Module 3: Electrode Systems Dept of Chemistry RRCE, Bengaluru-74

MODULE - 3

. What are the different types of electrodes?

The following are the important types of electrode:

1) Metal-Metal ion electrodes: These are the electrodes where a metal is dipped in

solution containing its own ions. Eg: Cu|CuSO4, Zn|ZnSO4

2) Gas electrodes: These are the electrodes where gas is in contact with an inert metal like

platinum dipped in an ionic solution of gas molecules. Eg: Hydrogen electrode.

3) Metal –Metal insoluble salt electrodes: These are the electrodes where a metal will be in

contact with its insoluble salt.

Eg: Calomel electrode (Hg|Hg2Cl2|Cl-), Silver-Silver chloride electrode (Ag|AgCl|Cl-).

4) Ion selective electrodes: These are the electrodes which are sensitive to particular ionic

species and will develop a potential when a membrane is in contact with an ionic

solution. Eg: Glass electrode

5) Redox electrodes: These are the electrodes, where inert metal like Pt will be in contact with

oxidized and reduced species of the same metal in solution. Eg: Pt|Fe2+,Fe3+ and

Pt|Sn2+,Sn4+

6) Amalgam electrodes: An amalgam electrode is a modification of metal-metal ion

electrode. These are electrodes in which metal amalgam is dipped in the solution of its own

metal ions. Eg: Lead amalgam electrode (Pb-Hg/Pb2+).

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Module 3: Electrode Systems Dept of Chemistry RRCE, Bengaluru-74

3.What are reference electrodes? Explain the construction and working of calomel electrode.
Mention its advantages and applications. [Jan2018]

Reference electrodes: These are the electrodes of constant potential which are used to determine
electrode potential of other electrodes.
Construction: Pt wire
Calomel electrode is a secondary reference electrode.
 Calomel electrode consists of a glass tube. A small amount of
salt bridge
pure mercury is placed at the bottom of the glass tube and
it is covered by a paste of Hg and Hg2C l 2 ( calomel paste). KCl (Sat)

 The remaining portion of the glass tube is filled with

KCl solution of known concentration (saturated, 1N or 0.1N). Hg + Hg2Cl2
(Calomel Paste)
 A platinum wire is dipped into the mercury and is used Hg
to provide external electrical contact.
Calomel Electrode
The calomel electrode is represented as:
Hg|Hg2Cl2|KCl (sat,1N,0.1N)
Working: Calomel Electrode can act as anode or cathode depending on the nature of the
other electrode.

 At anode: 2Hg + 2Cl- Hg2Cl2 + 2e-

 At cathode: Hg2Cl2 + 2e- 2Hg + 2Cl-

 At equilibrium: Hg2Cl2 + 2e- 2Hg + 2Cl-

The electrode potential is given by
Ecal= E0 – 0.0591 log [Cl-]

From the above equation it is clear that as the concentration of Cl- ion increases, the
electrode potential decreases.
Applications:
 It is used as reference electrode in all potentiometric titrations.
 It is used as reference electrode with glass electrode in pH determination.
 It is used as a secondary reference electrode in the measurement of single electrode
potential.

Note: At 298K, the electrode potentials are as follows.

KCl concentration 0.1N 1N Saturate

d
Ecal (V) 0.33 0.28 0.2422
4 1

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Module 3: Electrode Systems Dept of Chemistry RRCE, Bengaluru-74

1. What are Ion selective electrodes and Explain the construction and working
of a Glass electrode [July 2012].
These are the Electrode which response only certain specific ions (H + ions) and ignoring all
other ions present in the solution is k n o w n a s I o n S e l e c t i v e E l e c t r o d e s .
Ex: Glass electrode

Pt wire

E1, C1 ( INTERNAL SOLUTION )

0.1 M HCl

Ag/AgCl Electrode

Thin Glass Membrane

Construction:

 A glass electrode consists of a long glass tube with a very thin

walled bulb at the bottom.
 The approximate composition of this glass is
72%SiO2, 22%Na2O and 6%CaO.
 This is made up of special type of glass of low melting
Point, high electrical conductivity and can sense H+ ions
up to a PH about 9.
 The bulb is filled with 0.1M HCl in which a silver–silver chloride
Electrode is dipped as an internal reference electrode which provides
external electrical contact.
 The glass electrode is represented as:
Ag / AgCl / HCl(0.1M) / glass membrane

Working: When the glass electrode is dipped in the test solution, the Na+ ions of the glass
membranes are exchanged for H+ ions of the test solution.

Na+Gl- + H+ ↔ H+Gl- + Na+

Pt wire
Membrane Solution Membrane Solution

E1, C1 ( INTERNAL SOLUTION )

0.1 M HCl

Ag/AgCl Electrode
+
H SOLUTION OF UNKNOWN P H Thin Glass Membrane
( E2. C2)
EXTERNAL SOLUTION

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Module 3: Electrode Systems Dept of Chemistry RRCE, Bengaluru-74

When a thin walled glass tube containing an internal solution is immersed in test solution
(unknown PH solution), a potential developed across the glass membrane at the interface is called
boundary potential.

Eb=E2-E1

Log10 ]

Log [C2] - Log10 [C1]

Log [C2] - Log10 [C1]

Log10 [C1] where, Log [C2]

On substituting the values for R, T, F & n = 1 and C1= H+

10 [H+]

10 PH

Eb = L - 0.0591 pH since pH = - log [H+]

The potential of the glass electrode EG has some of three components:

i) Eb boundary potential.
ii) Asymmetric potential Eassy. (when C1=C2, a small potential is developed , is called asymmetric
potential)
iii) Potential of the internal reference Ag – AgCl electrodes i.e, EAg/AgCl.

Therefore, EG = Eb + Eassy + EAg/AgCl.

Substitting Eb Value in above equation,
EG = L-0.0591 pH + Eassy + EAg/AgCl.
EG = E0G -0.0591 pH

Where E0G = L+ Eassy + EAg/AgCl. E0G is the combination of three constant terms.

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Module 3: Electrode Systems Dept of Chemistry RRCE, Bengaluru-74

2. Explain the Determination of pH of a solution using glass electrode. [Jun 12]

Principle: when a thin glass membrane is placed between two solutions of different
pH values, a potential difference arises across the membrane. The potential
difference varies as the pH of the two solutions varies. pH of one of these two
solutions is kept constant and therefore the electrode potential depend on pH of
the other solution ie experimental solutio n

Potentiometer

Pt wire
Ag wire

Glass Electrode
KCl (Sat)

Hg + Hg2Cl2 Ag/AgCl Electrode

(Calomel Paste)
Hg Solution of Unknown pH

The cell consists of a glass electrode as indicator electrode and saturated calomel
electrode (SCE) as reference electrode, the cell is immersed in a solution whose pH is to be
determined. The emf cell is measured using potentiometer.

The cell assembly is represented as:

Pt / Hg / Hg2Cl2 / Cl- // solution of unknown pH / glass membrane / HCl (0.1M) / AgCl / Ag

The emf of the cell is given by

Ecell = Ecathode – Eanode
Ecell = EG – ECalomel

E cell = (EoG – 0.0591 pH) – Ecalomel (Where EG = E0G – 0.0591 pH)

pH = (EoG – ECell – Ecal) / 0.0591

EoG value is calculated by dipping the glass electrode in a buffer solution of known
pH and which is coupled with SCE and measuring the Ecell.

