17.

Electrochemistry
What is Electrochemistry ?

• Electrochemistry is best defined as the study of the

interchange of chemical and electrical energy.
• It is primarily concerned with two processes that
involve oxidation–reduction reactions: the generation
of an electric current from a spontaneous chemical
reaction and the opposite process, the use of a current
to produce chemical change.

Oxidation–reduction reactions are a chemical reaction involving

transfer one or more electron from a substance to another one
17.1 Oxidation-Reduction Reactions

What is Oxidation-Reduction Reactions ?

• A type of chemical reaction that involves a

transfer of electrons between two substances.
• Any chemical reaction in which the oxidation
number of a molecule, atom, or ion changes by
gaining or losing an electron
Combustion reactions, which provide most of the energy to power our
civilization, also involve oxidation and reduction. An example is the reaction
of methane with oxygen:

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) + Energy

Even though none of the reactants or products in this reaction is ionic, the
reaction is still assumed to involve a transfer of electrons from carbon to
oxygen.

Oxidation States
The concept of oxidation states (also called oxidation numbers)
provides a way to keep track of electrons in oxidation–reduction reactions,
particularly redox reactions involving covalent substances.
The Characteristics of Oxidation–Reduction Reactions
Oxidation–reduction reactions are characterized by a transfer of electrons. In some
cases, the transfer occurs in a literal sense to form ions, such as in the reaction

Na(s) + Cl2(g) → NaCl(s)

0 +1
0 -1

Na(s) → Na+ (s) + e-

An increase in oxidation state (a loss of electrons)
0 +1

Cl2(g) + e- → Cl- (s)

A decrease in oxidation state (a gain of electrons)

0 -1
However, sometimes the transfer is less obvious. For example, consider the
combustion of methane (the oxidation state for each atom is given):

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

CH4(g) → CO2(g) + 8e-

An increase in oxidation state (a loss of electrons)
-4 +4

2O2(g) + 8e- → CO2(g) + 2H2O(g)

-4 -4 A decrease in oxidation state (a gain of electrons)

0 -8

We can now define some important terms.

 Oxidation is an increase in oxidation state (a loss of electrons).
 Reduction is a decrease in oxidation state (a gain of electrons).
 The oxidizing agent is a decrease in oxidation state (electron acceptor)
 The reducing agent is an increase in oxidation state (electron donor)
17.2 Balancing Oxidation–Reduction Equations
The Half-Reaction Method for Balancing Oxidation–Reduction
Reactions in Aqueous Solutions
For oxidation–reduction reactions that occur in aqueous solution, it is useful to
separate the reaction into two half-reactions: one involving oxidation and the
other involving reduction. For example, consider the unbalanced equation for the
oxidation–reduction reaction between cerium(IV) ion and tin(II) ion:

Ce4+ (aq) + Sn2+ (aq) → Ce3+ (aq) + Sn4+ (aq)

This reaction can be separated into a half-reaction involving the substance being
reduced, Ce4+ → Ce3+
(aq) (aq)

+4 +3

and one involving the substance being oxidized,

Sn2+ (aq) → Sn4+ (aq)

+2 +4
The Half-Reaction Method for Balancing Equations for Oxidation–Reduction
Reactions Occurring in Acidic Solution

1. Write separate equations for the oxidation and reduction half-

reactions.
2. For each half-reaction,
a. Balance all the elements except hydrogen and oxygen.
b. Balance oxygen using H2O.
c. Balance hydrogen using H+.
d. Balance the charge using electrons.
3. If necessary, multiply one or both balanced half-reactions by an
integer to equalize the number of electrons transferred in the two
half-reactions.
4. Add the half-reactions, and cancel identical species.
5 . Check that the elements and charges are balanced.
We will illustrate this method by balancing the equation for the reaction between
permanganate and iron(II) ions in acidic solution:

MnO4 -(aq) + Fe2+ (aq) →

Acid
Fe3+ (aq) + Mn2+ (aq)

1. Identify and write equations for the half-reactions. The oxidation states for the
half-reaction involving the permanganate ion show that manganese is reduced:
MnO4 -(aq) → Mn2+ (aq)
The other half-reaction involves the oxidation of iron(II) to iron(III) ion and is the
oxidation half-reaction:
Fe2+ (aq) →Fe3+(aq)
2. Balance each half-reaction. For the reduction reaction, we have
MnO4 -(aq) →
Mn2+ (aq)
a. The manganese is balanced.
b. We balance oxygen by adding 4H2O to the right side of the equation:

