Which of the following materials is a semiconductor?

Correct Option: C. Solution: Silicon is a well-known semiconductor.

What is the conductivity of pure water at 298 K?

Correct Option: A. Solution: The conductivity of pure water at 298 K is 3.5 x 10⁻⁵ S m⁻¹.

Electrochemistry

Objectives Notes Exam Tips Practice Qs Ask AI

Learning Objectives

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Learning Objectives

Describe an electrochemical cell and differentiate between galvanic and electrolytic cells.
Apply the Nernst equation for calculating the emf of a galvanic cell and define the standard potential of the cell.
Derive the relation between standard potential of the cell, Gibbs energy of cell reaction, and its equilibrium constant.
Define resistivity (ρ), conductivity (κ), and molar conductivity (Λ) of ionic solutions.
Differentiate between ionic (electrolytic) and electronic conductivity.
Describe the method for measurement of conductivity of electrolytic solutions and calculation of their molar conductivity.
Justify the variation of conductivity and molar conductivity of solutions with change in their concentration.
Define 1°m (molar conductivity at zero concentration or infinite dilution).
Enunciate Kohlrausch's law and learn its applications.
Understand quantitative aspects of electrolysis.
Describe the construction of some primary and secondary batteries and fuel cells.
Explain corrosion as an electrochemical process.

Revision Notes & Summary

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Electrochemistry Notes

Objectives

Describe an electrochemical cell and differentiate between galvanic and electrolytic cells.
Apply Nernst equation for calculating the emf of galvanic cells and define standard potential of the cell.
Derive relation between standard potential of the cell, Gibbs energy of cell reaction, and its equilibrium constant.
Define resistivity (ρ), conductivity (κ), and molar conductivity (Λ) of ionic solutions.
Differentiate between ionic (electrolytic) and electronic conductivity.
Describe the method for measurement of conductivity of electrolytic solutions and calculation of their molar conductivity.
Justify the variation of conductivity and molar conductivity of solutions with change in their concentration.
Define 1°m (molar conductivity at zero concentration or infinite dilution).
Enunciate Kohlrausch law and learn its applications.
Understand quantitative aspects of electrolysis.
Describe the construction of some primary and secondary batteries and fuel cells.
Explain corrosion as an electrochemical process.

Key Concepts

Electrochemical Cells

Galvanic Cell: Converts chemical energy into electrical energy.
Electrolytic Cell: Uses electrical energy to drive a non-spontaneous chemical reaction.

Conductivity and Molar Conductivity

Conductivity (κ): Measured in S m⁻¹; depends on the concentration of the electrolyte, nature of solvent, and temperature.
Molar Conductivity (Λ): Defined as Λ = κ/c, where c is the concentration.
- Increases with dilution for weak electrolytes.
- Decreases with dilution for strong electrolytes.

Faraday's Laws of Electrolysis

First Law: The amount of chemical reaction at an electrode is proportional to the quantity of electricity passed.
Second Law: The amounts of different substances liberated by the same quantity of electricity are proportional to their chemical equivalent weights.

Important Formulas

Formula	Description
R = ρ(l/A)	Resistance in terms of resistivity, length, and area of cross-section.
κ = 1/ρ	Conductivity is the inverse of resistivity.
Λ = κ/c	Molar conductivity defined in relation to conductivity and concentration.
E°cell = E°cathode - E°anode	Standard potential of the cell.
ΔG° = -nFE°	Gibbs energy related to cell potential.

Examples

Example 2.7: Calculate Aₘ for CaCl₂ and MgSO₄ using Kohlrausch's law.
Example 2.8: Calculate A° for HAc using known molar conductivities.

Applications

Batteries and Fuel Cells: Convert chemical energy into electrical energy; used in various devices.
Corrosion: An electrochemical phenomenon affecting metals.
Hydrogen Economy: Hydrogen as a renewable energy source through electrochemical processes.

Summary

Electrochemistry involves the study of chemical reactions that produce electrical energy and the use of electrical energy to drive chemical reactions. Key concepts include the types of electrochemical cells, conductivity, molar conductivity, and Faraday's laws of electrolysis.

Exam Tips & Common Mistakes

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Common Mistakes and Exam Tips in Electrochemistry

Common Pitfalls

Misunderstanding Electrochemical Cells: Students often confuse galvanic and electrolytic cells. Remember, galvanic cells generate electricity from spontaneous reactions, while electrolytic cells require an external voltage to drive non-spontaneous reactions.
Incorrect Application of Nernst Equation: Ensure you understand the conditions under which the Nernst equation is applied, particularly the concentrations of reactants and products.
Confusion Between Conductivity and Molar Conductivity: Conductivity refers to the ability of a solution to conduct electricity, while molar conductivity is the conductivity of a solution divided by its molarity. Be clear on these definitions and their units.
Forgetting to Consider Temperature Effects: Conductivity can vary with temperature. Always note the temperature at which measurements are taken, especially in calculations.

Exam Tips

Practice Calculating Standard Cell Potentials: Familiarize yourself with the standard electrode potentials and practice calculating the standard cell potentials for various reactions.
Understand Faraday's Laws of Electrolysis: Be prepared to apply Faraday's laws to calculate the amount of substance produced during electrolysis based on the charge passed.
Review the Law of Independent Migration of Ions: This law is crucial for understanding molar conductivity and its dependence on ion concentration.
Draw Diagrams for Galvanic and Electrolytic Cells: Visual representations can help clarify the processes occurring at the electrodes and the flow of current.
Memorize Key Formulas: Ensure you know the formulas for calculating conductivity, molar conductivity, and the Nernst equation. Familiarize yourself with the units involved.

