Chapter 7

7. Electrochemistry (6 hours)
7.1 Introduction
7.2 Electrical conductivity
7.3 Electrochemical Cells
7.3.1 Galvanic cells
7.3.2 Electrolytic cells
7.4 Quantitative aspects of electrolysis
7.5 Applications of Electrochemistry

7.1. Introduction
Electrochemistry is the branch of chemistry that deals with the interconversion of electrical
energy and chemical energy. That is the spontaneous chemical reaction that produces electricity
and a nonspontaneous process in which the application of electrical energy produces chemical
change. Thus, electrochemistry is the study of chemical reactions at the interface of an electrode
usually solid metal or semiconductor, and an ionic conductor, the electrolyte. It deals with
chemical reactions that produce electricity and the changes associated with the passage of
electrical current through matter. The reactions involve electron transfer, and so they are
oxidation-reduction (or redox) reactions. Electrochemical reactions which are redox reactions
involve two parts that occur at two electrodes
1. The oxidation reaction occurs in one cell (at the anode)
2. The reduction reaction occurs in another cell (at the cathode)
There are two kinds of electrochemical cells
i. Electrochemical cells containing nonspontaneous chemical reactions are called
electrolytic cells and
ii. Electrochemical cells containing spontaneous chemical reactions are called voltaic or
galvanic cells.
Electrode conventions in electrochemical cells
The cathode is the electrode at which reduction occurs. It is negative in the electrolytic cell and
positive in the Galvanic cells. The anode is the electrode at which oxidation takes place. It is
positive in an electrolytic cell and negative in Galvanic cells.

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Redox reactions
Electrochemical reactions involve chemical reactions that transfer electrons i.e., the redox
reactions. For example, consider the reaction between magnesium metal and hydrochloric acid as
a redox reaction:

Recall that the numbers above the elements are the oxidation numbers of
the elements. The loss of electrons by an element during oxidation is
marked by an increase in the
element’s oxidation number. On the contrary, in reduction, there is a
decrease in the oxidation number resulting from a gain of electrons by an
element. In the preceding reaction Mg metal is oxidized and H+ ions are
reduced; the Cl- ions are spectator ions.

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Balancing Oxidation-Reduction (Redox) Reactions
Because atoms are neither created nor destroyed in an ordinary chemical reaction, chemical
the equation must have an equal number of atoms of each element on the reactant and product
sides. In addition, the net electrical charges on the reactant side must be equal to the net
electrical charges on the product side. Balancing complex redox reactions is somewhat complex
and requires following some integrated steps. The following steps are followed to balance redox
reactions.
Example: Balance the following chemical equations by following the steps

Step 1: Write the unbalanced equation for the reaction in ionic form.

Step 2: Separate the equation into two half-reactions.

Step 3: Balance each half-reaction for the number and type of atoms and charges. For
reactions in an acidic medium, add H 2O to balance the O atoms and H
to balance the H atoms.
Oxidation half-reaction: The atoms are already balanced. To balance the charge, we
add an electron to the right-hand side of the arrow:

Reduction half-reaction: Because the reaction takes place in an acidic medium, we

add seven H2O molecules to the right-hand side of the arrow to balance the O atoms:

To balance the H atoms, we add 14 H+ ions on the left-hand side:

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There are now 12 positive charges on the left-hand side and only six positive charges
on the right-hand side. Therefore, we add six electrons on the left.

Step 4: Add the two half-reactions together and balance the final equation by inspection. The
electrons on both sides must cancel. If the oxidation and reduction half-reactions contain
different numbers of electrons, we need to multiply one or both half-reactions to equalize the
number of electrons.

Here we have only one electron for the oxidation half-reaction and six electrons for the reduction half-
reaction, so we need to multiply the oxidation half-reaction by 6 and write.

The electrons on both sides cancel, and we are left with the balanced net ionic equation:

Step 5: Verify that the equation contains the same type and numbers of atoms and the same charges on
both sides of the equation.

A final check shows that the resulting equation is “atomically” and “electrically” balanced

For reactions in a basic medium, we proceed through step 4 as if the reaction were carried out in an
acidic medium. Then, for every H+ ion we add an equal number of OH- ions to both sides of the equation.
Where H and OH- ions appear on the same side of the equation, we combine the ions to give H2O.

Example 2: Balance the following ionic equations in basic medium

Step 1: The unbalanced equation is

Step 2: The two half-reactions are

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Step 3: We balance each half-reaction for number and type of atoms and charges. Oxidation half-reaction:
We first balance the I atoms:

To balance charges, we add two electrons to the right-hand side of the equation:

Reduction half-reaction: To balance the O atoms, we add two H2O molecules on the right:

To balance the H atoms, we add four H ions on the left:

There are three net positive charges on the left, so we add three electrons to the same side to balance the
charges:

Step 4: We now add the oxidation and reduction half reactions to give the overall reaction. To equalize
the number of electrons, we need to multiply the oxidation half-reaction by 3 and the reduction half-
reaction by 2 as follows:

The electrons on both sides cancel, and we are left with the balanced net ionic equation:

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This is the balanced equation in an acidic medium. However, because the reaction is carried out in a basic
medium, for every H+ ion we need to add an equal number of OH- ions to both sides of the equation:

Finally, combining the H+ and OH- ions to form water, we obtain

Step 5: A final check shows that the equation is balanced in terms of both atoms and charges.

