Introduction to Electrolysis

Electrolysis is the process by which ionic substances are broken down into simpler
substances using electricity. It uses electric energy to force a non-spontaneous redox
reaction to occur.
Electrolysis occurs in an electrolytic cell. Essentially an electrolytic cell is made of an
electrolyte in a container in which two electrodes connected to an e.m.f source as
shown below

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 Introduction to Electrolysis
The electrolyte may be an aqueous solution or a molten ionic compound and
the electrode may be a metal or graphite.

Factors affecting Preferential Discharge

There are two considerations:
A. IF the electrolyte is molten [ MX ] (and carbon/graphite INERT electrodes
are used) the
electrolytic reactions are simple and straightforward.

During electrolysis,
M+ positive ions migrate to Cathode and is discharge: M + (l) + e ==> M (l)
X - negative ions migrate to Anode and is discharge: 2X - (l) - 2e ==> X2
(g)

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 Introduction to Electrolysis
B. IF the electrolyte is aqueous the electrolytic reactions become more complicated. Water
comes into the picture as water ionises to form H + and OH - ions. Selection of ions for
preferential discharge is based on the following three factors:

1. Position of the ION in the Electrochemical Series

Cations K Na Ca Mg Al Zn Fe Sn Pb H Cu Ag

Anions SO4 NO3 CO3 Cl Br I OH

Increasing ease of discharge

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 Note:
*If two or more positive ions migrate to the cathode, it is the ion lowest in the ECS
(i.e. of the least electropositive metal) that is discharged preferentially.
Examples Copper ions are discharged in preference to Hydrogen ions; Hydrogen ions are
discharged in preference to Sodium ions.
It will be recalled that the reactivity series gives the order of the tendency of metals to form
ions. Metals at the top of the series have the greatest tendency to form ions and are therefore
the most difficult to discharge at the cathode.

* If two or more negative ions migrate to the anode, the ion lower in the ECS is discharged
preferentially.

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 Introduction to Electrolysis

2. CONCENTRATION of the ions in the electrolyte

IMPORTANT NOTE: A very dilute solution of any electrolyte, on electrolysis, liberates
oxygen at the anode.
However, if the concentration of a particular ion is high, then this can alter the preferential
discharge.
For example, if dilute hydrochloric acid is electrolysed, hydrogen gas is produced at the cathode
and oxygen gas at the anode. However, when concentrated hydrochloric acid is electrolysed,
hydrogen gas is still liberated at the cathode, but chlorine gas (and NOT oxygen gas) is now
produced at the anode instead. The higher concentration of the chloride ions favor their
discharge over the hydroxide ions.
NOTE: Ions of Sodium, Potassium, Calcium, Sulfate and Nitrate ions are NOT discharged even
if they are in high concentrations.

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 Introduction to Electrolysis

3. The nature of the electrode

Usually, inert electrodes such as graphite or platinum are used for electrolysis. These electrodes
do not interfere with the reactions occuring at the surface of the electrode, they simply act as a
point of connection between the electrical circuit and the solution.
However, if metal electrodes are used in metal ion solutions they can get involved in the reactions
by dissolving as ions, leaving their electrons behind (this can only happen when the metal takes
the place of the anode, the positive electrode) - this is called electrode participation

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 Examples of Electrolysis

1. Electrolysis of molten Lead bromide using

platinum or graphite electrode
The ions present are Pb2+ and Br-
At cathode, Pb2+ is reduce to Pb(s)
Pb2+ + 2e- → Pb(s)
At the anode, Br- is oxidized to bromide gas
2Br- → Br2(l) + 2e-
The overall reaction is
PbBr2(l) → Pb(s) + Br2(l)
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 Examples of Electrolysis
2. Electrolysis of dilute H2SO4 using platinum
electrode
The ions present are H+(aq) and SO42-
At the cathode, H+ ion is reduced to hydrogen
gas
2H+(aq) + 2e- → H2(g)
At the anode, H2O is oxidized to oxygen
2H2O(l) → O2(g) + 4H+(aq) + 4e-
The overall reaction is
2H2O(l) → 2H2(g) + O2(g)
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 Examples of Electrolysis

3 Electrolysis of copper II sulphate solution using inert electrodes

The ions present are cations; Cu2+(aq), H+(aq) (from H2O) anions; SO42-(aq) and OH-(aq) (from H2O)
At the cathode
There is competition between Cu2+ and H+. As the H+ | H redox equilibrium appears higher in
the electrochemical series than the Cu2+ | Cu equilibrium, then Cu2+ are preferentially reduced
and copper metal is deposited at the electrode (a pink layer is observed)
Cu2+ + 2e Cu
At the anode
SO4- compete with OH- to release their electrons to the anode. The OH- are much better reducing
agents and are preferentially released as oxygen gas and water
4OH- - 4e 2H2O + O2

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 Examples of Electrolysis

Ions remaining in the solution

The ions that are removed from the solution,
then, are the copper ions and the hydroxide ions.
This means that the H+ and SO4- remain in the
solution - i.e sulphuric acid is also produced.