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Practice Test – MCQs, True/False

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Multiple Choice Questions

1.00 F

1.50 F

2.00 F

0.50 F

Correct Answer: D

Solution:

The moles of calcium produced are

\frac{20.0}{40.08} = 0.499 \, \text{mol}

. Since 1 mole of calcium requires 2 moles of electrons (2 Faradays), the Faradays required are

0.499 \times 2 = 0.998 \, \text{F}

, which is approximately 0.50 F.

68.49 Ω

68.49 kΩ

6.849 Ω

0.6849 Ω

Correct Answer: A

Solution:

The resistance

R

is given by

R = \frac{1}{K \times \text{cell constant}}

. Substituting the values,

R = \frac{1}{0.146 \times 0.1} = 68.49 \text{ Ω}

126.4 S cm² mol⁻¹

149.6 S cm² mol⁻¹

150.0 S cm² mol⁻¹

91.0 S cm² mol⁻¹

Correct Answer: A

Solution:

The limiting molar conductivity of NaCl is 126.4 S cm² mol⁻¹ as per Kohlrausch's law.

Reduction of Cu²⁺ to Cu

Oxidation of Cu to Cu²⁺

Reduction of Ag⁺ to Ag

Oxidation of H₂O to O₂

Correct Answer: B

Solution:

In an electrolytic cell, oxidation occurs at the anode. Thus, Cu is oxidized to Cu²⁺.

3.91 S m⁻¹

0.0248 S cm⁻¹

0.146 x 10⁻³ S cm⁻¹

5.9 x 10³ S m⁻¹

Correct Answer: A

Solution:

The conductivity of 0.1 M HCl solution at 298 K is given as 3.91 S m⁻¹.

0.025

0.05

0.10

0.20

Correct Answer: B

Solution:

The degree of dissociation

\alpha

is given by

\alpha = \frac{\Lambda_m}{\Lambda_m^0}

. Here,

\Lambda_m = 10

S cm² mol⁻¹ and

\Lambda_m^0 = 400

S cm² mol⁻¹. Thus,

\alpha = \frac{10}{400} = 0.025

, which corresponds to option b.

0.132

0.215

0.345

0.500

Correct Answer: A

Solution:

The degree of dissociation

\alpha

is calculated using

\alpha = \frac{\Lambda_m}{\Lambda_m^0}

. Thus,

\alpha = \frac{46.1}{349.6} = 0.132

Fe³⁺ (aq) + I⁻(aq) \rightarrow Fe²⁺ (aq) + I₂ (s)

Ag⁺ (aq) + Cu(s) \rightarrow Ag(s) + Cu²⁺ (aq)

Fe³⁺ (aq) + Br⁻(aq) \rightarrow Fe²⁺ (aq) + Br₂ (s)

Ag(s) + Fe³⁺ (aq) \rightarrow Ag⁺ (aq) + Fe²⁺ (aq)

Correct Answer: A

Solution:

The feasibility of a reaction can be predicted using the standard electrode potentials. Reaction (i) is feasible as it has a positive cell potential.

Copper

Glass

Silicon

Gold

Correct Answer: C

Solution:

Silicon is a well-known semiconductor.

The limiting molar conductivity of an electrolyte is independent of its concentration.

The limiting molar conductivity of an electrolyte is the sum of the individual conductivities of its ions.

The limiting molar conductivity of an electrolyte decreases with dilution.

The limiting molar conductivity of an electrolyte is proportional to its concentration.

Correct Answer: B

Solution:

Kohlrausch's law states that the limiting molar conductivity of an electrolyte is the sum of the individual contributions of the anion and cation of the electrolyte.

The temperature of the electrolyte

The quantity of electricity passed through the electrolyte

The pressure of the system

The volume of the electrolyte

Correct Answer: B

Solution:

Faraday's first law states that the amount of chemical reaction at an electrode is proportional to the quantity of electricity passed through the electrolyte.

0.0202

0.1233

0.202

0.512

Correct Answer: B

Solution:

The molar conductivity (\Lambda_m) is calculated as

\Lambda_m = \frac{7.896 \times 10^{-5} \text{ S cm}^{-1} \times 1000}{0.00241 \text{ mol L}^{-1}} = 32.76 \text{ S cm}^2 \text{ mol}^{-1}

. The degree of dissociation

\alpha = \frac{\Lambda_m}{\Lambda_m^0} = \frac{32.76}{390.5} = 0.0839

119.0 + 2 \times 76.3

2 \times 119.0 + 76.3

119.0 + 76.3

2 \times (119.0 + 76.3)

Correct Answer: A

Solution:

According to Kohlrausch's law, the limiting molar conductivity

\Lambda^0_m

for CaCl

_2

is given by

\Lambda^0_{Ca^{2+}} + 2 \times \Lambda^0_{Cl^-} = 119.0 + 2 \times 76.3 = 271.6

S cm

^2

mol

^{-1}

0.059 V

-0.059 V

0.590 V

-0.590 V

Correct Answer: D

Solution:

The potential of the hydrogen electrode is given by the Nernst equation: E = E^0 - 0.0591 \times \text{pH}. For pH = 10, E = 0 - 0.0591 \times 10 = -0.590 V.