MnO4 -(aq) → Mn2+ (aq) + 4H2O(l)

c. Next, we balance hydrogen by adding 8H+ to the left side:
8H+(aq)+ MnO4 -(aq) → Mn2+ (aq) + 4H2O(l)
d. All the elements have been balanced, but we need to balance the charge using
electrons. At this point we have the following overall charges for reactants and
products in the reduction half-reaction:

8H+(aq)+ MnO4 -(aq) → Mn2+ (aq) + 4H2O(l)

+2
+7

We can equalize the charges by adding five electrons to the left side:

5e- + 8H+(aq)+ MnO4 -(aq) → Mn2+ (aq) + 4H2O(l)

Both the elements and the charges are now balanced, so this represents the
balanced reduction half-reaction. The fact that five electrons appear on the
reactant side of the equation makes sense, since five electrons are required to
reduce MnO4 (Mn has an oxidation state of +7) to Mn2 (Mn has an oxidation
state of +2).
For the oxidation reaction,
Fe2+ (aq) → Fe3+ (aq)
the elements are balanced, and we must simply balance the charge:
Fe2+ (aq) → Fe3+ (aq)
+2 +3
One electron is needed on the right side to give a net +2 charge on both sides:
Fe2+ (aq) → Fe3+ (aq) + e-
3. Equalize the electron transfer in the two half-reactions. Since the reduction half
reaction involves a transfer of five electrons and the oxidation half-reaction
involves a transfer of only one electron, the oxidation half-reaction must be
multiplied by 5:

Fe2+ (aq) → Fe3+ (aq) + e- (x 5)

5Fe2+ (aq) → 5Fe3+ (aq) + 5e-

5e- + 8H+(aq)+ MnO4 -(aq) → Mn2+ (aq) + 4H2O (l) (x1)

5Fe2+ (aq) + 8H+(aq)+ MnO4 -(aq) → 5Fe 3+

(aq) + Mn2+ (aq) + 4H2O (l)

The elements and charges have been balanced

The Method of Oxidation State Change for Balancing Oxidation–Reduction
Reactions in Aqueous Solutions
For example, consider the unbalanced equation for the oxidation–reduction
reaction between cerium(IV) ion and tin(II) ion:

Ce4+ (aq) + Sn2+ (aq) → Ce3+ (aq) + Sn4+ (aq)

(1 OSU) (x 2 )
+4 +3
OSU : (Oxidation State Unit)
(2 OSU) (x 1 )
+2 +4

The overall balanced equation is;

2Ce4+ (aq) + Sn2+ (aq) → 2Ce3+ (aq) + Sn4+ (aq)
17.3 Galvanic Cells

Electrons should flow through the wire from Fe 2+ to MnO4 - . However, when we
construct the apparatus as shown, no flow of electrons is apparent. Why?
When we connect the wires from the two compartments, current flows for an instant and
then ceases. The current stops flowing because of charge buildups in the two
compartments. If electrons flowed from the right to the left compartment in the
apparatus as shown, the left compartment (receiving electrons) would become negatively
charged, and the right compartment (losing electrons) would become positively charged.
Creating a charge separation of this type requires a large amount of energy. Thus
sustained electron flow cannot occur under these conditions.
 However, we can solve this problem very simply. The solutions must be connected
so that ions can flow to keep the net charge in each compartment zero. This
connection might involve a salt bridge (a U-tube filled with an electrolyte) or a
porous disk in a tube connecting the two solutions (see Fig. 17.2). Either of these
devices allows ions to flow without extensive mixing of the solutions. When we
make the provision for ion flow, the circuit is complete. Electrons flow through the
wire from reducing agent to oxidizing agent, and ions flow from one compartment
to the other to keep the net charge zero.

A galvanic cell, a device in

which chemical energy is
changed to electrical energy.
Cell Potential

A galvanic cell consists of an oxidizing agent in one compartment that pulls electrons
through a wire from a reducing agent in the other compartment. The “pull,” or
driving force, on the electrons is called the cell potential (Ԑcell ) or the
electromotive force (emf) of the cell. The unit of electrical potential is the volt
(abbreviated V), which is defined as 1 joule of work per coulomb of charge
transferred.
17.4 Standard Reduction Potentials
The reaction in a galvanic cell is always an oxidation–reduction reaction that can be
broken down into two half-reactions. It would be convenient to assign a potential to
each half-reaction so that when we construct a cell from a given pair of half-reactions
we can obtain the cell potential by summing the half-cell potentials. For example, the
observed potential for the cell shown in Fig. 17.5(a) is 0.76 V, and the cell reaction is

