CHEMISTRY ELECTROCHEMISTRY

ELECTROCHEMISTRY

Conductivity of Solutions
What is Conductivity?

 Electrical resistance is represented by the symbol ‘R’, and it is measured in ohm (Ω).
 The electrical resistance of any object is directly proportional to its length (l) and
inversely proportional to its area of cross-section (A), i.e.

where the constant of proportionality ρ is called resistivity (specific resistance).

 The inverse of resistance, R, is called conductance, G, and we have the relation

where the constant  is called conductivity (specific conductance).

 The SI unit of conductance is Siemens, represented by the symbol ‘S’, and it is equal to
ohm−1 (also known as mho) or Ω−1. The SI unit of conductivity (  ) is S m−1.

Conductivity of Electrolytic (Ionic) Solutions

 Very pure water has small amounts of hydrogen and hydroxyl ions (~10 −7M) which
lend it very lowconductivity (3.5 × 10−5 S m−1).
 When electrolytes are dissolved in water, they dissociate to give their own ions in the
solution; hence, its conductivity also increases.

Electrolytic or ionic conductance:

Conductance of electricity by ions present in solutions
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Measurement of Conductivity of Ionic Solutions

 We first need to find the resistance of an ionic solution to measure conductivity.

 We face two problems when measuring the resistance of an ionic solution:
1. Passing direct current (DC) changes the composition of the solution
2. A solution cannot be connected to the bridge like a metallic wire or other solid
conductor
 The first difficulty is resolved by using an alternating current (AC) source of power. The
second problem is solved by using a specially designed vessel called conductivity cell.
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 Conductivity Cell:

 It consists of two platinum electrodes coated with platinum black (finely divided
metallic Pt is deposited on the electrodes electrochemically).
 These have area of cross-section equal to ‘A’ and are separated by distance ‘l’. The
resistance of such a column of solution is then given by the equation:

 The quantity ‘l/A’ is called cell constant and is denoted by the symbol G*. It depends
on the distance between the electrodes and their area of cross-section and has the
dimension [L−1].
 Hence, the cell constant G* is given by the equation:

 When the cell constant is determined, we can use it for measuring the resistance or
conductivity of any solution.
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 The set up for the measurement of the resistance is nothing but the well-known
Wheatstone bridge.

 It consists of two resistances R3 and R4, a variable resistance R1 and the conductivity
cell having the unknown resistance R2.
 The Wheatstone bridge is fed by an oscillator O (a source of AC power in the audio
frequency range 550−5000 cycles per second).
 P is a suitable detector (a headphone or other electronic device).
 The bridge is balanced when no current passes through the detector.
 Under these conditions,

 When the cell constant and the resistance of the solution in the cell are determined,
the conductivity of the solution is given by the equation:

Molar Conductivity

 Conductivity of solutions of different electrolytes in the same solvent and at a given

temperature differs due to
1. Charge and size of ions in which they dissociate
2. Concentration of ions or ease with which the ions move under a potential
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gradient
 Therefore, it becomes necessary to define a quantity called molar conductivity
denoted by the symbol (λm). It is related to the conductivity of the solution by the
equation:

 Unit of λm is in S m2 mol−1.
 Hence, molar conductivity can be given by the formula

Variation of Conductivity and Molar Conductivity with Concentration

 Both conductivity and molar conductivity change with the concentration of the
electrolyte.
 Conductivity always decreases with a decrease in concentration for both weak and
strong electrolytes. It is because the number of ions per unit volume which carry the
current in a solution decreases on dilution.
 Molar conductivity increases with a decrease in concentration. This is because the
total volume (V) of solution containing one mole of electrolyte also increases.
 The decrease in  on dilution of a solution is more than compensated by an increase in
its volume.

Molar conductivity (λm): Conductance of the electrolytic solution kept between the
electrodes of a conductivity cell at unit distance but having area of cross section
large enough to accommodate sufficient volume of solution which contains one
mole of the electrolyte.

 When concentration approaches zero, the molar conductivity is known as limiting

molar conductivity and is represented by the symbol λ .
 The variation in λm with concentration is different for strong and weak electrolytes.

 Strong Electrolytes:
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 λ increases slowly with dilution and can be represented by the equation:

 It can be seen that if we plot λm against c1/2, we obtain a straight line with intercept
equal to λom and slope equal to ‘−A’.

 The value of the constant ‘A’ for a given solvent and temperature depends on the type
of electrolyte.