32.76 S cm² mol⁻¹

390.5 S cm² mol⁻¹

48.15 S cm² mol⁻¹

7.896 S cm² mol⁻¹

Correct Answer: A

Solution:

The molar conductivity is calculated using the formula

\Lambda_m = \frac{K \times 1000}{c}

. Substituting the given values,

\Lambda_m = \frac{7.896 \times 10^{-5} \times 1000}{0.00241} = 32.76 \text{ S cm}^2 \text{ mol}^{-1}

1 mole

2 moles

3 moles

4 moles

Correct Answer: C

Solution:

The reaction for the reduction of Al

^{3+}

to Al is:

\text{Al}^{3+} + 3e^- \rightarrow \text{Al}

. Therefore, 3 moles of electrons are required to produce 1 mole of aluminum.

2.65 V

2.75 V

2.71 V

2.67 V

Correct Answer: A

Solution:

Using the Nernst equation:

E_{cell} = E^°_{cell} - \frac{0.059}{n} \log \left( \frac{[Mg^{2+}]}{[Cu^{2+}]} \right)

. Here,

n = 2

E_{cell} = 2.71 - \frac{0.059}{2} \log \left( \frac{0.001}{0.0001} \right) = 2.71 - 0.0295 \times 1 = 2.65 \, \text{V}

2.44 g

1.22 g

3.66 g

4.88 g

Correct Answer: A

Solution:

Total charge passed is

Q = I \times t = 5 \times 20 \times 60 = 6000 \, \text{C}

. Moles of electrons =

\frac{6000}{96500} = 0.0622 \, \text{mol}

. Since 2 moles of electrons deposit 1 mole of Ni, moles of Ni =

\frac{0.0622}{2} = 0.0311 \, \text{mol}

. Mass of Ni =

0.0311 \times 58.69 = 1.82 \, \text{g}

3.5 x 10⁻⁵ S m⁻¹

5.9 x 10³ S m⁻¹

6.2 x 10³ S m⁻¹

1.0 x 10⁻¹⁶ S m⁻¹

Correct Answer: A

Solution:

The conductivity of pure water at 298 K is 3.5 x 10⁻⁵ S m⁻¹.

126.4 S cm² mol⁻¹

425.9 S cm² mol⁻¹

91.0 S cm² mol⁻¹

390.5 S cm² mol⁻¹

Correct Answer: A

Solution:

The limiting molar conductivity of NaCl at infinite dilution is 126.4 S cm² mol⁻¹.

Yes, the reaction is feasible.

No, the reaction is not feasible.

The reaction is feasible only under acidic conditions.

The reaction is feasible only under basic conditions.

Correct Answer: A

Solution:

The standard cell potential for the reaction can be calculated as

E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.77 \text{ V} - 0.54 \text{ V} = 0.23 \text{ V}

. Since the cell potential is positive, the reaction is feasible.

As the sum of the individual contributions of the anion and cation.

As the product of the concentrations of the ions.

As the difference between the conductivity of the cation and anion.

As the square root of the concentration of the electrolyte.

Correct Answer: A

Solution:

Kohlrausch's law states that the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation.

-74.4 kJ/mol

-148.8 kJ/mol

+74.4 kJ/mol

+148.8 kJ/mol

Correct Answer: A

Solution:

The standard Gibbs energy change is given by

\Delta G^0 = -nFE^0

. Here, n = 1, F = 96500 C/mol, and E^0 = +0.77 V. Thus,

\Delta G^0 = -1 \times 96500 \times 0.77 = -74.4

kJ/mol.

1.83 g

2.44 g

3.05 g

4.88 g

Correct Answer: A

Solution:

The charge passed is

Q = I \times t = 5 \text{ A} \times 20 \times 60 \text{ s} = 6000 \text{ C}

. The number of moles of electrons is

\frac{6000}{96,485} \approx 0.0622 \text{ mol}

. Since Ni^{2+} requires 2 moles of electrons per mole of Ni, the moles of Ni deposited is

\frac{0.0622}{2} \approx 0.0311 \text{ mol}

. Mass of Ni deposited is

0.0311 \times 58.69 = 1.83 \text{ g}

1 Faraday

3 Faradays

5 Faradays

7 Faradays

Correct Answer: C

Solution:

The reduction of MnO₄ to Mn²⁺ involves a change of 5 electrons, thus requiring 5 Faradays of charge.

Feasible

Not feasible

Feasible only in acidic medium

Feasible only in basic medium

Correct Answer: B

Solution:

The cell potential for the reaction is calculated as E

_{cell}

= E

^\circ

(Fe

^{3+}

/Fe

^{2+}

) - E

^\circ

(Br

_2

/Br

^-

) = 0.77 V - 1.07 V = -0.30 V. Since the cell potential is negative, the reaction is not feasible under standard conditions.

0.131

0.105

0.092

0.132

Correct Answer: A

Solution:

The degree of dissociation

\alpha

is given by

\alpha = \frac{\Lambda_m}{\Lambda^0_m}

. Substituting the values,

\alpha = \frac{46.1}{349.6} = 0.131

Oxidation occurs

Reduction occurs

No reaction occurs

Electrons are emitted

Correct Answer: B

Solution:

In an electrolytic cell, reduction occurs at the cathode.

