C10 ELECTROCHEMISTRY

CHEMISTRY UNIT
KOLEJ MATRIKULASI MELAKA

SHARED BY MISS DALINA BINTI DAUD

CHAPTER 9

Electrochemistry

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10.1
GALVANIC CELL

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Learning Outcome

At the end of this topic, students should be able to:
(a) define

(i) oxidation
(ii) reduction
(iii) redox reaction
(iv) electrode potential
(b) draw a Voltaic / Galvanic cell.
(c) describe the operation of a voltaic cell.
(d) state the function of salt bridge.
(e) write half-cell equations and the overall cell reaction
equation.

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(f) write cell notation for a Galvanic cell.

(g) define :

i. standard reduction/electrode potential, E°.

ii. standard cell potential,E o
cell

(h) draw and explain the Standard Hydrogen Electrode (SHE).

(i) describe the method used to determine standard
reduction/electrode potential.

(j) use the standard reduction/electrode potential values to:

i. compare the relative strength of oxidising
agents or reducing agents.

ii. calculate standard cell potential using the E°

values.

iii. predict the spontaneity of a redox reaction.

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Electrochemistry

Electrochemistry Is the study of the relationships
between electricity and chemical reactions

• Chemical reactions involved in electrochemistry are :

Reduction REDOX REACTION
Oxidation

• Reduction and oxidation occur simultaneously.

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• A process in which a substance loses one or more
Oxidation electrons.

reaction

• A process in which a substance gains one or more
Reduction electrons.

reaction

• A reaction in which there is either a transfer of
electrons or a change in the oxidation numbers of the
Redox substances taking part in the reaction.
reaction

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REDOX Reaction

REDUCTION OXIDATION

• occur when a substance • occurs when a substance
gains electrons loses electrons

• The oxidation number of • The oxidation number
substance decreases of substance increases.

• Take place at cathode • Take place at anode

Remember… Remember…
RED CAT AN OX

= REDuction at CAThode = OXidation at ANode

Example: Example:

Cu2+ + 2e-  Cu Zn  Zn2+ + 2e-

Oxidation number decreases Oxidation number increases

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• Electrochemical reaction consists of reduction
and oxidation.

• These two reactions are called ‘half-cell reactions’
• The combination of 2 half-cell reactions are called

‘cell reaction’

Example

Reduction : Cu2+(aq) + 2e-  Cu(s) Half-cell

Oxidation : Zn(s)  Zn2+(aq) + 2e- reaction

Overall cell Cu2+(aq) + Zn(s)  Cu(s) + Zn2+(aq)

reaction :

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Electrochemical Cells

There are 2 type of cells

Galvanic Cell or Electrolytic Cell
Voltaic Cell

redox reaction Electricity is used
occur spontaneously to perform chemical
to generate electricity
reactions

Chemical Electrical Electrical Chemical
reaction Energy Energy reaction

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Component and Operation of
Galvanic cell

Metal electrode (M) dipped into an aqueous
solution of its ions (Mx+) in a container

Metal electrode (N) dipped into an aqueous
solution of its ions (Ny+) in an another
container

•The two electrodes are connected by a wire.
•The two solutions are connected by a salt bridge.
•A voltmeter is used to detect voltage generated.

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Galvanic cell

Voltmeter

M electrode N electrode

Mx+ Ny+

Mx+(aq) solution Ny+(aq) solution

Salt bridge 13

Salt bridge

-The two half-cells are connected by a salt bridge.

-Salt bridge is an inverted U tube containing a gel permeated
with solution of an inert electrolyte such as KCl, Na2SO4,
NH4NO3, NaNO3

The functions of a salt bridge

-Salt bridge helps to maintain electrical neutrality
-Completes the circuit by allowing ions carrying charge from the
salt bridge to move to the half-cells.

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How does the cell maintains its electrical neutrality?

Left Cell Right Cell

• M (s)  Mx+ (aq) + xe- • Ny+ (aq) + ye-  N (s)

• Mx+ ions enter the solution. • Ny+ ions leave the
solution to form N atom.
• Causing an overall of +ve
charge in the cell. • Causing an overall of -ve
charge.
• Anions from salt bridge
move into M half cell • Cations from salt bridge
move into N half cell

Electrical neutrality is maintained

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What happened if there is no salt bridge?