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 Examples of Electrolysis

Electrolysis of CuSO4 solution using copper electrodes

The ions present are cations ;Cu2+(aq), H+(aq) (from H2O) anions; SO42-(aq) and OH-(aq) (from H2O)
At cathode Cu2+ +2e_ → Cu(s) At the anode no ion is discharged (because electrode is active and
capable of ionization) Cu(s) →Cu2+(aq) +2e_ ie Cu2+ (aq) is produced at the anode .The overall
reaction is Cu(s) +Cu2+(aq) →Cu2+(aq) + Cu(s)
NB. colour of electrolyte stay constant; copper dissolve from the anode at the same rate as it
deposited the cathode.

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 Questions
Predict the products of electrolysis in each of the
following:
(i) An aqueous solution of AgNO3 with silver electrodes.
(ii) An aqueous solution of AgNO3with platinum
electrodes.
(iii) A dilute solution of H2SO4with platinum electrodes.
(iv) An aqueous solution of CuCl2 with platinum
electrodes.

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 Application of electrolysis

Application of electrolysis include

1. Electroplating of metals

2. Electro-refining of metals

3. Extraction of metals or Electro-metallurgy

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 Application of electrolysis
Electroplating is the process of electrolytic deposition of a metal on the surface of another metal;
the substrate may be another metal, a polymer, a ceramic or a composite.

Principal components of an electroplating process

The principal components are shown in the following figure.

DC power

Supply ne-
ne-

Anode + __ Cathode
__

Electroplated

Metal layer

Electrolyt
e

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 Electro-refining of metals (Purification of
metals)

Similar to the process of electro-deposition, electrolysis can be used to purifying metals that are
obtained from the ores. The process is known as electro-refining of metals. The metals that are
generally refined by this process are Zn, Ag, Ni, Cu, Pb, Al, etc.
Figure below is a conventionally used electrolytic cell for purifying metal ores, in this case the
metal is impure copper

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 Extraction of metals or Electro-metallurgy

Extraction of metals by the process of electrolysis is known as electro-metallurgy. This

process is used in case highly reactive metals such as sodium. An ore containing
sodium is used in a molten form. This forms the electrolyte. Anode and cathodes are
generally carbon rods or steel.
The Na atoms get attracted to the cathode of the cell and then the entire cathode with
its coating is stored for further use.

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 The quantitative aspect of Electrolysis

The relationship between chemical changes and the quantities of electrical energy was investigated
by Michael Faraday. His results are summarized into two laws named after him

Faraday's first law

It states that, “The mass of any substance deposited or liberated at any electrode is directly
proportional to the quantity of electricity passed.”i.e., W ∝ Q
Where,
W = Mass of ions liberated in gm,
Q = Quantity of electricity passed in Coulombs
= Current in Amperes (I) × Time in second (t)
∴ W ∝ I × t or W = Z × I × t
where, Z = constant, known as electrochemical equivalent (ECE) of the ion deposited.
When a current of 1 Ampere is passed for 1 second (i.e. Q = 1 ), then, W = Z

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 The quantitative aspect of Electrolysis

Thus, electrochemical equivalent (ECE) may be defined as “the mass of the ion deposited by
passing a current of one Ampere for one second (i.e., by passing Coulomb of electricity)”. It's
unit is gram per coulomb.

Coulomb is the unit of electrical charge.

96500Coulombs electrons = 1 mole electrons.
1 Coulomb = 6.023×1023/96500 = 6.85 × 1018 electrons,
or 1 electronic charge 1.6 × 10–19 Coulomb.