32.76 S cm² mol⁻¹

43.23 S cm² mol⁻¹

54.32 S cm² mol⁻¹

65.78 S cm² mol⁻¹

Correct Answer: B

Solution:

Molar conductivity,

\Lambda_m

, is given by

\Lambda_m = \frac{\text{conductivity}}{\text{concentration}} \times 1000

. Substituting the given values,

\Lambda_m = \frac{7.896 \times 10^{-5}}{0.00241} \times 1000 = 43.23 \text{ S cm}^2 \text{ mol}^{-1}

^{3+}

(aq) + I

^-

(aq) \rightarrow Fe

^{2+}

(aq) + I

_2

(s)

^{2+}

(aq) + I

_2

(s) \rightarrow Fe

^{3+}

(aq) + 2I

^-

(aq)

2Fe

^{2+}

(aq) + I

_2

(s) \rightarrow 2Fe

^{3+}

(aq) + 2I

^-

(aq)

^{3+}

(aq) + 2I

^-

(aq) \rightarrow Fe

^{2+}

(aq) + I

_2

(s)

Correct Answer: D

Solution:

For a reaction to be feasible, the cell potential

E^0_{cell}

should be positive. The reaction

Fe^{3+} + 2I^- \rightarrow Fe^{2+} + I_2

has

E^0_{cell} = 0.77 - 0.54 = 0.23 \, \text{V}

, which is positive, indicating feasibility.

0.219 cm⁻¹

0.146 cm⁻¹

0.098 cm⁻¹

0.073 cm⁻¹

Correct Answer: A

Solution:

Cell constant,

G^* = \text{Resistance} \times \text{Conductivity} = 1500 \times 0.146 \times 10^{-3} = 0.219 \text{ cm}^{-1}

0.12 S m⁻¹

0.14 S m⁻¹

0.16 S m⁻¹

0.18 S m⁻¹

Correct Answer: A

Solution:

According to the provided data, the conductivity of 0.01 M NaCl solution at 298 K is 0.12 S m⁻¹.

0.14 S m⁻¹

0.12 S m⁻¹

3.91 S m⁻¹

0.016 S m⁻¹

Correct Answer: A

Solution:

The conductivity of 0.01 M KCl solution at 298 K is given as 0.14 S m⁻¹.

Copper

Silver

Teflon

Graphite

Correct Answer: C

Solution:

Teflon is an insulator with a conductivity of 1.0 x 10⁻¹⁸ S m⁻¹, which is much lower than metals and graphite.

96500 C

193000 C

482500 C

289500 C

Correct Answer: B

Solution:

The reduction of MnO₄⁻ to Mn²⁺ involves a transfer of 5 electrons per ion. Therefore, the charge required is

5 \times 96500 \text{ C/mol} = 482500 \text{ C}

0.084

0.123

0.045

0.067

Correct Answer: A

Solution:

The degree of dissociation,

\alpha

, is calculated using

\alpha = \frac{\Lambda_m}{\Lambda^0_m}

. Here,

\Lambda_m = 32.8 \, \text{S cm}^2 \text{mol}^{-1}

and

\Lambda^0_m = 390.5 \, \text{S cm}^2 \text{mol}^{-1}

. Thus,

\alpha = \frac{32.8}{390.5} = 0.084

-45.54 kJ/mol

-22.78 kJ/mol

-11.39 kJ/mol

0 kJ/mol

Correct Answer: B

Solution:

The standard Gibbs energy change is calculated using

\Delta G^\circ = -nFE^\circ_{cell}

, where

n = 2

moles of electrons,

F = 96,485 \text{ C/mol}

, and

E^\circ_{cell} = 0.236 \text{ V}

. Thus,

\Delta G^\circ = -2 \times 96,485 \times 0.236 = -45,544 \text{ J/mol} = -45.54 \text{ kJ/mol}

Copper

Silver

Gold

Sodium

Correct Answer: B

Solution:

Silver has the highest conductivity at 298.15 K with a value of 6.2x10³ S m⁻¹.

1.98 g

3.17 g

0.95 g

2.38 g

Correct Answer: D

Solution:

The charge passed through the cell is given by Q = I \times t = 3 \times 30 \times 60 = 5400 C. The number of moles of electrons is Q / F = 5400 / 96500 = 0.056 mol. Since the reaction at the cathode is Cu²⁺ + 2e⁻ \rightarrow Cu, the moles of Cu deposited = 0.056 / 2 = 0.028 mol. Therefore, the mass of Cu deposited = 0.028 \times 63.5 = 1.778 g, which is closest to option d.

+1.10 V

-1.10 V

+0.42 V

-0.42 V

Correct Answer: A

Solution:

The standard cell potential is given by E°cell = E°(cathode) - E°(anode). Here, E°(cathode) = E°(Cu²⁺/Cu) = +0.34 V and E°(anode) = E°(Zn²⁺/Zn) = -0.76 V. Thus, E°cell = +0.34 - (-0.76) = +1.10 V.

Glass

Teflon

Silicon

Copper

Correct Answer: C

Solution:

Silicon is a semiconductor with a conductivity of 1.5 x 10⁻² S m⁻¹.

Only cations

Only anions

Both cations and anions

Neither cations nor anions

Correct Answer: C

Solution:

Kohlrausch's law states that the limiting molar conductivity of an electrolyte is the sum of the individual contributions of the anion and cation of the electrolyte.

The conductivity of a solution increases with dilution.

The molar conductivity of a solution decreases with dilution.

The conductivity of a solution is independent of concentration.

The molar conductivity of a solution increases with dilution.

Correct Answer: D

Solution:

The molar conductivity of a solution increases with dilution because the number of ions per unit volume increases, leading to a higher conductivity.

-2.93 V

0.80 V

0.79 V

-2.37 V

Correct Answer: A

Solution:

The standard electrode potential for the reaction K⁺/K is -2.93 V.

Metals with higher electrode potential are deposited first.