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 Module 3: Electrode Systems Dept of Chemistry RRCE, Bengaluru-74

4. What are concentration cells? Derive an expression for EMF of an concentration cell.
It is an electrochemical cell in which both the similar electrodes are dipped in respective
electrolytic solution but different concentration, a potential difference arises are called
concentration cell.
Derivation:

The above concentration cell can be represented as:

Zn| ZnSO4 (C1) || ZnSO4 (C2)| Zn

Concentration cell can be represented as

Zn / ZnSO4 (C1) // ZnSO4 (C2) / Zn

The Electrode reactions are

At anode Zn Zn2+ (C1) +2e-

At Cathode Zn2+ (C2) +2e- Zn

Net cell reaction: Zn2+ (C2) Zn2+ (C1)

C1 and C2 are concentrations of ZnSO4 solution at anode and cathode respectively & C2> C1.
Emf of a concentration cell can be calculated using Nernst Equation,
Ecell = Ecathode – Eanode

= E0 + 2.303RT log (C2) - E0 + 2.303RT log (C1)

.nF nF
= 2.303RT log(C2) - 2.303RT log (C1)
nF nF

Ecell = 2.303RT [log C2 – log C1]

nF

On substituting the values for R = 8.314 J/K/mol, T = 298K & F = 96500C

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 Module 3: Electrode Systems Dept of Chemistry RRCE, Bengaluru-74

PROBLEMS
1. A concentration cell was constructed by immersing two silver electrodes in 0.02M
and 2M AgNO3 solutions. Write the cell representation, cell reactions and calculate the

emf of the cell at 250C. (Jan 2017)

(i)Cell representation:
Ag|AgNO3 (0.02 M) || AgNO3 (2M)|Ag
(ii) Electrode reactions:

At anode: Ag Ag+ (0.02M) + e¯

At cathode: Ag+ (2M) + e¯ Ag

NCR: Ag+ (2M) Ag+ 0(0.02M)

(iii) The emf of the concentration cell:

n=1,

Ecell = 0.0591 log 100

Ecell = 0.0591 × 2
Ecell = 0.1182V

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Module 3: Electrode Systems Dept of Chemistry RRCE, Bengaluru-74

2. A cell is obtained by combining two Cd electrodes immersed in cadmium sulphate

solutions of 0.1M and 0.5M at 250C. Write the cell representation, cell reactions and
calculate the emf of the concentration cell. (June 2015)
(i) Cell representation:
Cd|CdSO4 (0.1M )|| CdSO4 (0.5M)|Cd
(ii) Electrode reactions:

At anode: Cd Cd2+ (0.1M) + 2e¯

At cathode: Cd2+ (0.5M) + 2e ¯ Cd

NCR: Cd2+ (0.5M) Cd2+ (0.1M)

(iii) The emf of the concentration cell:

Ecell = 0.0591 / n log [ C2 / C1]

Ecell = 0.0591 / 2 log [ 0.5 / 0.1]

Ecell = 0.02955 X log [ 5[

Ecell = 0.02955 X 0.6989
Ecell = 0.0207V
3. A concentration cell was constructed by immersing two silver electrodes in 0.01
M and 10 M AgNO3 solutions. Write the cell reactions and calculate the emf of the
concentration cell. [July 2005]
Solution: An electrode of higher concentration of Ag+ (10M) acts as cathode, while the

other silver electrode with Ag+ ion concentration (0.01M) acts as anode.
(i) Cell representation:
Ag / AgNO3 (0.01 M) // AgNO3 (10 M) / Ag (ii)
Electrode reactions:
+
At anode: Ag Ag (0.01M) + e¯
+
At cathode: Ag (10M) + e¯ Ag
+
Net cell reaction: Ag (10M) Ag+ (0.01M)
(iii)The emf of the concentration cell, Ecell = 0.0591 log C2
n C1
Ecell = 0.0591 log 10
1 0.01
= 0.0591 log 1000 = 0.0591 × 3 Ecell = 0.1773V

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Module 3: Electrode Systems Dept of Chemistry RRCE, Bengaluru-74