 Weak Electrolytes:

 Weak electrolytes such as acetic acid have a lower degree of dissociation at higher
concentrations. Hence, for such electrolytes, the change in λm with dilution is due to
1. Increase in the degree of dissociation
2. The number of ions in total volume of solution which contains 1 mol of
electrolyte
 In such cases, λ increases steeply on dilution, especially near lower
concentrations. Therefore, it cannot be obtained by extrapolating λm to zero
concentration.
 At infinite dilution, electrolytem dissociates completely (α = 1), but at such low
concentration, the conductivity of the solution is so low that it cannot be measured
accurately. Therefore, λ for weak electrolytes is obtained by using Kohlrausch’s
law of independent migration of ions.
 Thus, at any concentration c, if α is the degree of dissociation, then it can be
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approximated to the ratio of molar conductivity, λm, at the concentration c to
limiting molar conductivity λ .

 But we know that for a weak electrolyte,

 By putting the value of α in the above equation, we get the equation:

Numerical

1) The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500Ω.
What is the cell constant if conductivity of 0.001M KCl solution at 298K is 0.146 × 10-3

Scm-1?
Answer :

Given Data:
R =1500Ω
=0.146 × 10-3 Scm-1

G* = ?
Solution :
The cell cons tant is given by the equation,

Cell constant=G*=R 
= 1500 × 0.146 × 10-3
= 0.219 m-1
Hence, cell constant of given conductivity cell is 0.219 m-1 .
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2) Conductivity of 0.00241 M acetic acid is 7.896 × 10 -5 Scm-1. Calculate its molar

conductivity and if λ for acetic acid is 390.5 Scm2 mol-1, what is its dissociation
constant?
Answer:

Hence, the molar conductivity of given solution is 327.63104 Sm2 mol1

and dissociation cons tant is 1.696 1011 molL1.
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Important Questions
Multiple Choice questions-
Question 1. If the conductivity and conductance of a solution is same then its cell constant is
equal to:

(a) 1

(b) 0

(c) 10

(d) 1000

Question 2.The units of conductivity are:

(a) ohm-1

(b) ohm-1 cm-1

(c) ohm-2 cm² equiv-1

(d) ohm-1 cm²

Question 3.The resistance of 0.1 N solution of acetic acid is 250 ohm, when measured in a cell
of cell constant 1.15 cm-1. The equivalent conductance (in ohm-1 cm² equivalent-1) of 0.1 N
acetic acid is

(a) 18.4

(b) 0.023

(c) 46

(d) 9.2

Question 4.In infinite dilution of aqueous solution of BaCl2, molar conductivity of Ba 2+ and Cl–
ions are = 127.32 S cm²/mol and 76.34 S cm2/mol respectively. What is A°m for BaCI2 at same
dilution?

(a) 280 S cm² mol-1

(b) 330.98 S cm² mol-1

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(c) 90.98 S cm² mol-1

(d) 203.6 S cm² mol-1

Question 5.The specific conductance of 0.1 M NaCl solution is 1.06 × 10 -2 ohm-1 cm-1. Its molar
conductance in ohm-1 cm² mol-1 is

(a) 1.06 × 10²

(b)1.06 × 10³

(c) 1.06 × 104

(d) 53

Question 6.The limiting molar conductivities A° for NaCl, KBr and KCl are 126, 152 and 150 S
cm² mol-1 respectively. The A° for NaBr is

(a) 278 S cm² mol-1

(b) 976 S cm² mol-1

(c) 128 S cm² mol-1

(d) 302 S cm² mol-1

Question 7. λ(CICH2COONa) = 224 ohm-1 cm² gm eq-1, λ(NaCl) = 38.2 ohm-1 cm² gm eq-1. λ(HCl)
= 203 ohm-1 cm² gm eq-1, what is the value of λ(CICH2COOH)?

(a) 288.5 ohm-1 cm² gm eq-1

(b) 289.5 ohm-1 cm² gm eq-1

(c) 388.8 ohm-1 cm² gm eq-1

(d) 59.5 ohm-1 cm² gm eq-1

Question 8. The limiting molar conductivities of HCl, CH3COONa and NaCl are respectively 425,
90 and 125 mho cm² mol-1 at 25 °C. The molar conductivity of 0.1 M CH3COOH solution is 7.8
mho cm² mol-1 at the same temperature. The degree of dissociation of 0.1 M acetic acid
solution at the same temperature is
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(a) 0.10

(b) 0.02

(c) 0.15

(d) 0.03

Question 9.The values of limiting ionic conductance of H and HCOO– ions are respectively 347
and 53 S cm² mol-1 at 298 K. If the molar conductance of 0.025 M methanoic acid at 298 K is 40
S cm² mol-1, the dissociation constant of methanoic acid at 298 K is

(a) 1 × 10-5

(b) 2 × 10-5

(c) 1.5 × 10-4

(d) 2.5 × 10-4

Question 10.The ionisation constant of a weak electrolyte is 2.5 × 10 -5 and molar conductance
of its 0.01 M solution is 19.6 S cm² mol-1. The molar conductance at infinite dilution (S cm²
mol-1) is

(a) 402

(b) 392

(c) 306

(d) 39.2

Very Short Question:

Question 1. Can you store AgCl solution in Zinc pot?