Metals with lower electrode potential are deposited first.

All metals deposit simultaneously regardless of electrode potential.

Deposition order is independent of electrode potential.

Correct Answer: A

Solution:

In an electrolytic cell, metals with higher electrode potential (more positive) are reduced and deposited at the cathode first.

0.1233

0.321

0.456

0.789

Correct Answer: A

Solution:

The degree of dissociation,

\alpha

, is given by

\alpha = \frac{\Lambda_m}{\Lambda_m^0}

. Thus,

\alpha = \frac{48.15}{390.5} = 0.1233

The reaction is feasible.

The reaction is not feasible.

The reaction is at equilibrium.

The feasibility cannot be determined from the given information.

Correct Answer: B

Solution:

A negative standard electrode potential indicates that the reaction is not feasible under standard conditions.

5 Faraday

3 Faraday

2 Faraday

1 Faraday

Correct Answer: A

Solution:

The reduction of MnO_4^- to Mn^{2+} involves a change in oxidation state from +7 to +2, which is a gain of 5 electrons. Therefore, 5 Faraday of charge is required.

124 S cm

^2

mol

^{-1}

248 S cm

^2

mol

^{-1}

12.4 S cm

^2

mol

^{-1}

2480 S cm

^2

mol

^{-1}

Correct Answer: A

Solution:

Molar conductivity,

\Lambda_m

, is given by

\Lambda_m = \frac{\kappa}{c}

, where

\kappa = 0.0248 \, \text{S cm}^{-1}

and

c = 0.20 \, \text{mol L}^{-1}

. Therefore,

\Lambda_m = \frac{0.0248}{0.20} \times 1000 = 124 \, \text{S cm}^2 \text{mol}^{-1}

K, Mg, Cr, Hg, Ag

Ag, Hg, Cr, Mg, K

K, Cr, Mg, Ag, Hg

Ag, Cr, Hg, Mg, K

Correct Answer: B

Solution:

The reducing power of a metal is inversely related to its standard electrode potential. The more negative the potential, the stronger the reducing agent. Therefore, the order is Ag < Hg < Cr < Mg < K.

0.132

0.111

0.091

0.072

Correct Answer: A

Solution:

The degree of dissociation,

\alpha

, is given by

\alpha = \frac{\Lambda_m}{\Lambda^0_m}

. Substituting the given values,

\alpha = \frac{46.1}{349.6} = 0.132

Gold

Copper

Silver

Iron

Correct Answer: C

Solution:

Silver has the highest conductivity among the given metals with a value of 6.2 x 10³ S m⁻¹.

266.0 S cm² mol⁻¹

271.6 S cm² mol⁻¹

276.0 S cm² mol⁻¹

281.6 S cm² mol⁻¹

Correct Answer: A

Solution:

According to Kohlrausch's law,

\Lambda^0_{\text{MgSO}_4} = \Lambda^0_{\text{Mg}^{2+}} + \Lambda^0_{\text{SO}_4^{2-}} = 106.0 + 160.0 = 266.0 \text{ S cm}^2 \text{ mol}^{-1}

32.8 \text{ S cm}^2 \text{ mol}^{-1}, 1.75 \times 10^{-5} \text{ mol L}^{-1}

32.8 \text{ S cm}^2 \text{ mol}^{-1}, 1.78 \times 10^{-5} \text{ mol L}^{-1}

32.8 \text{ S cm}^2 \text{ mol}^{-1}, 1.80 \times 10^{-5} \text{ mol L}^{-1}

32.8 \text{ S cm}^2 \text{ mol}^{-1}, 1.82 \times 10^{-5} \text{ mol L}^{-1}

Correct Answer: B

Solution:

Molar conductivity, \Lambda_m = \frac{7.896 \times 10^{-5} \text{ S cm}^{-1} \times 1000 \text{ cm}^3}{0.00241 \text{ mol}} = 32.8 \text{ S cm}^2 \text{ mol}^{-1}. Degree of dissociation, \alpha = \frac{\Lambda_m}{\Lambda^0_m} = \frac{32.8}{390.5} = 0.084. Dissociation constant, K_a = c\alpha^2 = 0.00241 \times (0.084)^2 = 1.78 \times 10^{-5} \text{ mol L}^{-1}.

Aluminium

Copper

Iron

Zinc

Correct Answer: A

Solution:

Aluminium has the highest reducing power and will displace the others from their salt solutions.

0.74 V

0.91 V

-0.74 V

-0.91 V

Correct Answer: A

Solution:

The standard cell potential is calculated using the standard electrode potentials: E^0_{cell} = E^0_{cathode} - E^0_{anode}. Assuming standard electrode potentials for Cr^{3+}/Cr and Cd^{2+}/Cd, the cell potential is calculated as E^0_{cell} = 0.74 V.

1 Faraday

2 Faradays

3 Faradays

4 Faradays

Correct Answer: C

Solution:

To reduce 1 mol of Al³⁺ to Al, 3 Faradays of charge are required because the reduction involves 3 electrons.

Only reaction (i) is feasible

Only reaction (ii) is feasible

Both reactions are feasible

Neither reaction is feasible

Correct Answer: A

Solution:

For reaction (i), the E°cell is positive, indicating a feasible reaction. For reaction (ii), the E°cell is negative, indicating it is not feasible.

