reaction method
+1
+3
Fe(s) + Cu (aq) → Fe (aq) + Cu(s)
Reaction is broken into two half reactions, oxidation and
reduction half reactions.
+3
Oxidation:
Fe
→ Fe + 3e
+1
Reduction:
Cu + 1e → Cu
Half reactions are balanced separately and same number
of electrons appear in each half reaction.
+3
Oxidation:
Fe
→ Fe + 3e
+1
Reduction:
3Cu + 3e → 3Cu
Then two half reactions are summed up to get a balanced
net equation.
+3
-
Oxidation:
Fe
→ Fe + 3e
+1
Reduction:
3Cu + 3e → 3Cu
+1
+3
Fe(s) + 3Cu (aq) → Fe (aq) + 3Cu(s)
Example 5
Balance the following redox reaction in acidic medium by half
reaction method.
+2
-2
+
+3
+3
Fe + Cr2O7 + H → Fe + Cr + H2O
Solution
+2
+3
Oxidation:
Fe → Fe
-2
+3
Reduction:
Cr2O7 → Cr
+2
+3
-
Fe → Fe + 1e
e- balanced
-2
+3
Cr2O7 → 2Cr
Cr atoms balanced
-2
+3
Cr2O7 → 2Cr + 7H2O O atoms balanced
-2
+
+3
Cr2O7 + 14H → 2Cr + 7H2O H atoms balanced
-2
+
+3
Cr2O7 + 14H + 6e → 2Cr + 7H2O
e- balanced
+2
+3
6Fe → 6Fe + 6e
-2
+
+3
Cr2O7 + 14H + 6e → 2Cr + 7H2O e- are equal
+2
-2
+
+3
+3
6Fe + Cr2O7 + 14H → 6Fe + 2Cr + 7H2O
Example 6
Balance the following redox reaction in basic medium by half
reaction method.
-1
+2
H2S + MnO4 → S + Mn
Solution
Oxidation:
H2S → S
-1
+2
Reduction:
MnO4 → Mn
+
Oxidation: H2S → S + 2H H atoms balanced
+
H2S → S + 2H + 2e e- balanced
-1
+2
Reduction:
MnO4 → Mn + 4H2O O atoms
balanced
-1
+
+2
MnO4 + 8H → Mn + 4H2O H atoms balanced
-1
+
+2
MnO4 + 8H + 5e → Mn + 4H2O e- balanced
+
5x
5H2S → 5S + 10H + 10e
-1
+
+2
2x
2MnO4 + 16H + 10e → 2Mn + 8H2O
+
5H2S → 5S + 10H + 10e
-1
+
+2
2MnO4 + 16H + 10e → 2Mn + 8H2O
-1
+
+2
5H2S + 2MnO4 + 6H → 2Mn + 8H2O + 5S
The reaction is balanced in acidic medium, to make it in basic
+
medium, same number of OH ion with H ion is added to the
both side of reaction.
-1
+2
5H2S + 2MnO4 + 6H2O → 2Mn + 8H2O + 5S + 6OH
-1
+2
5H2S + 2MnO4 → 2Mn + 2H2O + 5S + 6OH
Example 7
Balance the following redox reaction in acidic medium by half
reaction method.
-2
Cl2 + SO2 → SO4 + Cl
Example 8
Balance the following redox reaction in basic medium by half
reaction method.
+2
+
Zn + NO3 → Zn + NH4
3. Balancing Redox Reactions
By Change in Oxidation State Method
The given equation is divided into two half reactions.
Al + Br2 → AlBr3
+3
Oxidation:
Al → Al + 3e
x2
Reduction:
Br2 + 2e → 2Br
x3
+3
-
2Al → 2Al + 6e
3Br2 + 6e → 6Br
+3
Net Reaction:
2Al + 3Br2 → 2Al + 6Br
or
2Al + 3Br2 → 2AlBr3
Example 8
Balance the following redox reaction by the change in
-3
-3
oxidation state method. IO3 + AsO3 → I + AsO4
Solution
-3
-3
IO3 + AsO3 → I + AsO4
+5
+3
+5
I + 6e → I
As → As +2e
+5
1x
I + 6e → I
+3
+5
3x
3As → 3As +6e
+5
+3
+5
I + 6e → I
3As → 3As +6e
+5
+3
+5
I + 3As → I + 3As
coefficients are transferred to main equation.