2H+(aq) + Zn (s) → Zn2+ (aq) + H2(g)

For this cell, the anode compartment contains a zinc metal electrode with Zn+2
and SO42- ions in aqueous solution. The anode reaction is the oxidation half-
reaction:
Zn (s) → Zn2+(aq) + 2e-
The zinc metal, in producing Zn+2 ions that go into solution, is giving up electrons,
which flow through the wire. For now, we will assume that all cell components are in
their standard states, so in this case the solution in the anode compartment will
contain 1 M Zn2+.
The cathode reaction of this cell is the reduction half-reaction :
2H+(aq) + 2e- → H2(g)
The cathode consists of a platinum electrode (used because it is a chemically inert
conductor) in contact with 1 M H+ ions and bathed by hydrogen gas at 1 atm. Such
an electrode, called the standard hydrogen electrode, is shown in Fig. 17.5(b).
Although we can measure the total potential of this cell (0.76 V), there is no way to
measure the potentials of the individual electrode processes. Thus, if we want
potentials for the half-reactions (half-cells), we must arbitrarily divide the total
cell potential. For example, if we assign the reaction
2H+ + 2e- → H2
Where, [H+] = 1 M and P H2 = 1 atm
a potential of exactly zero volts, then the reaction
Zn → Zn2+ + 2e-
Volt ?
will have a potential of 0.76 V because Ԑo cell = Ԑo (H+ → H2) + Ԑo (Zn → Zn+2)

0.76 V 0V 0.76 V

In fact, by setting Ԑo = 0 Volt for the half-reaction of 2H+ + 2e- → H2 , we can assign
values to all other half-reactions.
For example, the measured potential for the cell shown in Fig. 17.6 is 1.10 V
The cell reaction is Zn (s) + Cu2+(aq) + → Zn2+ (aq) + Cu (s)

Which can divided into the half-reactions

Anode : Zn → Zn2+ + 2e-

Cathode : Cu2+ + 2e- → Cu
Then, Ԑo cell = Ԑo (Zn → Zn+2) + Ԑo (Cu2+ → Cu)
1.10 V = 0.76 V + Ԑo (Cu2+ → Cu)
Ԑo (Cu2+ → Cu) = (1.10 - 0.76) V = + 0.34 V

The scientific community has universally accepted the half-reaction potentials based
on the assignment of zero volts to the process 2H+ + 2e- → H2

Standard reduction potentials are the values corresponding to

reduction half-reactions with all solutes at 1 M and all gases at 1 atm
 Oxidizing agent

Reducing agent
Oxidizing agent
Combining two half-reactions to obtain a balanced oxidation–reduction reaction
often requires two manipulations:

1. One of the reduction half-reactions must be reversed (since redox reactions must
involve a substance being oxidized and a substance being reduced). The half-
reaction with the largest positive potential will run as written (as a reduction),
and the other half-reaction will be forced to run in reverse (will be the oxidation
reaction). The net potential of the cell will be the difference between the two.
Since the reduction process occurs at the cathode and the oxidation process
occurs at the anode, we can write;
Ԑo cell = Ԑo (cathode) - Ԑo (anode)

Because subtraction means “change the sign and add,” in the examples done
here we will change the sign of the oxidation (anode) reaction when we reverse it
and add it to the reduction (cathode) reaction.
 2. Since the number of electrons lost must equal the number gained, the half-
reactions must be multiplied by integers as necessary to achieve the balanced
equation. However, the value of Ԑo is not changed when a half-reaction is
multiplied by an integer. Since a standard reduction potential is an intensive
property (it does not depend on how many times the reaction occurs), the
potential is not multiplied by the integer required to balance the cell reaction.

Consider a galvanic cell based on the redox reaction

Fe3+(aq) + Cu (s) → Cu2+ (aq) + Fe2+ (aq)

The pertinent half-reactions are

Fe3+(aq) + e- → Fe2+ (aq) Ԑo = 0.77 V (1)

Cu2+(aq) + 2e- → Cu (s) Ԑo = 0.34 V (2)

To balance the cell reaction and calculate the standard cell potential, reaction (2)
must be reversed:
Cu → Cu2+ + 2e- - Ԑo = -0.34 V (3)
Note the change in sign for the Ԑo value. Now, since each Cu atom produces two
electrons but each Fe3+ ion accepts only one electron, reaction (1) must be
multiplied by 2:

2Fe3+(aq) + 2e- → 2Fe2+ (aq) Ԑo = 0.77 V (4)

Note that Ԑo is not changed in this case.