124 S cm² mol⁻¹

248 S cm² mol⁻¹

12.4 S cm² mol⁻¹

24.8 S cm² mol⁻¹

Correct Answer: B

Solution:

Molar conductivity,

\Lambda_m = \frac{\text{Conductivity}}{\text{Concentration}} = \frac{0.0248}{0.20} \times 1000 = 248 \text{ S cm}^2 \text{ mol}^{-1}

1 Faraday

2 Faradays

3 Faradays

0.5 Faraday

Correct Answer: A

Solution:

To reduce 1 mol of Cu²⁺ to Cu, 2 moles of electrons are required, which corresponds to 1 Faraday of charge.

0.219 cm^{-1}

0.219 m^{-1}

0.146 cm^{-1}

0.146 m^{-1}

Correct Answer: A

Solution:

The cell constant (G*) is calculated using the formula

G^* = \text{Conductivity} \times \text{Resistance} = 0.146 \times 10^{-3} \text{ S cm}^{-1} \times 1500 \Omega = 0.219 \text{ cm}^{-1}

True or False

Correct Answer: True

Solution:

The given data shows that as the concentration of sodium chloride increases, its conductivity also increases. For example, at 0.001 M, the conductivity is 1.237 S m⁻¹, and at 0.100 M, it is 106.74 S m⁻¹.

Correct Answer: True

Solution:

During the electrolysis of CuCl₂, Cu²⁺ ions are reduced at the cathode to form copper metal.

Correct Answer: True

Solution:

In an electrolytic cell, the anode is positively charged because it is connected to the positive terminal of the external voltage source, causing oxidation to occur at the anode.

Correct Answer: True

Solution:

As a solution is diluted, the number of ions per unit volume decreases, leading to a decrease in conductivity.

Correct Answer: False

Solution:

The conductivity of a material depends on temperature, as well as the nature of the material and pressure.

Correct Answer: True

Solution:

As a solution is diluted, the number of ions per unit volume decreases, leading to a decrease in conductivity.

Correct Answer: True

Solution:

In an electrolytic cell, reduction occurs at the cathode, where cations gain electrons and are reduced.

Correct Answer: True

Solution:

Kohlrausch's law allows the calculation of the limiting molar conductivity of an electrolyte by summing the individual contributions of its ions.

Correct Answer: False

Solution:

The excerpt does not provide the standard electrode potential for this specific reaction. Instead, it gives the potential for a different cell reaction.

Correct Answer: True

Solution:

As acetic acid is diluted, its degree of dissociation increases, leading to a significant increase in molar conductivity.

Correct Answer: True

Solution:

The excerpt indicates that the molar conductivity of KCl decreases as concentration increases, which is typical for strong electrolytes due to ion pairing at higher concentrations.

Correct Answer: True

Solution:

The molar conductivity of a solution generally decreases with an increase in concentration due to the decrease in the degree of dissociation.

Correct Answer: True

Solution:

The table in the excerpt lists the conductivity of pure water at 298.15 K as 3.5 \times 10^{-5} \text{ S m}^{-1}.

Correct Answer: False

Solution:

The conductivity of pure water is 3.5 x 10⁻⁵ S m⁻¹, which is lower than that of 0.01 M KCl solution, which is 0.14 S m⁻¹.

Correct Answer: True

Solution:

In an electrolytic cell, the cathode is the site of reduction, where copper ions gain electrons to form copper metal.

Correct Answer: True

Solution:

Resistivity is the inverse of conductivity. Therefore, as resistivity increases, conductivity decreases, and vice versa.

Correct Answer: True

Solution:

Kohlrausch's law states that the limiting molar conductivity of an electrolyte is the sum of the individual contributions of the anion and cation of the electrolyte.

Correct Answer: False

Solution:

The conductivity of pure water is 3.5 x 10⁻⁵ S m⁻¹, while the conductivity of a 0.01 M NaCl solution is 0.12 S m⁻¹. Therefore, the conductivity of pure water is lower than that of the NaCl solution.

Correct Answer: False

Solution:

For weak electrolytes like acetic acid, the molar conductivity increases with dilution because the degree of dissociation increases.

Correct Answer: False

Solution:

In an electrolytic cell, oxidation occurs at the anode, while reduction occurs at the cathode.

Correct Answer: True

Solution:

As shown in the data, the conductivity of sodium chloride increases with increasing concentration, from 1.237 S m⁻¹ at 0.001 M to 106.74 S m⁻¹ at 0.100 M.

Correct Answer: True

Solution:

The limiting molar conductivity,

\Lambda_m^0

, is defined as the molar conductivity at infinite dilution, where the electrolyte is completely dissociated, and is therefore independent of concentration.

Correct Answer: False

Solution:

The standard electrode potential for the reaction between Fe³⁺ and I⁻ is not given as positive in the provided excerpts, suggesting it may not be spontaneous.

Correct Answer: False

Solution:

The conductivity of pure water is 3.5x10⁻⁵ S m⁻¹, while the conductivity of 0.01 M NaCl is 0.12 S m⁻¹, which is higher.

Correct Answer: True

Solution:

For weak electrolytes, dilution increases the degree of dissociation because more ions are formed as the concentration decreases.

Correct Answer: False

Solution:

The oxidation of 1 mol of H₂O to O₂ requires more electricity in terms of coulombs than the oxidation of 1 mol of FeO to Fe₂O₃.

Correct Answer: True

Solution:

In an electrolytic cell, Cu²⁺ ions gain electrons at the cathode and are reduced to form copper metal.

Correct Answer: True

Solution:

Resistivity and conductivity are inversely related; resistivity is the resistance of a material with a specific dimension, while conductivity measures how easily electricity flows through it.