-3
-3
IO3 + 3AsO3 → I + 3AsO4
Example 9
Balance the following redox reaction by the change in
oxidation state method.
ClO3 + N2H4 → NO + Cl
Example 10
Balance the following redox reaction in acidic medium by the
change in oxidation state method.
-2
P4 → PH3 + HPO3
2. Electrochemical Cells
An electrochemical cell is a device that produces an
electric current from energy released by a spontaneous redox
reaction. This kind of cell includes the Galvanic cell or Voltaic
cell.
Electrochemical cells have two conductive electrodes (the
anode and the cathode). Anode is defined as the electrode
where oxidation occurs and the cathode is the electrode
where the reduction takes place. In between these electrodes
there is the electrolyte, which contains ions that can freely
move.
The half reactions for the Zn-Cu cell;
+2
Zn(s) → Zn (aq) + 2e
Oxidation(Anode)
+2
Cu (aq) + 2e → Cu(s)
Reduction(Cathode)
+2
+2
Zn(s) + Cu (aq) → Zn (aq) + Cu(s)
And
+2
+2
Zn-Cu cell is represented as; Zn| Zn (1M) ||Cu (1M)| Cu
The function of salt bridge is to connect half cells together
and to avail the electrical current.
Electrons migrate from anode to cathode through external
circuit.
+2
-
Zn → Zn + 2e
+
2H + 2e → H2
+
+2
Zn(s) + 2H (aq) → H2 + Zn (aq)
ox = ?
red = 0
cell = 0.76 V
o
o
o
 cell
  reduction
  oxidation
The standard cell potential is 0.76 V. Therefore the standard
oxidation potential of Zn electrode is
ε ocell = εHo + / H + ε oZn/Zn+2
2
Example 11
Following spontaneous reaction occurs when metallic
magnesium is immersed in a copper (II)sulfate solution:
Mg(s) + CuSO4(aq) → Cu(s) + MgSO4(aq)
A. Write half cell reactions,
B. Draw a cell diagram for a voltaic cell in which this
reaction takes place.
Solution
+2
Mg → Mg + 2e
oxidation(Anode)
+2
Cu + 2e → Cu
reduction (Cathode)
+2
+2
Mg + Cu → Mg + Cu net reaction
o
1. Standard Electrode Potential( ε )
• Electrode potential is the difference in electrical
potential between the metal strip and the solution in
a half cell. The larger the electrode potential the
greater the energy required to move an electron
from the metal strip to the solution.
• The unit of electrical potential is volt, and is used to
show potential difference between two half-cells.
• In order to determine electrode potential energies of
elements, hydrogen is used as a reference electrode.
It is called Standard Hydrogen Electrode (SHE).
• The potential of SHE is accepted as zero at standard
o
conditions (1M, 25 C). The pressure of hydrogen gas
produced by the cell is 1 atm.
o
• The standard electrode potential is denoted by ε
o
and standard cell potential is represented by ε cell.
+
2H (aq) + 2e  H2(g) The cell reaction
Zn-SHE Cell
0.76 = 0 + ε oZn/Zn+2
ε oZn/Zn+2 = 0.76V
By international convention a standard electrode potential
measures the tendency for reduction process of an electrode.
+2
Zn + 2e → Zn
Zn+2/Zn = - 0.76V
+2
+
Zn | Zn (1M) || H (1M) | H2(1atm) | Pt
Example 12
Calculate the standard cell potential of a cell composed of
standard zinc and chromium electrodes. What is the reaction
taking place in the cell?
Solution
By using standard reduction potentials table, we get
following reactions;
+2
o
Zn
+ 2e → Zn
 red = -0.76 V
+3
o
Cr + 3e → Cr
 red = -0.74 V
+3
Since the reduction potential of Cr is greater than that of
+2
+3
Zn , then Cr ions will be reduced, and metallic zinc will be
+2
oxidized to Zn ions.
+2
o
Zn → Zn
+ 2e
 ox = +0.76 V
+3
o
Cr + 3e → Cr
 red = -0.74 V
When electrons are balanced;
+2
o
3Zn → 3Zn
+ 6e
 ox = +0.76 V
+3
o
2Cr + 6e → 2Cr
 red = -0.74 V
+3
+2
o
o
o
3Zn + 2Cr → 3Zn
+ 2Cr  cell =  ox +  red
= +0.76 + (-0.74) = +0.02 V
o
Positive standard cell potential,  cell, indicates the reaction
takes place spontaneously, if it is negative, the reaction is non
spontaneous.