4. The emf of the cell Cu/ CuSO4(0.01 M) //CuSO4(x M) / Cu is 0.0295V at 250C. Find the
value of x. [July2013]

Solution: Ecell = 0.0591 log C2

n C1

0.0295 = 0.0591 log x / 0.01

2
0.0295 X 2 = 0.0591 logx / 0.01

0.059 / 0.0591 = log x /0.01

0.9983 = logx - (log 0.01)

logx = 0.9983 + log0.01

logx = 0.9983 – 2 = -1.007

x= Antilog (-1.007) = 0.0996M

5 Represent the cell formed by the coupling of two Cu electrodes immersed in CuSO4
solutions concentration of cupric ions in one electrode system in 100 times more concentrated
than other. Write the cell reaction and calculate the potential,
at 300K. [July 2007]
Solution:
(i)Cell representation:
Cu/CuSO4 (x M ) // CuSO4 (100x M)/Cu
(ii) Electrode reactions:
2+
At anode: Cu →Cu (x M) + 2e¯
At cathode: Cu2+ (100x M) + 2e¯ → Cu

2+
Net Chemical reaction: Cu (100x M) → Cu2+ (x M)

(iii)The emf of the concentration cell,

Ecell = 2.303RT log C2
nF C1
Ecell = 2.303 × 8.314 × 300 log 100x
2 × 96500 x

= 0.0297 log 100

= 0.0297 × 2 Ecell = 0.0595V

6. A concentration cell was constructed by immersing two silver electrodes in 0.05M and 1M AgNO3
solutions. Write the cell reactions and calculate the emf of the concentration cell. (July 2006)

Solution:
(i) Cell representation:
Ag/AgNO3 (0.05 M) // AgNO3 (1M)/Ag
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Module 3: Electrode Systems Dept of Chemistry RRCE, Bengaluru-74

(ii) Electrode reactions:

+
At anode: Ag Ag (0.05M) + e¯
+
At cathode: Ag (1M) + e¯ Ag
+
Net cell reaction: Ag (1M) Ag+ (0.05M)

(iii)The emf of the concentration cell, Ecell = 0.0591 log C2

n C1
Ecell = 0.0591 log 1/0.05
1
EcelL= 0.0591 log(20), = 0.0591 × 1.3010

Ecell = 0.0768V

7. A spontaneous galvanic cell tin/tin ion (0.024M)//tin ion (0.064M)/tin develops an emf of
0.0126V at 250C. Calculate the valency of tin (Jan 2006).

Solution: The emf of the concentration cell

Ecell = 0.0591 log C2
n C1
0.0126 = 0.0591 log 0.064
n 0.024

n= 0.0591 log 0.064

0.0126 0.024

n= 4.690 log 2.6666 = 4.690 × 0.4259

n=1.99 ≈ 2 (n is the valency of tin) Valency of tin is 2. Tin is divalent.

8. The emf of the cell Ag/AgNO3(C1) // AgNO3(C2= 0.2M) /Ag is 0.8V at 250C. Find the
value of C1. [June2012]

Solution: The emf of the concentration cell

Ecell = 0.0591 log C2

n C1

0.8 = 0.0591 log 0.2/ C1

1
0.8/0.0591 = Log 0.2 / C1

0.2/C1 = Antilog (0.8/0.0591) = Antilog (13.54)

0.2/C1 = 3.467 x 1013

C1 = 0.2/3.467×1013 C1= 5.77×10-15 M

9. Calculate the concentration of H+ in the cell Pt ,H2 (1 atm)|H+(10-6M) || H+ (C2=?)|

H2 (1 atm), Pt at 250C. Given Ecell= 0.1182V (Nov 2000).

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Module 3: Electrode Systems Dept of Chemistry RRCE, Bengaluru-74

Solution: The emf of the concentration cell

Ecell = 0.0591 log C2

n C1

0.1182 = 0.0591 log C2

1
10-6
0.1182 / 0.0591 = log C2 – log (10-6)

2 = log C2 – (-6) = log C2 + 6

log C2 = 2-6 = -4
C2 = antilog (-4) = 1 x 10-4 M
C2 = 1 x 10-4 M

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