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 The quantitative aspect of Electrolysis
Faraday's second law
It states that, “When the same quantity of
electricity is passed through different
electrolytes, the
masses of different ions liberated at the
electrodes are directly proportional to their
chemical
equivalents (Equivalent weights).” i.e.,
W1/W2 = E1/E2 or Z1It/Z2It or Z1/Z2 =
E1/E2 (∴
W = ZIt)
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 The quantitative aspect of Electrolysis

Thus the electrochemical equivalent (Z) of

an element is directly proportional to its
equivalent weight (E), i.e., E ∝ Z or E = FZ
or E = 96500 × Z
where, F = Faraday constant = 96500 C
mol–1
So, 1 Faraday = 1F =Electrical charge
carried out by one mole of electrons.
1F = Charge on an electron × Avogadro's
number
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 The quantitative aspect of Electrolysis

Stoichiometry of Electrolytic Process

Step 1
Convert current and time to quantity of charge in coulombs
Current (amps) x time(s) = total charge transferred in coulombs ie Q = It
Step2
Convert quantity of charge in coulombs to moles of electrons, n(e-)
Coulombs(Q) /(96500coulombs/mole-) = mole- ( n(e-)
Step3
Using a half reaction determine the number of moles of the substance that can be reduced/oxidized
by the moles of electrons from step2
Mole e- x ( mole substance/mole-) = mol. substance

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 The quantitative aspect of Electrolysis

Worked Examples
EX1
Calculate the quantity of electric charge, in coulomb, represented by:
a a current of 6.00 A flowing for 25.0 minutes
b a current of 25 mA flowing for 2.0 days
Soln
a Q = It
Q = 6.00 A (25.0 60) s
= 9.00 103 C
b Q = It
Q = (25 10–3) A (2 24 60 60) s
= 4.3 103 C

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 The quantitative aspect of Electrolysis

EX2
The electrolyte in an electrolytic cell used to chromium-plate bicycle frames contains Cr3+ ions. If
the cell operates at 30.0 A for 25.0 minutes, calculate:
a the quantity of charge that passes through the cell
b the amount, in mol, of electrons that pass through the cell, in mol
c the mass of metal deposited on the frames
Soln.
a Q = It

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 The quantitative aspect of Electrolysis

Q = 30.0 A (25 60) s

= 4.50 104 C
Q
b n(e–) =
F
– 4.50 10 4 C
n(e ) =
96 500 C mol  1
= 0.466 mol
c Step 1 Write the equation.
Cr3+(aq) + 3e–  Cr(s)
Step 2 Use stoichiometry to calculate the amount of Cr.
n(Cr) 1
=
n (e  ) 3
0.4663
n(Cr) =
3
= 0.1554 mol
Step 3 Calculate the mass of Cr.
m(Cr) = 0.1554 mol 52.00 g mol–1
= 8.08 g

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 The quantitative aspect of Electrolysis

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 Worked Examples
Soln

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 Worked Examples

EX3
Calculate the standard cell potential produced by a galvanic cell consisting of a nickel
electrode in contact with a solution of Ni2+ ions and a silver electrode in contact with a

Ni2+(aq) + 2e- ⇌ Ni(s) E° = -0.26 V (must be flipped) Ag+(aq) + e- ⇌ Ag(s) E° =

solution of Ag+ ions. Which is anode and which is the cathode?

0.80 V

ANODE: Ni(s) ⇌ Ni2+(aq) + 2e- E° = 0.26 V

CATHODE: Ag+(aq) + e- ⇌ Ag(s) E° = 0.80 V
E° = 1.06 V

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 Worked Examples
EX4
Sketch the galvanic cells based on the following overall reactions. Show the direction of electron
flow, and identify the cathode and anode. Give the overall balanced equation. Assume that all
concentrations are 1.0 M And that All partial pressures are 1.0 atm.

Soln.
A typical galvanic cell diagram is:

The diagram for all cells will look like this. The contents of each half-cell compartment will
be identified for each reaction, with all solute concentrations at 1.0 M and all gases at 1.0 atm.

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 Voltaic cells at nonstandard conditions (The
Nernst Equation)
The Nernst equation will have to be applied under nonstandard conditions:

RT
Ecell  E cell  ln Q
nF

where Ecell = nonstandard cell potential in volts

E˚cell = standard cell potential in volts
R = constant, 8.314 J/mol·K
T = temperature in Kelvin
n = moles of electrons transferred
F = Faraday’s constant = 96485 C/mol eˉ
note: C represents the coulombs (unit of charge)
Q = the reaction quotient expression

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 Voltaic cells at nonstandard conditions (The
Nernst Equation)

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