Practice
BalanceExercise: Balancechemical
the following the following equation using
equations, for the the
reaction in an acidic
ion-electron medium by the half-reaction
method.
2−¿ ¿
2+¿ +SO ¿
2−¿ →Mn 4
¿
−¿+ SO3 ¿
a . MnO 4
(in acidic medium)
method:
2−¿+ Bi¿
3+ ¿→[Sn (OH )6 ] ❑ ¿
¿
b . Sn 2+¿+Bi (Basic medium)

7.2. Electrical Conductivity

After completing this subunit, you will be able to:

 define electrical conductivity;
 explain metallic conductivity;
 explain electrolytic conductivity;
 distinguish between metallic and electrolytic conduction;
 distinguish between weak and strong electrolytes;
 use conductivity apparatus to test the conductivity of substances.
Electrical conductivity is the capacity of a substance to transmit electricity. The materials that
allow the passage of electricity through them are called electrical conductors. For example,
Metals. On the other hand, substances that hinder the passage of electricity are called insulators.
For example, non-metals. Metals are good conductors of electricity because they allow electrons
to flow through the entire piece of material. Thus, electrons flow like a “sea of electrons”
through metals. On the contrary, non-metals are non-conductors because they don’t allow the
following electrons through them. Conductivity in solutions/ molten form is also possible.

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Depending on the nature of the particles responsible for the flow of electric charges through
conductors, electrical conductivity can be classified as metallic conductivity or electrolytic
conductivity.

a) Metallic conductivity
Metallic conductivity refers to the transmission of electric current through metals. This
transmission is directly related to the structure of metals. In atoms of metals, the valence
electrons are bounded very loosely to their respective nuclei and move very easily throughout the
metal. This means metals contain electrons that do not have fixed positions and are relatively
free to move. These electrons are called free electrons or mobile electrons or delocalized
electrons. Thus, the structure of the metals can be regarded as a series of positively charged
metal ions, or cations, in a sea of negatively charged electrons. Hence, metallic conductivity is
also called electronic conductivity.
Non-metals are generally non-conductors of electricity because they do not have freely
moving electrons. Graphite is a form of carbon in which the carbon atoms are bonded in trigonal
planar fashion to the three other carbon atoms, to form inter-connected hexagonal rings,
electrons move freely through the hexagonal layers, making graphite a good conductor of
electricity.
b) Electrolytic conductivity
Electrolytes are substances that transmit electricity in a molten state or in an aqueous solution.
Based on their degree of ionization or the extent to which they produce anions and cations,
electrolytes can be classified as strong electrolytes or weak electrolytes. Strong electrolytes
ionize almost completely in aqueous solutions. Weak electrolytes ionize only slightly. When an
electrical potential is applied through an electrolyte solution, the positive ions (cations) move in
one direction and the negative ions (anions) move in the opposite direction. This movement of
ions through the electrolyte, brought about by the application of electricity, is called electrolytic
conductivity. Hence, the charge carriers in electrolytic conductivity are ions (anions and cations).

The substances that do not transmit electricity either in solution or in a molten state are called
non-electrolytes. Ionic compounds are non-conductors of electricity in the solid state. This is

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 because their ions are held at fixed positions and cannot move. Other examples of non-
electrolytes include sugar, ethanol, oil, benzene, and liquid nitrogen.

Table 7.1: Comparison of metallic and electrolytic conductions

Metallic conduction Electrolytic conduction
The current is solely carried by the electrons in the
conduction band The current is carried by both cations and anions
Specific conductance is low; they are moderately good
The velocity of the electrons is very large conductors
During the passage of the current no chemical
reaction occurs; only a heating effect is produced Ionic velocities are much smaller than electron velocities
The specific conductance of many metals is quite Passage of current brings about chemical reactions;
high, they are very good conductors of electricity heat is also evolved
Temperature co-efficient in general is negative
(alloys show complex behavior) The temperature co-efficient is positive
Ohm’s law applies Ohm’s law applies
Conductance is measured by a-c source; d-c current
Conductance may be measured by d-c or a-c current can be used only by elaborate arrangements

7.3 Electrochemical Cells

An electrochemical cell is a device capable of either generating electrical energy from chemical
reactions or use electrical energy to cause chemical reactions. The electrochemical cells which
generate an electric current are called voltaic or galvanic cells and those that generate chemical
reactions, via electrolysis are called electrolytic cells.