Question 2. Define the term – standard electrode potential?

Question 3. What is electromotive force of a cell?

Question 4. Can an electrochemical cell act as electrolytic cell? How?

Question 5. Single electrode potential cannot be determined. Why?

Question 6. What is SHE? What is its electrode potential?

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Question 7. What does the positive value of standard electrode potential indicate?

Question 8. What is an electrochemical series? How does it predict the feasibility of a certain
redox reaction?

Question 9. Give some uses of electrochemical cells?

Question 10. State the factors that affect the value of electrode potential?

Short Questions:
Question 1. What is the cell potential for the cell at

Question 2. Calculate for the reaction

R = 8.314 J/K

Question 3. Calculate Equilibrium constant K for the reaction

+0.34v.

Question 4. For what concentration of will the emf of the given cell be zero at

if the concentration of is 0.1 M?

Question 5. Calculate the standard free energy change for the cell- reaction.

How is it related to the equilibrium

constant of the reaction? ,

Question 6. How much charge is required for the following reductions:

(i) 1 mol of to Al.
(ii) 1 mol of to Cu.
(iii) 1 mol of to .

Question 7. How much electricity in terms of Faraday is required to produce

(i) 20.0 g of Ca from molten .
(ii) 40.0 g of Al from molten
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Question 8. How much electricity is required in coulomb for the oxidation of
(i) 1 mol of to .
(ii) 1 mol of FeO to

Question 9. A solution of is electrolysed between platinum electrodes using a

current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?

Question 10. Depict the galvanic cell in which the reaction takes place. Further show:
(i) Which of the electrode is negatively charged?
(ii) The carriers of the current in the cell.
(iii) Individual reaction at each electrode.

Long Questions:
Question 1. Explain construction and working of standard Hydrogen electrode? (b) Write any
two differences between amorphous solids and crystalline solids.

Question 2.

The molar conductivity of 0.025 mol L-1 methanoic acid is 46.1 S . Calculate its
degree of dissociation and dissociation constant. Given = 349.6 S
and

Question 3.Explain how rusting of iron is envisaged as setting up of an electrochemical cell.

Question 4 Calculate the standard cell potentials of galvanic cells in which the following
reactions take place:

Question 4. Write the Nernst equation and emf of the following cells at 298 K:

Question 5.Define conductivity and molar conductivity for the solution of an electrolyte.
Discuss their variation with concentration.

Assertion and Reason Questions:

1. In these questions, a statement of assertion followed by a statement of reason is given.
Choose the correct answer out of the following choices.

a) Assertion and reason both are correct statements and reason is correct explanation for
assertion.
b) Assertion and reason both are correct statements but reason is not correct explanation
for assertion.
c) Assertion is correct statement but reason is wrong statement.
d) Assertion is wrong statement but reason is correct statement.
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Assertion: At the end of electrolysis using platinum electrodes, an aqueous solution of copper
sulphate tums colourless.

Reason: Copper in CuSO4 is converted to Cu(OH)2 during the electrolysis.

2. In these questions, a statement of assertion followed by a statement of reason is given.

Choose the correct answer out of the following choices.

a) Assertion and reason both are correct statements and reason is correct explanation for
assertion.
b) Assertion and reason both are correct statements but reason is not correct explanation
for assertion.
c) Assertion is correct statement but reason is wrong statement.
d) Assertion is wrong statement but reason is correct statement.

Assertion: Zinc displaces copper from copper sulphate solution.

Reason: Eº of zinc is -0.vV and that of copper is +0.34V.