Correct Answer: False

Solution:

The conductivity of pure water is 3.5 x 10⁻⁵ S m⁻¹, which is lower than the conductivity of 0.1 M HAc, which is 0.047 S m⁻¹.

Correct Answer: True

Solution:

Kohlrausch's law of independent migration of ions states that the limiting molar conductivity of an electrolyte is the sum of the individual contributions of the anion and cation, which is independent of concentration.

Correct Answer: True

Solution:

Electrolysis is used in industrial processes to purify metals, such as converting impure copper into high purity copper.

Correct Answer: False

Solution:

The oxidation of 1 mol of H₂O to O₂ and the reduction of 1 mol of Al³⁺ to Al both involve the transfer of 2 moles of electrons. However, the reduction of Al³⁺ to Al requires more electricity due to the higher charge of Al³⁺.

Correct Answer: True

Solution:

For strong electrolytes, the molar conductivity increases with dilution due to the decrease in ion-ion interactions.

Correct Answer: False

Solution:

The conductivity of pure water is much lower than that of a 0.1 M HCl solution. Pure water has a conductivity of 3.5 \times 10^{-5} \text{ S m}^{-1}, while 0.1 M HCl has a conductivity of 3.91 \text{ S m}^{-1}.

Correct Answer: True

Solution:

Kohlrausch's law of independent migration of ions states that the limiting molar conductivity of an electrolyte is the sum of the contributions of its ions.

Correct Answer: True

Solution:

According to Kohlrausch's law of independent migration of ions, the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation.

Correct Answer: False

Solution:

The conductivity of 0.00241 M acetic acid is 7.896 x 10⁻⁵ S cm⁻¹, which is lower than the conductivity of 0.01 M KCl, which is 0.14 S m⁻¹.

Correct Answer: True

Solution:

The resistance R is given by the formula R = \rho \frac{l}{A}, where \rho is the resistivity, l is the length, and A is the area of cross-section.

Correct Answer: False

Solution:

Faraday's first law of electrolysis states that the amount of chemical reaction at an electrode is directly proportional to the quantity of electricity passed through the electrolyte.

Correct Answer: True

Solution:

Molar conductivity decreases with an increase in concentration because the ions are closer together, leading to increased interactions and decreased mobility.

Correct Answer: False

Solution:

The production of aluminum requires more electricity because it involves a higher molar mass and more electrons per atom in the reduction process compared to calcium.

Correct Answer: True

Solution:

For strong electrolytes, molar conductivity decreases with an increase in concentration due to ion interactions.

Correct Answer: False

Solution:

The conductivity of pure water is 3.5 x 10⁻⁵ S m⁻¹, which is much lower than the conductivity of 0.01 M KCl, which is 0.14 S m⁻¹.

Correct Answer: True

Solution:

During the electrolysis of copper sulfate solution with copper electrodes, copper ions are reduced and deposited at the cathode.

Correct Answer: True

Solution:

Faraday's second law states that the amounts of different substances liberated by the same quantity of electricity passing through the electrolytic solution are proportional to their chemical equivalent weights.

Correct Answer: True

Solution:

Faraday's first law states that the amount of chemical reaction at an electrode during electrolysis is proportional to the quantity of electricity passed through the electrolyte.

Correct Answer: True

Solution:

Kohlrausch law states that the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation.

Correct Answer: True

Solution:

The data provided shows that as the concentration of sodium chloride increases, its conductivity also increases.

Correct Answer: True

Solution:

Resistivity is the inverse of conductivity, as resistivity is the measure of how strongly a material opposes the flow of electric current, while conductivity measures how easily electricity flows through a material.

Correct Answer: True

Solution:

According to Kohlrausch's law of independent migration of ions, the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its constituent ions.

Correct Answer: False

Solution:

Acetic acid is a weak electrolyte and partially dissociates in solution, resulting in lower conductivity compared to the fully dissociated strong electrolyte sodium chloride.

Correct Answer: True

Solution:

According to the provided excerpt, the conductivity of 0.00241 M acetic acid is indeed 7.896 \times 10^{-5} \text{ S cm}^{-1}.

Correct Answer: True

Solution:

Faraday's first law states that the amount of chemical reaction at an electrode during electrolysis is proportional to the quantity of electricity passed.

Correct Answer: False

Solution:

In an electrolytic cell, copper ions are reduced and deposited at the cathode, not the anode.

Correct Answer: True

Solution:

The standard electrode potential for the reaction between Fe³⁺ and I⁻ is positive, suggesting that the reaction is feasible.

Correct Answer: False

Solution:

The conductivity of a semiconductor increases with an increase in temperature as more charge carriers are generated.

Correct Answer: True

Solution:

The given standard electrode potential for the reaction is indeed 0.236 V at 298 K, as stated in the provided excerpts.

Correct Answer: False

Solution:

The conductivity of pure water is 3.5 \times 10^{-5} S \ m^{-1}, while the conductivity of 0.1 M HCl is 3.91 \ S \ m^{-1}. Therefore, the conductivity of 0.1 M HCl is higher.

Correct Answer: True

Solution:

During electrolysis, Cu²⁺ ions are reduced at the cathode to form copper metal.