7.3.1 Galvanic cells

By the end of this section, you will be able to:
 Use cell notation to describe galvanic cells
 Describe the basic components of galvanic cells
Galvanic cells, also known as voltaic cells, are electrochemical cells in which spontaneous oxidation-
reduction reactions produce electrical energy. The cell must be designed in such a way that oxidation
occurs at one electrode (anode) with reduction at the other electrode (cathode). The electrons produced at
the anode must be transferred to the cathode, where they are consumed. To do this, the electrons move
through an external circuit, where they do electrical work. In writing the equations, it is often convenient
to separate the oxidation-reduction reactions into half-reactions to facilitate balancing the overall equation
and to emphasize the actual chemical transformations.

To understand how a voltaic cell operates, let us start with some simple cells that are readily made in the
general chemistry laboratory.

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The Zn-Cu2+ Cell

When a piece of zinc metal is added to a water solution containing Cu 2+ ions, some zinc dissolves
and some copper metal forms on the surface of the remaining zinc. The following net redox
reaction takes place: Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq)
It is an oxidation-reduction reaction (redox reaction) in which electrons are transferred from zinc atoms to
copper ions. In this reaction, copper metal plates out on the surface of the zinc. The blue color of the
aqueous Cu2+ ion fades as it is replaced by the colorless aqueous Zn 2+ ion (Figure 7.2). Clearly, this redox
reaction is spontaneous; it involves electron transfer from a Zn atom to a Cu 2+ ion.

Figure 7.1 Zn-cu2+ spontaneous redox reaction. The blue of the Cu 2+ ion in the solution fades (from left to right) as
the ion is reduced to metallic Cu that deposits on the zinc strip in the solution
The half-reaction equations are:
Oxidation half-reaction (Anode): Zn(s) → Zn2+ (aq) + 2e–
Reduction half-reaction (Cathode): Cu2+(aq) + 2e– → Cu(s)

The electrons flow directly from the zinc metal atoms to the copper ions without doing useful
work. If the half-reactions could be made to occur in separate compartments, and the electrons
were transferred through an external wire from one compartment to the other, it could produce a
Galvanic cell capable of doing electrical work.

A Galvanic cell was invented by the British Chemist John Daniell in 1836. It is called the Daniell cell. A
diagram of this cell is as shown in Figure 7.2.

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 Figure 7.2 Daniell cell.

One compartment of the Daniell cell contains a zinc electrode immersed in a solution of zinc
sulfate (ZnSO4(aq)). The other compartment contains a copper electrode in a solution of copper
sulfate (CuSO4(aq)). When the two electrodes are connected by a wire, the zinc atoms give up
electrons, forming positive ions:
Zn(s) → Zn2+(aq) + 2e–
The zinc ions repel each other and enter the solution. The negative electrons also repel each other
and travel through the external wire to the copper electrode, where they are accepted by copper
ions from the surrounding solution:
Cu2+(aq) + 2e– → Cu(s)
The resulting copper atoms deposit as the copper metal on the surface of copper electrode. The
Daniell cell reaction is identical to the reaction that occurs when zinc metal is dipped directly
into copper sulphate solution (zinc atoms are oxidized and copper ions are reduced), but the cell
is constructed so that the electrons pass through an external circuit where they can do useful
work.

The solutions in the two halves of a Galvanic cell must be connected in order to complete the
circuit, but they must not mix. The connection can be made through a salt bridge (see Figure
7.3).

A salt bridge is an inverted U-tube filled with an electrolyte, such as NH 4NO3 or KCl, chosen so
that it does not interfere with the operation of the cell. Regardless of how the cell is constructed,

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the solutions in each compartment remain more or less separated. Furthermore, because ions can
move into and out of the salt bridge, the solutions remain electrically neutral.

How does the salt bridge maintain the electroneutrality of a solution?

Consider the zinc compartment. The oxidation of zinc atoms tends to build up the concentration
of positive zinc ions in a solution around the zinc electrode. Some of the charge is neutralized by
negative ions flowing out of the salt bridge and enter into the zinc compartment, and the rest of
zinc ions flow out of the compartment into the salt bridge.

Now consider the copper compartment. The reduction of copper ions tends to decrease the
concentration of positive copper ions around the copper electrode. The resulting negative charge
is neutralized, partially by positive ions flowing from the salt bridge and enter into the copper
compartment, and partially by negative sulphate ions flowing out of the compartment. If the flow
of ions did not occur, a charge difference would build up between the compartments, and the
reaction would stop. Every Galvanic cell has an oxidation half-reaction and a reduction half-
reaction. The compartments in which these reactions occur (and their associated electrodes) are
referred to as half-cells.

What are the anode and the cathode in a Galvanic cell?