Case Study Questions:

1. The concentration of potassium ions inside a biological cell is at least twenty times higher than the
outside. The resulting potential difference across the cell is important in several processes such as
transmission of nerve impulses and maintaining the ion balance. A simple model for such a
concentration cell involving a metal M is,

M(s) | M+(aq.; 0.05 molar) || M+(aq; 1 molar) | M(s)

The following questions are multiple choice questions. Choose the most appropriate answer:

(i) For the above cell,

a) Ecell < 0;ΔG > 0

b) Ecell > 0;ΔG < 0
c) Ecell < 0;ΔG° > 0
d) Ecell > 0;ΔG° < 0

(ii) If the 0.05 molar solution of M+ is replaced by a 0.0025 molar M+ solution, then the magnitude
of the cell potential would be:

a) 130mV
b) 185mV
c) 154mV
d) 600mV
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(iii) The value of equilibrium constant for a feasible cell reaction is:

a) <1
b) =1
c) >1
d) Zero

(iv) What is the emf of the cell when the cell reaction attains equilibrium?

a) 1
b) 0
c) >1
d) <1

(v) The potential of an electrode change with change in:

a) Concentration ofions in solution.

b) Position of electrodes.
c) Voltage of the cell.
d) All of these.

2. All chemical reactions involve interaction of atoms and molecules. A large number of
atoms/molecules are present in a few gram of any chemical compound varying with their
atomic/ molecular masses. To handle such large number conveniently, the mole concept was
introduced. All electrochemical cell reactions are also based on mole concept. For example, a
4.0 molar aqueous solution of NaCl is prepared and 500mL of this solution is electrolysed. This
leads to the evolution of chlorine gas at one of the electrode. The amount of products formed
can be calculated by using mole concept.

The following questions are multiple choice questions. Choose the most appropriate answer:

(i) The total number of moles of chlorine gas evolved is:

a) 0.5
b) 1.0
c) 1.5
d) 1.9

(ii) If cathode is a Hg electrode, then the maximum weight of amalgam formed from this
solution is:

a) 300g
b) 446g
c) 396g
d) 296g
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(iii) The total charge (coulomb) required for complete electrolysis is:

a) 186000
b) 24125
c) 48296
d) 193000

(iv) In the electrolysis, the number of moles of electrons involved are:

a) 2
b) 1
c) 3
d) 4

(v) In electrolysis of aqueous NaCl solution when Pt electrode is taken, then which gas is
liberated at cathode?

a) H2gas
b) Cl2gas
c) O2gas
d) None of these.

Answers key
MCQ answers:
1. Answer: (a) 1

2. Answer: (b) ohm-1 cm-1

3. Answer: (c) 46

4. Answer: (a) 280 S cm² mol-1

5. Answer: (a) 1.06 × 10²

6. Answer: (c) 128 S cm² mol-1

7. Answer: (c) 388.8 ohm-1 cm² gm eq-1

8. Answer: (b) 0.02

9. Answer: (d) 2.5 × 10-4

10. Answer: (b) 392

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Very Short Answers:
1. No. We can’t store AgCl solution in Zinc pot because standard electrode potential of Zinc is less
than silver..
2. When the concentration of all the species involved in a half-cell is unity, then the electrode
potential is called standard electrode potential.

3. Answer: Electromotive force of a cell is also called the cell potential. It is the difference
between the electrode potentials of the cathode and anode.

4. Answer: Yes, An electrochemical cell can be converted into electrolytic cell by applying an
external opposite potential greater than its own electrical potential.

5. Answer: A single half cell does not exist independently as reduction and oxidation occur
simultaneously therefore single electrode potential cannot be measured.

6. Answer: SHE stands for standard Hydrogen electrode. By convention, its electrode potential is
taken as 0 (zero).

7. Answer: The positive value of standard electrode potential indicates that the element gets
reduced more easily than ions and its reduced form is more stable than Hydrogen gas.

8. The arrangement of metals and ions in increasing order of their electrode potential values is
known as electrochemical series.The reduction half reaction for which the reduction potential is
lower than the other will act as anode and one with greater value will act as cathode. Reverse
reaction will not occur.

9. Electrochemical cells are used for determining the

 pH of solutions

 solubility product and equilibrium constant

 in potentiometric titrations

10.Factors affecting electrode potential values are –

 Concentration of electrolyte

 Temperature.

Short Answers:
1. Answer
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The cell reaction is

Nernst Equation –

= (-0.44v – (-0.74v) –

= 0.3V – 0.0394V

= +0.2606 V

2. Answer:

The half-cell reactions are

Anode:

Cathode:

Nernst Equation

= (- 0.403 – (-0.763) –

= 0.36V – 0.0798V = 0.4398V

= - 8488 J mol-1

3. Answer:

From the reaction, n =2

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= + 0.34v – (-0.76v) = 1.10V