V

M(s) N(s)

Mx+ (aq) Mx+ e e Ny+ (aq)
e
e e Ny+
e

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The voltmeter shows deflection in a few seconds
then stopped.

Reason: At anode, when M rod dissolves, the
concentration of Mx+ ion is slightly higher than
negative ion.

The reaction stops because the increase in
positive charge is not neutralized by the
negative charge which is actually being supplied
by the salt bridge.
A difference in charge in one half-cell prevent
the flow of electron. Thus, the circuit is not
complete.

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Example:
Consider the Daniel Cell below,

a) Give the observations of both electrodes.
b) Explain how does the cell maintains its electrical

neutrality.
c) Write the overall cell equation of the reaction.

Voltmeter

e- flow

Zn electrode Zn2+ Cu2+ Cu electrode
Anode (–) Cathode (+)
Salt
ZnSO4(aq) bridge CuSO4(aq)
Solution (1.0M) Solution (1.0 M)

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ANSWER:

a) Observations :

i Zinc electrode dissolves

A brown solid is deposited on the copper

ii electrode and the blue colour of copper(II)

solution fades

How is this explained?

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i) At the zinc electrode :

Oxidation occurs at the Zn electrode
Zn (s)  Zn2+ (aq) + 2e-

Zinc is oxidized to zinc ion, Zn2+

Zinc solid dissolves.

Zn2+ ions enter Zn2+ solution.
Zn(s) is the negative electrode since it is a source of
electrons  anode.

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ii) At the copper electrode:

The electrons from the Zn metal move out through the
wire enter the Cu metal.

Cu2+ ions of the solution accept electrons.
Cu2+ (aq) + 2e-  Cu (s)

Copper is deposited on the electrode

Hence, the blue colour of copper(II) sulphate fades

Reduction occurs at the Cu electrode.

Cu is the positive electrode  cathode

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b) How does the cell maintains its electrical neutrality?

Left Cell Right Cell

Zn (s)  Zn2+ (aq) + 2e- Cu2+ (aq) + 2e-  Cu (s)

Zn2+ ions enter the solution. Cu 2+ ions leave the solution.
Causing an overall of +ve Causing an overall of
-ve charge.
charge.

Electrical neutrality is maintained

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Zn (s)  Zn2+ (aq) + 2e- Cu2+ (aq) + 2e-  Cu (s)

ANODE (-) CATHODE (+)

E = +1.10 V

e- flow

Zn Cl- K+ Cu

ZnSO4(aq) e Cu2+ CuSO4(aq)
Zn2+ e
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Salt bridge
(KCl)

c) Overall cell equation

Anode : Zn (s)  Zn2+ (aq) + 2e-
Cathode : Cu2+ (aq) + 2e-  Cu (s)

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

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Cell notation

A standard notation to represent an electrochemical cell.

Zn(s) + Cu2+(aq)  Cu(s) + Zn2+(aq)

Also can be represented as: Phase boundary

Salt bridge

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

anode cathode

• The electrods are placed at the beginning and at the end of
the cell diagram
• The salt bridge is represented by ‘ || ’
• The phase boundary is represented by ‘ | ’
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Cell notation

Salt bridge Phase boundary

Zn (s) | Zn2+ (aq, 1M) || Cu2+ (aq, 1M) | Cu (s)

anode cathode

Mnemonic tips !
ABC

Anode I Bridge I Cathode

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Exercise

For the cell below, write the reaction at anode and cathode
and the overall cell reaction.

Zn (s) | Zn2+(aq) || Cr3+ (aq) | Cr (s)

Answer: 3 Zn(s) → 3Zn2+(aq) + 62ee-- X 3

Anode :

Cathode : 2 Cr3+(aq) + 63ee- →2 Cr(s) X2

Overall cell 3Zn(s) + 2Cr3+(aq) + → 3Zn2+ (aq) +2Cr(s)

reaction:

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Cell Potential (Ecell)

The difference in electrical
potential between the anode and
cathode
Also known as cell voltage and
electromotive force (emf)

Measured by a voltmeter

Acts as ‘electrical pressure’ that
pushes electron through the wire.

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Cell Potential (Ecell)

The cell potential measured under standard
conditions is known as standard cell

potential (Eocell )

Standard conditions are : 25C
: 1 atm
•Temperature : 1.0 M
•Gas pressure
•Concentration of solution

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At standard condition

Eocell = Eocathode - Eoanode
or

Eocell = Eored + Eoox

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Electrode Potential (Ered or Eox)

• The potential difference produced between the
electrode and its electrolyte in a half cell.