Electrochemistry

Learning Objectives

Learning Objectives

Revision Notes & Summary

Electrochemistry Notes

Objectives

Key Concepts

Electrochemical Cells

Conductivity and Molar Conductivity

Faraday's Laws of Electrolysis

Important Formulas

Examples

Applications

Summary

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips in Electrochemistry

Common Pitfalls

Exam Tips

AI Learning Assistant

Practice Test – MCQs, True/False

Multiple Choice Questions

Q1.For the electrolysis of molten CaCl2CaCl_2CaCl2​, how many Faradays are required to produce 20.0 g of calcium? (Atomic mass of Ca = 40.08 g/mol)

Correct Answer: D

Q2.The conductivity of a 0.01 M KCl solution at 298 K is 0.146 S m−1^{-1}−1. If the cell constant is 0.1 cm−1^{-1}−1, what is the resistance of the solution?

Correct Answer: A

Q3.According to Kohlrausch's law, what is the limiting molar conductivity of NaCl?

Correct Answer: A

Q4.In an electrolytic cell, which of the following reactions occurs at the anode?

Correct Answer: B

Q5.What is the conductivity of a 0.1 M HCl solution at 298 K?

Correct Answer: A

Q6.The molar conductivity of a 0.01 M solution of a weak acid HA is measured to be 10 S cm² mol⁻¹. If the limiting molar conductivity of HA is 400 S cm² mol⁻¹, what is the degree of dissociation (α\alphaα) of HA?

Correct Answer: B

Q7.What is the degree of dissociation for methanoic acid if its molar conductivity is 46.1 S cm² mol⁻¹ and the limiting molar conductivity is 349.6 S cm² mol⁻¹?

Correct Answer: A

Q8.Using the standard electrode potentials, which of the following reactions is feasible?

Correct Answer: A

Q9.Which of the following materials is a semiconductor?

Correct Answer: C

Q10.According to Kohlrausch's law, which of the following statements is true?

Correct Answer: B

Q11.According to Faraday's first law of electrolysis, the amount of chemical reaction at an electrode is proportional to:

Correct Answer: B

Q12.The conductivity of a 0.00241 M acetic acid solution at 298 K is 7.896 \times 10^{-5} \text{ S cm}^{-1}. If the molar conductivity at infinite dilution for acetic acid is 390.5 \text{ S cm}^2 \text{ mol}^{-1}, what is the degree of dissociation (\alpha) of acetic acid in this solution?

Correct Answer: B

Correct Answer: A

Q14.What is the potential of a hydrogen electrode in contact with a solution whose pH is 10?

Correct Answer: D

Q15.What is the molar conductivity of 0.00241 M acetic acid if its conductivity is 7.896 x 10⁻⁵ S cm⁻¹?

Correct Answer: A

Q16.For the electrolysis of molten Al2_22​O3_33​, how many moles of electrons are required to produce 1 mole of aluminum?

Correct Answer: C

Correct Answer: A

Q18.In an electrolytic cell, a current of 5 amperes is passed for 20 minutes through a solution of Ni(NO3)2Ni(NO_3)_2Ni(NO3​)2​. What mass of nickel is deposited at the cathode? (Atomic mass of Ni = 58.69 g/mol, 1 Faraday = 96500 C/mol)

Correct Answer: A

Q19.What is the conductivity of pure water at 298 K?

Correct Answer: A

Q20.What is the limiting molar conductivity of NaCl at infinite dilution?

Correct Answer: A

Q21.Using the standard electrode potentials, predict if the reaction between Fe^{3+} (aq) and I^- (aq) is feasible. Given: E^\circ (Fe^{3+}/Fe^{2+}) = +0.77 V, E^\circ (I_2/I^-) = +0.54 V.

Correct Answer: A

Q22.According to Kohlrausch's law, how is the limiting molar conductivity of an electrolyte represented?

Correct Answer: A

Q23.If the standard electrode potential for the reaction Fe³⁺ + e⁻ \rightarrow Fe²⁺ is +0.77 V, what is the standard Gibbs energy change (ΔG0\Delta G^0ΔG0) for the reduction of 1 mole of Fe³⁺ ions? (F = 96500 C/mol)

Correct Answer: A

Q24.Consider the electrolysis of a solution of Ni(NO_3)_2 using platinum electrodes with a current of 5 amperes for 20 minutes. What mass of nickel is deposited at the cathode? (Molar mass of Ni = 58.69 g/mol, 1 Faraday = 96,485 C/mol)

Correct Answer: A

Q25.How much charge is required to reduce 1 mol of MnO₄ to Mn²⁺?

Correct Answer: C

Correct Answer: B

Q27.What is the degree of dissociation (α\alphaα) of a 0.025 mol L−1^{-1}−1 methanoic acid solution with a molar conductivity of 46.1 S cm2^22 mol−1^{-1}−1, given that Λm0\Lambda^0_mΛm0​ for methanoic acid is 349.6 S cm2^22 mol−1^{-1}−1?

Correct Answer: A

Q28.In an electrolytic cell, what happens at the cathode?

Correct Answer: B

Q29.The conductivity of a 0.00241 M acetic acid solution is 7.896 x 10⁻⁵ S cm⁻¹. Calculate its molar conductivity.

Correct Answer: B

Q30.Which of the following reactions is feasible based on the given standard electrode potentials? EFe3+/Fe2+0=0.77 V, EI2/I−0=0.54 VE^0_{Fe^{3+}/Fe^{2+}} = 0.77 \, \text{V}, \; E^0_{I_2/I^-} = 0.54 \, \text{V}EFe3+/Fe2+0​=0.77V,EI2​/I−0​=0.54V.

Correct Answer: D

Q31.What is the cell constant if the resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 Ω and its conductivity is 0.146 x 10⁻³ S cm⁻¹?

Correct Answer: A