The electrode where oxidation takes place is called the anode, and the electrode where reduction
takes place is the cathode. In the Daniell cell, the zinc electrode is the anode and the copper
electrode is the cathode. The anode sends electrons through the connecting wire (it is the electron
source or negative terminal of the cell). The cathode receives electrons, so it is a positive
terminal. When the terminals of a Galvanic cell are connected by an external circuit, there will
always be a flow of electrons from anode to cathode through the external circuit. Note that the
electrode polarities in Galvanic cells are the reverse of those in an electrolytic cell

7.3.1.1. Cell Notation

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A cell notation allows us to describe a Galvanic cell without drawing a diagram. It is a shorthand
representation of a Galvanic cell. For example, a Daniell cell in which the electrolyte
concentrations are each 1 molar is represented as;

The anode is written on the left. The cathode on the right and concentrations and other data are
given in parentheses. The vertical line (|) indicates phase boundaries, and the double vertical line
(||) indicates a salt bridge or a porous partition.

The cell notation for the Daniell cell tells us that the zinc anode is dipped into a 1 M solution of
zinc ions, the copper cathode is dipped into a 1 M solution of copper ions, and the two half-cells
are separated by a salt bridge or a porous partition. If the two half-cells were in direct contact, the
double vertical line would be replaced by a single line. The notation for any Galvanic cell has the
following form:
anode | anode electrolyte (M) || cathode electrolyte (M) | cathode
Examples:
1. When chlorine gas is bubbled through an aqueous solution of NaBr, chloride ions and liquid
bromine are the products of the spontaneous reaction using a Pt electrode.
a) Draw a sketch of the cell, labeling the anode, the cathode, and the direction of electron
flow.
b) Write the half-reaction that takes place at the anode and at the cathode.
c) Write a balanced equation for the cell reaction.
d) Write an abbreviated notation for the cell.
To summarize the discussion of the structure of voltaic cells,
 a voltaic cell consists of two half-cells. They are joined by an external electric circuit through
which electrons move and a salt bridge through which ions move.
 each half-cell consists of an electrode dipping into a water solution. If a metal participates in
the cell reaction, either as a product or as a reactant, it is ordinarily used as the electrode;
otherwise, an inert electrode such as platinum is used.
 in one half-cell, oxidation occurs at the anode; in the other, reduction takes place at the
cathode. The overall cell reaction is the sum of the half-reactions taking place at the anode
and cathode.

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7.3.1.2 Standard Reduction Potentials
The driving force behind the spontaneous reaction in a voltaic cell is measured by the cell
voltage, which is an intensive property, independent of the number of electrons passing through
the cell. Cell voltage depends on the nature of the redox reaction and the concentrations of the
species involved; for the moment, we’ll concentrate on the first of these factors.
The standard cell voltage (Eo) (also referred to as the standard voltage) for a given cell is that
measured when the current flow is essentially zero, all ions and molecules in solution are at a
concentration of 1 M, and all gases are at a pressure of 1 atm. To illustrate, consider the Zn-H +
cell. Let us suppose that the half-cells are set up in such a way that the concentrations of Zn 2+ and
H+ are both 1 M and the pressure of H2(g) is 1 atm. Under these conditions, the cell voltage at
very low current flow is +0.762 V. This quantity is referred to as the standard voltage and is
given the symbol Eo.
Zn(s) + 2H+(aq, 1 M) → Zn2+(aq, 1 M) + H2(g, 1 atm), Eo = +0.762 V
Any redox reaction can be split into two half-reactions, oxidation and reduction. It is possible to
associate the standard voltages, standard oxidation voltage (), and standard reduction voltage
(), with the oxidation and reduction half-reactions. The standard voltage for the overall reaction,
Eo, is the sum of these two quantities.
o
Ecell = (cathode) – (anode)
To illustrate, consider the reaction between Zn and H + ions, for which the standard voltage is
+0.762 V.

Figure 4.8 Standard Hydrogen Electrode

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There is no way to measure the standard voltage for a half-reaction; only E o can be measured
directly. To obtain values, the value zero is arbitrarily assigned to the standard voltage for the
reduction of H+ ions to H2 gas called the standard hydrogen electrode (SHE):

Using this convention, it follows that the standard voltage for the oxidation of Zn in the
spontaneous reaction in Fig above and the reduction of H+:

i. The Zn2+ | Zn electrode is the anode, and the SHE is the cathode.
ii. The cell voltage is 0.76 V.
By using the defined standard reduction potential of H +, we can determine the standard reduction
potential for Zn2+ | Zn half-reaction:

As soon as one half-reaction voltage is established, others can be calculated. For example, the
standard voltage for the Zn-Cu2+ cell is found to be +1.101 V. Knowing that for zinc is +0.762
V, it follows that

Standard half-cell voltages are ordinarily obtained from a list of standard voltages such as those
in Table 7.2. The unit for the electric potential is the volt. Each standard potential listed is the
standard voltage for a given half-reaction, that is, standard potential = E o
Table 7.2 Standard Reduction Potential in Aqueous Solution at 25°C.