Cu2+(aq) + 2e → Cu(s) Eored = +0.34 V
Zn2+(aq) + 2e → Zn(s) Eored = -0.76 V

Standard reduction
potential

• The more positive the value of standard reduction
potential, the stronger the attraction for electrons &
becomes cathode.

• Standard reduction potential of copper half-cell
is more positive compared to zinc.

• Copper half-cell becomes cathode. 31

• The potentials of half cells, Ered or Eox cannot be
measured on their own.

• Therefore, Standard Hydrogen Electrode (SHE)
with Eored = 0.00 V is used as a standard reference
half-cell to get the value of electrode potential of
any half cells.

• It can act as anode or cathode based on the value
of Eo of the other half cell.

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Standard Hydrogen Electrode (SHE)

SHE is made up of a platinum electrode, immersed in an
aqueous solution of H+ (1 M) and bubbled with hydrogen
gas at 1 atm and 25oC.

H2(g, 1 atm)

H+ (aq,1 M) Platinum
electrode

• The half-equation for SHE:
2H+(aq) + 2e ⇌ H2(g) Eo = 0.00 V

The standard reduction potential of SHE is 0.00 V
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Example 1:

Standard electrode potential of Zinc half cell
(SHE as cathode)

e- flow Eo = +0.76 V H2(g, 1 atm)
Zn(s) H+(aq, 1M)
V
Pt electrode
Salt bridge
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Zn2+ (aq, 1M)

Anode (oxidation): Zn (s) Zn2+ (aq) + 2e-
Catho=de (reduction):
2e- + 2H+ (aq) H2 (g)

+ Cell reaction : Zn (s) + 2H+ (aq)  Zn2+(aq) + H2 (g)

Eocell = +0.76 V
Eoox = ?
Eored = 0.00 V
Eocell = Eoox + Eored
+0.76 V = Eoox + 0.00 V
Eoox = +0.76 V

EoZn2+/Zn = -0.76 V

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Example 2:

Standard electrode potential of Copper half cell
(SHE as anode)

Eo = +0.34V

V e- flow

Cu(s) H2 (g, 1 atm)

H+(aq,1M)

Cu2+(aq, 1M) Pt(s)

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Anode (oxidation): H2 (g) 2H+(aq) + 2e-

Cathode (reduction): 2e- + Cu2+ (aq) Cu (s)

Cell reaction: H2(g) + Cu2+(aq) Cu (s) + 2H+(aq)

Eocell = +0.34 V
Eoox = 0.00 V
Eored = ?

Eocell = Eoox + Eored
+0.34 V = 0.00 V + Eored
Eored = +0.34 V

EoCu2+/Cu = +0.34 V

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The cell notation for SHE is either:

 H+(aq, 1M) | H2(g, 1 atm) | Pt(s) when it is cathode

Example :
Zn (s) | Zn2+ (aq, 1 M) || H+ (aq, 1 M) | H2 (g, 1 atm) | Pt (s)

 Pt(s) | H2(g, 1 atm) | H+ (aq, 1M) when it is anode

Example :
Pt (s) | H2 (g, 1 atm) | H+ (aq, 1 M) || Cu2+ (aq, 1 M) | Cu (s)

In either case, E0 of SHE remains 0.00 V

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Also known as Standard Reduction Potential
It is the potential difference produced between the
electrode and its electrolyte in a half cell measured at 25oC,
the pressure of 1 atm (for gases), and the concentration of
electrolyte is 1M.

Standard The sign of Eo changes when the reaction is reversed
Electrode Changing the stoichiometric coefficients of a half-cell
Potentials reaction does not change the value of Eo

(Eo)

A series of reduction reactions at standard condition which
is arranged in the order of their electrode potentials
produce The Standard Reduction Potential Series

The Standard Reduction Potential, Eo listed below refers to
the reduction reaction in a half cell.

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Standard Electrode Potentials (Eo)

Standard reduction potential = Standard
electrode potential

The potential difference produced
between the electrode and its electrolyte

in a half cell.

At 25oC, the pressure is 1 atm (for
gases), and the concentration of

electrolyte is 1M.