Example 13
Find the standard potential of the cell
2+
–
Cu(s) | Cu || Cl | AgCl(s) | Ag(s)
and predict the direction of electron flow when the two
electrodes are connected.
Solution
The net reaction corresponding to this cell will be
–
2+
2AgCl(s) + Cu(s) → 2 Ag(s) + 2Cl (aq) + Cu (aq)
o
o
o
 cell =  ox +  red = -0.34 + 0.8 = 0.46 V
Example 14
For each pair of species, choose the better reducing agent.
Ag(s) or Sn(s) , given:
+
o
Ag (aq) + e- → Ag(s)
 red = 0.799 V
2+
o
Sn (aq) + 2e- → Sn(s)
 red = -0.136 V
Solution
According to reduction potentials, Ag(s) will be reduced,
being an oxidizing agent, and Sn(s) will be oxidized, being
reducing agent.
Example 15
Write the cell reactions and calculate the cell potentials for
the following cells constructed by
+2
+2
+2
A. Sn/Sn and Pb/Pb
B. Mg/Mg and Cl2/Cl
Example 16
Calculate the standard cell voltage for the following reactions
assuming standard conditions.
+3
+2
A. 3Fe + 2Al → 3Fe + 2Al
+2
+3
B. 2Al + 3Zn → 2Al + 3Zn
Example 17
Determine whether the given reaction will occur
spontaneously.
o
o
 Ag+/Ag = + 0.8 V and  Ni+2/Ni = - 0.25 V
+
+2
2Ag + Ni → 2Ag + Ni
Example 18
Which of the following reactions are spontaneous under
standard conditions? (Refer to Appendix D on page 188)
+2
+2
A. Sn + Cu → Cu + Sn
+3
+
B. 2Al + 3H2 → 2Al + 6H
Example 19
A voltaic cell is set up as follows: An aluminum electrode is
+3
dipped into a beaker containing a solution of 1 M Al . A silver
+
is dipped into a solution of 1 M H in another beaker. A salt
bridge is used to join the beakers. The two electrodes are
connected to a voltmeter. (Refer to Appendix D on page 188)
A. What is the reaction occurring in the cell?
B. What is the voltmeter reading?
2. The Effect of Concentration on Cell Voltages
Effect of concentration changes on cell potential can be
explained by Le Chatelier’s Principle.
Any factor which forces the reaction in forward direction will
cause an increase in the cell potential or vice versa.
When the reaction reaches to equilibrium the cell voltage
becomes zero, and the concentrations of ions are equilibrium
concentrations.
The cell potential can be calculated for non standard
conditions by Nernst Equation.
0.0592
logQ
n
n is the number of electrons transferred in the balanced
redox
reaction, and Q is the reaction quotient.
o
 cell   cell
-
Example 20
+2
Calculate the cell voltage of a Mg-Zn cell where [Mg ]=0.1 M
+2
and [Zn ]=1 M. (Refer to Appendix D on page 188)
Solution
Possible half cell reactions and overall reaction are;
+2
o
Zn
+ 2e → Zn
 red = - 0.76 V
+2
o
Mg + 2e → Mg
 red = - 2.36 V
According to half cell potentials Zn will be reduced and Mg
will be oxidized.
+2
o
Zn
+ 2e → Zn
 red = - 0.76 V
+2
o
Mg → Mg + 2e
 ox = +2.36 V
+2
+2
o
Zn (aq) + Mg → Zn + Mg (aq)  cell = +2.32 + (-0.76) =
+1.56 V
o
 cell = cell
-
0.0592
[Mg 2 ]
0.0592
0.1
log
=1.56log
 1.5895 V
n
[Zn 2 ]
2
1
Example 21
+3
Calculate the cell voltage of a Al-Cu cell where [Al ]=0.01 M
+2
and [Cu ]=10 M. (Refer to Appendix D on page 188)
Example 22
What conditions increase the cell potential for the reaction,
+2
+1
+3
Fe (aq) + Ag (aq) → Fe (aq) + Ag(s)
(Refer to Appendix D on page 188)
Example 23
What is the value of equilibrium constant Kc for the reaction
between silver ions and iron (II) ions in an aqueous solution at
o
+
+3
25 C?
Ag (aq) + Fe(aq) ↔ Ag(s) + Fe (aq)
Solution
+1
o
Ag + e → Ag
 red = + 0.80 V
+2
+3
o
Fe → Fe + e
 ox = - 0.77 V
+
+3
o
Ag (aq) + Fe(aq) ↔ Ag(s) + Fe (aq)  cell = 0.03 V
Because the reaction is at equilibrium cell is zero.