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15
In calculations using the above table, be sure to understand these points about the information in
Table 7.2:
1. The values apply to the half-cell reactions as read in the forward (left to right) direction.
2. The more positive is the greater tendency for the substance to be reduced. For example, the
reaction

has the highest Eo value for all of the half-cell reactions. Thus, F 2 is the strongest oxidizing agent
because it has the greatest tendency to be reduced. At the other extreme is the reaction,

which has the most negative value. Thus, Li + is the weakest oxidizing agent, because it is the
most difficult species to be reduced. Alternatively, Li metal is the strongest reducing agent since
it gets oxidized most readily.

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3. Under standard state conditions, any species on the left of a given cell reaction reacts
spontaneously with a species that appears on the right of any half-cell reaction that is located
below it in Table 7.2. This principle is sometimes called the diagonal rule. In the case of the
Daniell cell,

We see that the substance on the left of the first half-cell reaction is Cu2+, and on the right in the
second half-cell reaction is Zn. Therefore, Zn spontaneously reduces Cu2+ to form Zn2+ and Cu.

4. Changing the stoichiometric coefficients of a half-cell reaction does not affect the value of,
because electrode potentials are intensive properties. This means that the value is unaffected
by the size of the electrodes or the amount of solutions present but is dependent on the
concentration of the solutions and the pressure of a gas (if any).
For example,
I2(s) + 2e– → 2I– (1M); = +0.53 V
but Eo does not change if we multiply the half-reaction by 2:
2I2(s) + 4e– → 4I– (1M); = +0.53 V
5. The sign of changes, but its magnitude remains the same when we reverse a reaction.
6. The more positive the reduction potential, the greater the tendency to accept electrons. As a
result, when two half-cells are coupled, the reaction with higher (more positive) reduction
potential proceeds as reduction, while the other proceeds as oxidation.
Example
1. What is the standard emf (Eocell) of a Galvanic cell which is 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 at 25°C?
a) Given the following overall reaction,
Cu2+(1 M) + Mg(s) → Cu(s) + Mg2+(1 M)
Sketch the Galvanic cell based on the reaction.
b) Identify the cathode and anode.
c) Show the direction of electron flow through the external circuit.

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 Electromotive Force (emf)

When electrons move through a wire, they encounter resistance from localized atoms in their
paths. The driving force that allows the electrons to overcome this resistance and move around
the circuit is called electromotive force (emf). The emf in a Galvanic cell comes from the redox
reaction that pushes electrons from the anode to a cathode through the external circuit. Electrons
on the negative electrode repel each other and have more potential energy than electrons on the
positive electrode. This emf overcomes the external resistance through the wire and do work.
Electrical work is
used for starting cars, running watches, radios, and computers etc.

Electrical work (W) is the product of the emf of the cell and the total charge (in
coulombs) that passes through the cell:

W=Q×E

where Q is the charge and E is electrical potential.

Electrical energy = coulombs × volts = joules

The total charge is determined by the number of moles of electrons (n) that pass through
the circuit.
By definition, Q = nF

where F, the Faraday constant, is the total electrical charge contained in 1 mole of
electrons.
1 F = 96,500 C/mol e–
Because 1 J = 1 C × 1 V,

We can also express the unit of Faraday as 1 F = 96, 500 J/V. mol e–.
The measured emf is the maximum voltage that the cell can achieve. This value is used
to calculate the maximum amount of electrical energy that can be obtained from the
chemical reaction. This energy is used to do electrical work (Wele), so

where, Wmax is the maximum amount of work that can be done.

What does the negative sign indicate?

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The change in free energy (∆G) represents the maximum amount of useful work that can be
obtained from a reaction.

Both n and F are positive quantities, and ∆G is negative for a spontaneous process, so Ecell must
be positive. Therefore, a positive Ecell value corresponds to a negative ∆G value, which is the
condition for spontaneity. For standard state conditions,

Now we can relate E°cell to the equilibrium constant (K) of a redox reaction. The standard free
energy change, ∆G°, for a reaction is related to its equilibrium constant, as follows.

If we substitute ∆G° by – nFE°cell it becomes + nFE°cell = + RT ln K

When T = 298 K, the equation can be simplified by substituting for R and F.

Thus, if any one of the three quantities ∆G°, K or E°cell is known, the other two quantities can
be calculated, using the equations ∆G° = – RT ln K, ∆Go = – nFE°cell

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Let us summarize the relationships among ∆G°, K, and E°cell and characterize the spontaneity of
a redox reaction.

Table 7.3 Relationship among ∆Go, K and E°cell

What is the effect of changing the concentration of reactants on cell potential?