The sign of Eo changes when the reaction
is reversed

Changing the stoichiometric coefficients
of a half-cell reaction does not change
the value of Eo 40

For example: E0 = +1.36 V
E0 = +1.36 V
Cl2(g) + 2e-  2Cl-(aq) E0 = -1.36 V
½Cl2(g) + e-  Cl-(aq)
Cl-(aq)  ½Cl2(g) + e-

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• Eo value reflects the tendency of ions to accept
electron (to be reduced)

• The more positive the value of Eo, the greater
is the ability of the ions to be reduced.

Ag+(aq) + e  Ag(s) Eo = +0.80 V
Fe2+(aq) + 2e  Fe(s) Eo = 0.44 V

• Therefore, the stronger its tendency to act as an

oxidizing agent.

⇒ Ag+ is a stronger oxidizing agent than

Fe2+

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Oxidising agent → on the left of the half cell equation
Reducing agent → on the right of the half cell equation

Example :

Oxidising Reducing
agent agent

Increasing Ag+(aq) + e  Ag(s) Eo = +0.80 V
strength as Fe2+(aq) + 2e  Fe(s) Eo = 0.44 V

oxidising
agent

Increasing 43
strength as

reducing
agent

Example 1

Arrange the 3 elements in order of increasing strength of
reducing agents.

X3+ + 3e-  X E0 = -1.66 V
Y2+ + 2e-  Y E0 = -2.87 V
L2+ + 2e-  L E0 = +0.85 V

Answer: L < X < Y

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Example 2

For the cell below, write the half-cell equation for the
reaction at anode and cathode and the overall cell
equation of the reaction.

Cell notation

Zn (s) | Zn2+(aq) || Cr3+ (aq) | Cr (s)

Answer : 3 Zn(s) → 3 Zn2+(aq) + 6e- X3
2 Cr3+(aq) + 6e- → 2 Cr(s) X2
Anode :
Cathode :

Overall cell 3Zn(s) + 2Cr3+(aq) → 3Zn2+ (aq) +2Cr(s)
reaction:

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Example 3

Calculate the standard cell potential for the following
reaction.

Cl2(g) + 2Br –(aq)  Br2(l) + 2Cl(aq)
Given:

EoCl2|Clˉ = +1.36 V ; EoBr2|Brˉ = +1.07 V

Answer:

Eocell = Eocathode  Eoanode OR Eocell = Eored + Eo ox
= +1.36  (+1.07) = +1.36 + (-1.07)
= +0.29 V
= +0.29 V
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Example 4

Calculate the Eocell for the reaction :

Mg(s) | Mg2+(aq) || Sn4+(aq),Sn2+(aq) | Pt(s)
Given :

Mg2+(aq) + 2e → Mg(s) Eo = -2.38 V
Sn4+(aq) + 2e → Sn2+(aq) Eo= +0.15 V
ANSWER:

Anode : Mg(s) → Mg2+(aq) + 2e
Cathode : Sn4+(aq) + 2e →
Mg(s) + Sn4+(aq) → Sn2+(aq)
Overall : Mg2+(aq) + Sn2+(aq)

Eocell = Eored + Eo ox or Eocell = Eocathode - Eoanode

= +2.37 + 0.13 =+0.13- (-2.37)

= +2.5 V =+2.5V 47

Example 5

A cell is set up between a chlorine electrode and a
hydrogen electrode

Pt(s) | H2(g, 1 atm) | H+(aq, 1M) || Cl-(aq, 1M) | Cl2(g, 1atm) | Pt(s)
Eocell = +1.36 V

(a) Draw a labeled diagram of the respective cell.

(b) Determine the standard reduction potential for Cl2 .

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Answer: E0cell =1.36V

a) Cell diagram V

e- flow -+

H2 (g, 1 atm) Cl2 (g,1 atm)
Pt(S)
Pt(s)
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H+(aq, 1M) Cl-(aq,1M)

Answer:

a) Determine the standard reduction potential for Cl2

Reduction (cathode) : Cl2(g) + 2e-  2Cl-(aq)
Oxidation(anode) : H2(g)  2H+(aq) + 2e-

Overall : Cl2(g) + H2(g) 2Cl-(aq) +2H+(aq)

Given, Eocell =+1.36 V

Eocell = Eocathode - Eoanode
+1.36 = Eocathode – 0.00
Eocathode = +1.36 V

So the standard reduction potential for Cl2 is: 50
Eo = +1.36 V