0.0592
0.0592
o
log K c   cell
=
log K c
1
1
0.0592
0.03 
log K c  K c  3.21
1
o
0= cell
-
3. Concentration Cells
A concentration cell is any voltaic cell in which two half-cells
consist of identical electrodes with different solution
concentrations.
For such a cell its cell potential under standard conditions,
o
 cell, is zero
+2
As the reaction proceeds, [Cu ] increases at the anode, and
decreases at the cathode.
+2
-
Cu(s)→ Cu (0.05 M) + 2e
+2
Cu (1 M) + 2e → Cu(s)
+2
+2
Cu (1 M) → Cu (0.05 M)
 ox = - 0.34 V
o
 red = + 0.34 V
o
 cell = 0.00 V
thought of as galvanic cell.
When iron contacts with water it tends to oxidize as follows;
+2
Fe → Fe + 2eanode
2H2O(l) + 2e- → 2OH (aq) + H2 cathode
+2
Fe + 2H2O(l) → Fe + 2OH (aq) + H2
More serious problem is caused when iron is in contact with
both water and oxygen.
+2
2Fe → 2Fe + 4eanode
2H2O(l) + O2 + 4e- → 4OH
cathode
+2
2Fe + 2H2O(l) + O2 → 2Fe + 4OH
+2
+3
Fe is then oxidized to Fe and forms Fe(OH)3. It readily
releases water to form Fe2O3.
2Fe(OH)3 → Fe2O3 + 3H2O
The composition of rust is Fe2O3.xH2O
o
0.0592
[Cu 2 (0.05M )]
log
2
[Cu 2 (1 M )]
0.0592
0.05
=0 log
 0.0385 V
2
1
o
 cell = cell
-
Example 24
An electrochemical consists of two hydrogen electrodes. One
is
standard hydrogen electrode and the other is hydrogen
electrode immersed in a solution of 0.1M HCl solution. The
two
cells are joined by a salt bridge. Write the overall cell reaction
and calculate cell potential.
Solution
+1
o
Anode H2(g) → 2H (0.1 M) + 2e
 ox = 0.00 V
+1
o
Cathode 2H (1 M) + 2e → H2(g)
 red = 0.00 V
+1
+1
o
Cell
2H (1 M) → 2H (0.1 M)
 cell = 0.00 V
0.0592
[H  (0.1M )]
log 1
2
[H (1 M )]
0.0592
0.1
=0 log
 0.0296 V
2
1
o
 cell = cell
-
Example 25
An electrochemical consists of two zinc electrodes. One is
immersed in a solution of 0.001M ZnSO4 and the other is in
10
M ZnSO4 solution. Calculate cell potential.
4. Corrosion
Many metals corrode when exposed to air or water. It can be
To prevent corrosion;
• Painting
• Plating with a thin layer of less oxidized
substance.
• Galvanizing with zinc.
• Cathodic protection, or sacrificial anode.
3. Electrolysis and Electrolytic Cells
The reactions in which electricity is used to cause a non
spontaneous reaction to occur are called electrolysis reaction.
The apparatus in which an electrolysis reaction occurs is
called electrolytic cell.
+2
Zn + 2e → Zn
Cathode
+2
Cu(s) → Cu + 2e
Anode
+2
+2
Zn + Cu → Zn + Cu
o
o
o
 cell =  ox +  red = -0.34 + (-0.77) = -1.10 V
1. Electrolysis of Water
2H2O + 2e → H2 + OH
Cathode
+
H2O → 1/2O2 +2H + 2e
Anode
H2O → H2(g) + 1/2O2(g)
o
o
o
 cell =  ox +  red = -1.23 + (-0.83) = -2.06 V
Pure water does not conduct electricity. To provide ion flow
Na2SO4 or H2SO4 is added to water.