A change in concentration affects the change of cell potential which is a result of free energy
change. According to chemical thermodynamics

Where Q is the reaction quotient, and Q was used to calculate the effect of concentration on ∆G.
Since ∆G = – nFEcell and ∆Go = – nFEocell , the equation becomes – nFE
cell = – nFEocell + RT ln Q

By giving the constant values

The Q increases, which means that Ecell decreases. Eventually, the cell reaction reaches
equilibrium. At equilibrium, there is no net transfer of electrons, so Ecell = 0 and Q = K, where K
is the equilibrium constant.
Let’s consider the following reaction to seeing the dependence of Ecell on concentration

The equation for this cell at 25°C can be written as:

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Note that the concentration of liquids and solids is constant and does not appear in
the equilibrium constant expression. If the ratio [Zn2+]/[Cu2+] is less than 1, then
log ([Zn2+]/[Cu2+]) is a negative number. Therefore, the second term on the right-hand
side of the preceding equation is positive. Under this condition, Ecell is greater than
the standard cell potential. If the ratio is greater than 1, Ecell is smaller than Eocell.
7.3.2 Electrolytic cells
After completing this section, you will be able to:
 define the term electrolysis;
 define the terms electrode, anode, cathode, electrolyte, anion, and cation;
 describe an electrolytic cell;
 draw and label a diagram of an electrolytic cell;
 define the terms half-cell reaction and cell reaction;
 write the oxidation half-reaction, reduction half-reaction, and cell reaction for the
electrolysis of molten or fused electrolytes;
 perform an activity to show the electrolysis of molten electrolytes.

Electrolysis is a process in which electrical energy is used to produce chemical changes. This
process is carried out in an electrochemical cell known as an electrolytic or electrolysis cell. A
typical electrolysis cell contains a source of direct electric current, an electrolyte, and connecting
wires that join the source to the electrodes.

Electrodes are strips of metal or graphite that allow electrons to leave or enter the electrolytes.
They can be chemically active or inert. Active electrodes directly take part in reactions.
Examples include zinc and magnesium. Inert electrodes do not directly take part in chemical
reactions. They only serve to transfer electrons. Examples include platinum and graphite.

The electrode connected to the positive terminal of the source is positively charged and is called
the anode. It is the electrode through which electrons leave the cell. The electrode connected to
the negative terminal of the source is negatively charged and is called the cathode. It is the
electrode through which electrons enter the cell.

During electrolysis, the ions of the electrolyte migrate to the electrodes of the opposite charge.
The positive ions are attracted to the cathode and are called cations. Since the cathode has excess
electrons, the cations will discharge by gaining electrons. This process of gaining electrons is
called reduction. The negative ions are attracted by the positive electrode (anode) and are called
anions. These ions are discharged by losing electrons at the anode. This process of losing
electrons is called oxidation. Thus, the cathode is the electrode at which reduction occurs and the
anode is the electrode at which oxidation takes place.

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The reaction that takes place at each electrode is known as a half-cell reaction. Oxidation half-
reactions occur at the anode and reduction half-reactions at the cathode. The net reaction that
takes place in the electrolytic cell is known as a cell reaction. This overall reaction is also
referred to as an oxidation-reduction reaction or redox reaction. So electrolysis is a process in
which electric energy is used to bring about an oxidation-reduction reaction. It is also defined as
the decomposition of an electrolyte, using electricity. The process of electrolysis includes
electrolyzing aqueous solutions of electrolytes. However, at this level, electrolysis of molten
electrolytes will be discussed.

Electrolysis of Molten (Fused) Electrolytes

When ionic solids melt, they dissociate into positive and negative ions that are not
held in fixed positions. To understand the chemical reactions that occur during
electrolysis, consider a hypothetical electrolyte, MX, that dissociate into M+ and X–.

During electrolysis, the cations, M+ ions, move toward the cathode, gain one electron each and
become M atoms. The anions, X– ions, move toward the anode, lose one electron and become X
atoms. The reaction at each electrode and the entire reaction in the electrolytic cell are
represented by the following equations.

Figure 7.6 Electrolysis of fused Electrolyte, MX.

Here, it is very important to realize that the number of electrons gained by the cations at the
cathode is exactly equal to the number of electrons lost by anions at the anode during

22
electrolysis. Reduction and oxidation cannot occur separately. Oxidation is always accompanied
by reduction and there cannot be a reduction in the absence of oxidation, and vice versa.
Example
1. write the electrode, and cell reactions and show by diagram for electrolysis of fused lead
bromide (PbBr2)
2. Consider the electrolysis of KI and NaCl
c) Identify ions that migrate towards the anode.
d) Identify ions that migrate toward the cathode.
e) Write down the half-reactions at the anode and cathode and cell reactions.
f) Write the substances produced at the electrodes.
Preferential Discharge
What does preferential discharge mean? What are the factors that affect the preferential discharge of
ions?