At 298k,

Log kc =

4. Answer:

5. Answer :

6. Answer
(i)
Therefore, Required charge = 3 F

= 289461 C
(ii)
Therefore, Required charge = 2 F

= 192974 C
(iii)
i.e.,
Therefore, Required charge = 5 F

= 482435 C

7. Answer:
.(i) According to the question,
Electricity required to produce 40 g of calcium = 2 F
Therefore, electricity required to produce 20 g of calcium =
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=1F
(ii) According to the question,

Electricity required to produce 27 g of Al = 3 F

Therefore, electricity required to produce 40 g of Al =

= 4.44 F

8. Answer:

(i) According to the question,

Now, we can write:

Electricity required for the oxidation of 1 mol of to =2F

= 192974 C

(ii) According to the question

Electricity required for the oxidation of 1 mol of FeO to =1F

= 96487 C

9. Answer :

Given,
Current = 5A
Time = = 1200 s
Therefore,

= 6000 C
According to the reaction,

Nickel deposited by = 58.71 g

Therefore, nickel deposited by 6000 C

= 1.825 g
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Hence, 1.825 g of nickel will be deposited at the cathode.

10.Answer :

The galvanic cell in which the given reaction takes place is depicted as:

(i) Zn electrode (anode) is negatively charged.

(ii) Ions are carriers of current in the cell and in the external circuit, current will flow from
silver to zinc.
(iii) The reaction taking place at the anode is given by,

The reaction taking place at the cathode is given by,

Long Answers:
1. Answer:

Construction: SHE consists of a platinum electrode coated with platinum black. The
electrode is dipped in an acidic solution and pure Hydrogen gas is bubbled through it. The
concentration of both the reduced and oxidized. Forms of Hydrogen is maintained at
unity i.e) pressure of gas is 1 bar and concentration of Hydrogen ions in the solution is 1
molar.

Working – The reaction taking place in SHE is At 298 K , the emf of the cell constructed by
taking SHE as anode and other half-cell as cathode, gives the reduction potential of the
other half cell whereas for a cell constructed by taking SHE as anode gives the oxidation
potential of other half cell as conventionally the electrode potential of SHE is zero.

2. Answer:
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= 349.6 + 54.6 =
Now, degree of dissociation:

= 0.114 (approximately)
Thus, dissociation constant:

3. Answer:
In the process of corrosion, due to the presence of air and moisture, oxidation takes place
at a particular spot of an object made of iron. That spot behaves as the anode. The
reaction at the anode is given by,
Electrons released at the anodic spot move through the metallic object and go to another
spot of the object.
There, in the presence of ions, the electrons reduce oxygen. This spot behaves as the
cathode. These H+ ions come either from , which are formed due to the dissolution of
carbon dioxide from air into water or from the dissolution of other acidic oxides from the
atmosphere in water.
The reaction corresponding at the cathode is given

The overall reaction is:

Also, ferrous ions are further oxidized by atmospheric oxygen to ferric ions. These ferric ions
combine with moisture, present in the surroundings, to form hydrated ferric oxide
i.e., rust.
Hence, the rusting of iron is envisaged as the setting up of an electrochemical cell.
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4. Answer (i)

The galvanic cell of the given reaction is depicted as:

Now, the standard cell potential is
= 0.40 – (–0.74)
= +0.34 V

In the given equation,

n=6
F = 96487 C mol – 1
= +0.34 V
Then,
= –196833.48 CV
= –196833.48 J
= –196.83 kJ v
Again,

= 34.496
Therefore, K = antilog (34.496)

(ii)

The galvanic cell of the given reaction is depicted as:

Now, the standard cell potential is
= 0.80 – 0.77
= 0.03 V
Here, n = 1.
Then,
=
= –2894.61 J
= –2.89 kJ
Again,
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= 0.5073
Therefore, K = antilog (0.5073)
= 3.2 (approximately)

5. Answer (i) For the given reaction, the Nernst equation can be given as:

= 2.7 – 0.02955
= 2.67 V (approximately)

(ii) For the given reaction, the Nernst equation can be given as:

= 0.44–0.02955(–3)
= 0.52865 V
= 0.53 V (approximately)

(iii) For the given reaction, the Nernst equation can be given as:

=
= 0.14–0.062
= 0.078 V
= 0.08 V (approximately)

(iv) For the given reaction, the Nernst equation can be given as:
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= –1.09–0.02955(0.0453+7)
= –1.09–0.208
= –1.298 V

Assertion and Reason Answers:

1. (c) Assertion is correct statement but reason is wrong statement.

Explanation:

Cu2+ ions are deposited as Cu.

2. (a) Assertion and reason both are correct statements and reason is correct explanation for
assertion.

Case Study Answers:

1. Answer :
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2. Answer :
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