Now we can obtain the balanced cell reaction by summing the appropriately
modified half-reactions (eq. 3) and (eq. 4):

2Fe3+(aq) + 2e- → 2Fe2+ (aq) Ԑo (cathode) = +0.77 V

Cu → Cu2+ + 2e- - Ԑo (anode) = -0.34 V

2Fe3+(aq) + Cu (s) → Cu2+ (aq) + 2Fe2+ (aq) Ԑo cell = Ԑo (cathode) - Ԑo (anode)

= (0.77-0.34)V = + 0.43 V
17.5 Electrolysis
What is Electrolysis ?

• The process of electrolysis involves forcing a current through a cell to

produce a chemical change for which the cell potential is negative; that
is, electrical work causes an otherwise nonspontaneous chemical
reaction to occur.
• An electrolytic cell, a device in which electrical energy is changed to
chemical energy.
• Electrolysis has great practical importance; for example, charging a
battery, producing aluminum metal, and chrome plating an object are all
done electrolytically

What is the difference between a galvanic cell and an electrolytic cell ?

To illustrate the difference between a galvanic cell and an electrolytic cell, consider the cell shown in Fig.
17.19(a) as it runs spontaneously to produce 1.10 V. In this galvanic cell the reaction at the anode is Zn (s)
→ Zn2+(aq) + 2e- whereas at the cathode the reaction is Cu2+ + 2e- → Cu
Now we will consider the stoichiometry of electrolytic processes, that is, how much
chemical change occurs with the flow of a given current for a specified time.
Suppose we wish to determine the mass of copper that is plated out when a current
of 10.0 amps (an ampere [amp], abbreviated A, is 1 coulomb of charge per second)
is passed for 30.0 minutes through a solution containing Cu2+.

Plating means depositing the neutral metal on the electrode by reducing the metal
ions in solution. In this case each ion requires two electrons to become an atom of
copper metal: Cu2+ (aq) + 2e- → Cu (s)

This reduction process will occur at the cathode of the electrolytic cell.
To solve this stoichiometry problem, we need the following steps:

Current Quantity of Moles Moles Mass

(A) and 1 charge in 2 numbers of 3 numbers 4 of Copper
time coulombs electrons of Copper in gram
(seconds)
1. Since an amp is a coulomb of charge per second, we multiply the current by the time
in seconds to obtain the total coulombs of charge passed into the Cu2+ solution at
the cathode:
𝑪
𝑪𝒐𝒖𝒍𝒐𝒎𝒃𝒔 𝒐𝒇 𝑪𝒉𝒂𝒓𝒈𝒆 = 𝒂𝒎𝒑𝒔 𝒙 𝒔𝒆𝒄𝒐𝒏𝒅𝒔 = 𝒙 𝒔
𝒔
𝑪 𝒔
=𝟏𝟎 𝒔 𝒙 𝟑𝟎 𝒎𝒊𝐧 𝐱 𝟔𝟎 𝒎𝒊𝒏

= 𝟏𝟖 𝐱𝟏𝟎𝟑 C

2. Since 1 mole of electrons carries a charge of 1 faraday, or 96,485 coulombs, we can

calculate the number of moles of electrons required to carry 18 x103 coulombs of
charge:
1 𝑚𝑜𝑙 𝑒 −
18 x103 C x = 1.87 x 10-1 mol e-
96,485 𝐶

This means that 0.187 mole of electrons flowed into the Cu+2 solution.
3. Each Cu+2 ion requires two electrons to become a copper atom. Thus each mole of
electrons produces ½ mole of copper metal:

1.87 x 10-1 mol e- x 1 mol Cu/2 mol e- = 9.35 x 10-2 mol Cu

4. We now know the moles of copper metal plated onto the cathode, and we can
calculate the mass of copper formed:

9.35 x 10-2 mol Cu x 63.546 g/mol Cu = 5.94 g Cu

Or we can use First Law of Faraday for electrolysis :

Where , W = mass of substances in gram

W = e i t/F e = Ar/valence (equivalent mass)
i = current (A)
t = time (seconds)
F = Faraday constant = 96500
1 F = 1 mol of electron
Galvanic cell :
It is useful for producing batteries
Electrolysis :
 Used to place a thin coating of metal onto steel
 Used to produce pure metals such as aluminum and copper