In an Electrolysis cell when more than one types of ion are present in a cell, the electrodes select
one to discharge. The process is called preferential discharge. It is affected by the nature of the
electrodes, the positions of the ions in the electrochemical series and the concentration of the
ions in the electrolyte. Let us discuss these factors one by one.
1. Nature of the electrodes
How do electrodes affect the preferential discharge of ions?
Inert electrodes, like graphite or platinum, do not affect the product of electrolysis, but reactive
or active electrodes, like copper, can affect the product of electrolysis. For example, in the
electrolysis of copper sulphate solutions, using graphite electrodes, oxygen gas is liberated at the
anode and copper metal is deposited at the cathode, as shown below.
CuSO4(aq) —→ Cu2+(aq) + SO42–(aq)

However, if the electrolysis of copper sulphate is performed using copper electrodes, the copper
electrode at the anode dissolves and copper metal will be deposited at the cathode, as indicated
below.

2. The position of the ions in the electrochemical series

The reactivity series or activity series is an empirical series of metals, in order of "reactivity"
from highest to lowest. It is used to summarize information about the reactions of metals with
acids and water, single displacement reactions and the extraction of metals from their ores. The
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electrochemical series is built up by arranging various redox equilibria in order of their standard
electrode potentials (redox potentials). The reactivity series is sometimes quoted in the reverse
order of standard electrode potentials:
Li > K > Sr > Ca > Na > Mg > Al > Zn > Cr > Fe > Cd > Co > Ni > Sn > Pb > H > Cu > Ag >
Hg > Pt > Au
The positions of lithium and sodium are changed on such a series; gold and platinum are also
inverted, although this has little practical significance as both metals are highly unreactive.
Standard electrode potentials offer a quantitative measure of the power of a reducing agent,
rather than the qualitative considerations of other reactivity series. However, they are only valid
for standard conditions. In particular, they only apply to reactions in aqueous solution.
3. Concentration of the ions in the electrolyte
If an electrolyte contains a higher concentration of ions that are higher in the electrochemical
series than of those that are lower, then the higher ions get discharged in preference to the lower
ones.
For example, a solution of sodium chloride in water contains two types of anions i.e., the
chloride (Cl–) ions and the hydroxide (OH–) ions. The hydroxide ions are lower in the
electrochemical series than the chloride ions. But if the concentration of chloride ions is much
higher than that of the hydroxide ions, then the chloride ions get discharged first.
Electrolysis of Some Selected Aqueous Solutions
1a. Electrolysis of concentrated sodium chloride solution (brine solution)
A concentrated solution of sodium chloride (brine solution) contains Na+, Cl–, H+ and
OH– ions. However, the concentrations of H+ and OH– are very small. Why? When
a potential difference is established across the two electrodes, Na+ and H+ ions move
towards the cathode and Cl– and OH– ions move toward the anode. At the cathode,
the H+ ions are discharged in preference to the Na+ ions. Similarly, Cl– ions are discharged
at the anode in preference to the OH– ions because the concentration of Cl– ions is very high.

Since the Na+ and OH– ions remain in the solution, the solution yields NaOH
1b. Electrolysis of Dilute Sodium Chloride Solution

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1. Write the cathode, anode, and the cell reaction of the electrolysis of dilute sodium chloride
solution and
a) What is the pH of the solution?
b) What is the difference between the concentrated NaCl (aq) concentrated case?
2a. Electrolysis of Dilute Sulphuric Acid Solution
Which ions are discharged at the cathode and anode?
The dilute sulphuric acid solution contains H+, OH–, and SO42– ions. When a potential
difference is applied across the two electrodes (anode and cathode), only H+ ions
migrate toward the cathode, and OH– and SO42– ions migrate toward the anode.
At the cathode, H+ ions are discharged, and at the anode, OH– ions are discharged in
preference to SO42– ions because the hydroxide ion is below the SO42– ion in the
electrochemical series.

Note that the electrolysis of dilute NaCl and dilute H2SO4 solutions results in the decomposition
of water to oxygen and hydrogen gases.

3a. Electrolysis of Copper (II) Sulphate, Using Inert Electrodes

Which ions are liberated at the cathode and anode if copper sulphate solution is electrolysed,
using platinum (inert) electrodes?

Copper sulfate solution contains Cu2+, H+, SO42–, and OH– ions. When a potential difference is
established between the two electrodes, Cu2+ and H+ ions migrate toward the cathode, and SO42–
and OH– ions migrate toward the anode.
At the cathode, the Cu2+ ions are discharged in preference to the H + ions, because copper is
below hydrogen in the electrochemical series. Similarly, at the anode, the OH – ions are
discharged in preference to the SO42– ions.

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Since copper ions are deposited at the cathode, and hydroxide ions at the anode, the solution
becomes acidic due to the formation of H2SO4.