2. Electrolysis of Molten NaCl
+
2Na + 2e → 2Na(l)
Cathode
2Cl → Cl2(g) + 2e
Anode
+
2Na + 2Cl → 2Na(l) + Cl2(g)
o
o
o
 cell =  red +  ox = - 2.71 + (-1.36)
= - 4.07 V
3. Electrolysis of Aqueous Na2SO4 Solution
+
-2
+
In the aqueous Na2SO4 solution, there are Na , SO4 , H and
OH ions. Possible oxidation and reduction reactions are;
+
o
Cathode Na + 1e → Na
 red = -2.71 V
o
H2O + 1e → 1/2H2 + OH
 red = -0.83 V
-2
-2
o
Anode 2SO4 → S2O8 + 2e
 ox = -2.01 V
+
- o
H2O → 1/2O2 + 2H + 2e  ox = -1.23 V
At cathode H2O is reduced and at anode H2O is oxidized.
o
H2O + 1e → 1/2H2 + OH
 red = -0.83 V
+
- o
H2O → 1/2O2 + 2H + 2e  ox = -1.23 V
o
H2O → H2 (g) + 1/2O2(g)  cell = -0.83 + (-1.23) = -2.06 V
4. Purification of Metals and Electroplating
Electro refining method is used to purify metals by the input
of electrical current.
During electrolysis, copper passes into solution from the
anodes, (leaving the impurities, normally containing silver,
gold and platinum) as an anode slime, which sinks to the
bottom of the cell. At the cathode, copper (II) ions are
discharged and the pure copper sheet becomes coated with
an increasingly thick layer of very pure copper.
Electroplating is a process that uses electrical current to
reduce cations of a desired material from a solution and coat
an object with a thin layer of the material, such as a metal.
Chromo-plating is very common plating for steel parts to
increase the resistant of materials against to corrosion.
4. Quantitative Aspects of Electrolysis
Michael Faraday invented the relationship between the
amount of electricity that passes through the electrolytic cell
and the amount of a substance undergoing a chemical
change. This law is known as Faraday’s Law.
+
Ag (aq) + 1e  Ag(s)
1 mol of e  1 mol of Ag(108 g)
2 mol of e  2 mol of Ag(2x108 g)
+2
Cu (aq) + 2e  Cu(s)
2 mol of e  1 mol of Cu(63.5 g)
-
• The charge of 1 mol e is named as Faraday, F.
• 1 mol e = 96500 Coulombs(C) = 1 F.
Charge = Current . time or Q = I . t
Example 26
+2
Calculate the mass of Zn produced by the reduction of Zn
ions at the cathode, using 1.4 A for 20 min. (Zn: 65.41)
Solution
+2
Zn (aq) + 2e  Zn(s)
Q  1.4x20x60  1680 C charge must pass.
1 mol e is 96500 C
2 mol e- produce 1 mol Zn
x mol e 1680 C
0.0174 mol ex mol Zn
x  0.0174 mol e
x = 0.0087 mol Zn will deposit.
mZn = 0.0087x65.41 = 0.5694 g
Example 27
In the electrolysis of sodium sulfate solution the anode
reaction may be written
+
2H2O → 4H + O2 + 4e
If a current of 2.5 A is passed through the solution for 2
o
hours, what volume of oxygen gas measured at 27 C and 1
atm pressure is released?.
Solution
+
2H2O → 4H + O2 + 4e
Q  2.5x2x60x60  18000 C charge passed.
1 mol e is 96500 C
4 mol e- produce
x mol e 18000 C
x  0.1865 mol e0.1865 mol e-
1 mol O2
x mol O2
x = 0.047 mol O2 will evolve.
PV = nRT  1xV = 0.047x0.082x(27+273)
V = 1.147 L
Example 28
Two electrolytic cells containing molten salts of CuCl2 and XCl3
are connected in series and electrolyzed for a while. At the
end of electrolysis, it is found that 1.92 g of Cu is deposited at
the cathode of the first cell and 3.94 g of X at the cathode of
the other cell. Calculate the atomic weight of element X.
(Cu:63.54)
Solution
Because the cell are series same charge passes through the
cells.
+2
+3
Cu (aq) + 2e → Cu(s) and X (aq) + 3e → X(s)
1.92 g
 0.03 mol
63.54 g / mol
2 mol e produces 1 mol Cu
x mol e 0.03mol Cu
x  0.06 mol enCu 
0.02 mol =
3 mol e- produce 1 X
0.06 mol e- x mol X
x = 0.02 mol X.
3.94 g
 MMx  197 g/mol
MMx