3a. Electrolysis of Copper (II) Sulphate Solution, Using Copper Electrodes

2. Write the cathode, anode, and the cell reaction of the electrolysis of CuSO4(aq) using a
copper active electrode
c) What is the pH of the solution?
d) What is the difference between the CuSO4(aq) using inactive electrode?

7.4. QUANTITATIVE ASPECTS OF ELECTROLYSIS

After completing this subunit, you will be able to:

 state Faraday’s first law of electrolysis;
 write the mathematical expressions for Faraday’s first law of electrolysis;
 do calculations related to Faraday’s first law of electrolysis;
 state Faraday's second law of electrolysis;
 write the mathematical expressions for Faraday’s second law of electrolysis; and
 do calculations related to Faraday’s second law of electrolysis.

Is it possible to calculate the amount of product formed at the electrodes? Is the amount of
substance produced at electrodes related to the amount of electricity?
The quantitative treatment of electrolysis was developed primarily by Michael Faraday in the
year 1834. Different laws were proposed by Faraday
7.4.1. Faraday's First Law of Electrolysis
Faraday's First Law states that “the amount of substance consumed or produced at one of the
electrodes in an electrolytic cell is directly proportional to the amount of electricity that passes
through the cell”.
If m is the mass of a substance librated or deposited at an electrode due to the passage of charge
Q, then according to Faraday's first law of electrolysis,

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Where z is a constant of proportionality and is called electrochemical equivalent of the
substance. It has the same charge which passes due to a steady current I flowing for time t, then
the above equation can be written as:
m=z×I×t
=z×Q
MIt
m=
nF
where, m is mass of the substance deposited or liberated, I is current in amperes, t is time in
seconds, F is Faradays constant, n is the number of moles of electrons lost or gained, and M is
the molar mass of the substance.

Example: How many moles of chlorine, magnesium and aluminum are formed when 2, 2 and 3
moles of charge are passed through three different solution containing chloride, magnesium and
aluminum ions respectively?

In an electrolysis experiment, we generally measure the current in amperes (A) that pass through
an electrolytic cell in a given period of time. By definition, 1 coulomb of charge is transferred
when a 1 ampere current flows for 1 second:
1C=1A×1s
The charge on 1 mole of electrons is 96,500 C, which is obtained by multiplying 1 mol (6.02×
1023 electrons) with the charge of electron (1.602×10–19 C).

Charge of 1 mol of electrons = 6.02×1023 electrons×1.602×10–19

C= 96, 485 C ≈ 96,500 C = 1 F (one faraday). Therefore, 1F = 96,500 C.mol–1.

7.4.2. Faraday's Second Law of Electrolysis

Faraday's second law of electrolysis states that “when the same quantity of charge (Q) is passed
through different electrolytes, then the masses of different substances deposited (m 1, m2, m3, ---)
at the respective electrodes will be directly proportional to their equivalent masses (E 1, E2, E3...).”
m∝E
or, m1 ∝ E1, m2 ∝ E2, and m3 ∝ E3 etc.
Replacing the proportionality by a proportionality constant k, the equations become:
M1/E1 = M2/E2 = M3/E3 = .... = k

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The law can be illustrated by passing the same quantity of electric current through three solutions
containing H2SO4, CuSO4 and AgNO3, connected in series, as shown below.

Figure 7.5 Solutions connected in series to a battery.

7.5. APPLICATIONS OF ELECTROCHEMISTRY
7.5.1. Industrial Application of Electrolysis
Electrolysis can be industrially applicable for;
a. Electroplating and Electrorefining of Metals
b. Extraction of Metals
c. Manufacture of Nonmetals and Some Compounds
Would you elaborate the applications of electrolysis for above applications with examples
7.5.2. Industrial Application of Galvanic cells
Batteries
A battery is a Galvanic cell, or a series of combined Galvanic cells, that can be used as a source
of direct current at a constant voltage. The principal advantages offered by batteries are
convenience and portability, rather than efficient use of energy.

Based on their rechargeability, we can classify batteries as primary and secondary. Primary
batteries use redox reactions that cannot be returned to their original state by recharging, so when
the reaction is complete, the batteries become “dead” and must be discarded. Secondary batteries
are often called storage batteries, or rechargeable batteries. The reactions in these batteries can be
reversed, which means that the batteries can be recharged. They have the advantage of being
more cost-efficient in the long term although the individual batteries are more expensive.

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Some common primary batteries
i. The Dry Cell Battery
ii. Alkaline Battery
Some common secondary batteries
i. Nickel-Cadmium Battery
ii. The Lead Storage Battery
Fuel cells
Write the cathode, anode, the cell reaction for the above all. Give the potential difference
produced
Corrosion of Metals

 What are the causes of corrosion?

 Can you give more examples of materials that can be damaged by corrosion from your
environment?
 Can you write the anode and cathode reactions of iron rusting?
 How iron rusting will be prevented?

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