2nd PUC Chemistry Question Bank Chapter 3 Electrochemistry

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Karnataka 2nd PUC Chemistry Question Bank Chapter 3 Electrochemistry

Question 1.
What is standard electrode potential?
Answer:
The potential developed between the interface of metal and the solution containing its ions of unit concentration is called standard electrode potential.

Question 2.
What is electrochemical ceil?
Answer:
A device used to convert chemical energy produced in redox reaction into electrical energy is called electrochemical cell or galvanic cell or voltaic cell.

Question 3.
What is cell potential?
Answer:
The cell potential is the difference between the electrode potentials of the cathode and anode in electrochemical cell. It is also called emf (electromotive force) of the cell when no current is drawn through the cell.
E_cell = E_Cathode – E_Anode

Question 4.
Explain with diagram the construction, cell reactions and symbolic representation of the Daniel cell.
Answer:
Daniel cell is a electrochemical cell. It consists of two half cells or two electode systems.
Zinc rod dipped in 1M ZnSO₄ forms Anode. In this electrode, Zinc rod is -vely charged with respect to the solution and zinc undergoes oxidation.

Copper rod dipped in 1M CuSO₄ forms Cathode. In this electrode, copper rod is +vely charged with respect to the solution and copper ions undergoes reduction.

The metallic rods in the vessel are connected through ammeter by means of insulated wire. The solutions in the two beakers are connected by means of salt bridge.

When the circuit is completed electrons flows from zinc rod (anode) to copper rod (cathode) in the external circuit. Therefore, current flows from copper to zinc.

Question 5.
With diagram, explain the construction and working of standard hydrogen electrode.
Answer:

The standard hydrogen electrode consists of a platinum electrode coated with platinum black.
The electrode is dipped in an acidic solution containing 1MH⁺ ions. Pure hydrogen gas at one bar pressure is bubbled through it.
Standard hydrogen electrode (SHE) is represented as

Potential is developed when equilibrium attained between H+ ions of the solution and H, gas on the interface of the electrode.
H⁺(aq) + e^– → \(\frac { 1 }{ 2 }\)H₂(g)
The electrode potential for standard hydrogen electrode (SHE) is taken as zero at all temperature.

Question 6.
Draw a neat diagram of SHE and write its half cell reaction and E® value.
Answer:

The standard hydrogen electrode consists of a platinum electrode coated with platinum black.
The electrode is dipped in an acidic solution containing 1MH⁺ ions. Pure hydrogen gas at one bar pressure is bubbled through it.
Standard hydrogen electrode (SHE) is represented as

Potential is developed when equilibrium attained between H+ ions of the solution and H, gas on the interface of the electrode.
H⁺(aq) + e^– → \(\frac { 1 }{ 2 }\)H₂(g)
The electrode potential for standard hydrogen electrode (SHE) is taken as zero at all temperature.

Question 7.
Draw a neat labelled diagram of standard hydrogen electrode.
Answer:

The standard hydrogen electrode consists of a platinum electrode coated with platinum black.
The electrode is dipped in an acidic solution containing 1MH⁺ ions. Pure hydrogen gas at one bar pressure is bubbled through it.
Standard hydrogen electrode (SHE) is represented as

Potential is developed when equilibrium attained between H+ ions of the solution and H, gas on the interface of the electrode.
H⁺(aq) + e^– → \(\frac { 1 }{ 2 }\)H₂(g)
The electrode potential for standard hydrogen electrode (SHE) is taken as zero at all temperature.

Question 8.
Draw a neat labelled diagram of SHE and write its symbolic representation.
Answer:

The standard hydrogen electrode consists of a platinum electrode coated with platinum black.
The electrode is dipped in an acidic solution containing 1MH⁺ ions. Pure hydrogen gas at one bar pressure is bubbled through it.
Standard hydrogen electrode (SHE) is represented as

Potential is developed when equilibrium attained between H+ ions of the solution and H, gas on the interface of the electrode.
H⁺(aq) + e^– → \(\frac { 1 }{ 2 }\)H₂(g)
The electrode potential for standard hydrogen electrode (SHE) is taken as zero at all temperature.

Question 9.
What is electrochemical series?
Answer:
Arrangement of various electrode systems in the decreasing order of their standard reduction potential is called electrochemical series.

Question 10.
Give the list of reduction potential of some important electrode systems.
Answer:

Question 11.
Write the Nernst equation for an electrode system to calculate standard reduction potential.
Answer:

Question 12.
Write Nernst equation at 298 K to calculate reduction potential.
Answer:

Question 13.
Write the Nernst equation for Daniel cell.
Answer:
Cell reaction in Daniel cell is

At 298 K (25° C)

Question 14.
How E_cell of the Daniel cell affected by increasing the concentration of Cu²⁺ and Zn²⁺ ?
Answer:
As [Cu²⁺] increases E_cell increases and as [Zn²⁺] increases E_cell decreases.

Question 15.
Give Nernst equation to calculate electrode potential for any concentration of ions for the following electrochemical cell.
Answer:
Ni(s) | Ni²⁺ (aq) | Ag(aq) | Ag^–
The cell reaction is Ni + 2Ag⁺(aq) → Ni²⁺(aq) + 2Ag

Question 16.
Give the relation between equilibrium constant of the reaction and standard potential of the cell.
Answer:

E°_(cell) → Standard potential of the cell
R → Gas constant (8.314 J K^-1 mol^-1).
T → Absolute temperature.
n → No. of electrons involved.
F → Faraday constant (96500C mol^-1‘)
K_c → Equilibrium constant.
At 298 K or 25°C

Question 17.
Give relation between standard Gibbs energy of the reaction (Δ_rG°) and standard potential of the cell.
Answer:
Δ_r G° = -nFE°_cell
Δ_r G° → Standard Gibbs energy of the reaction.
n → No. of electrons involved in the reaction.
E°_cell → Standard potential of the cell.

Question 18.
Among cell potential (E_cell) and Gibbs free energy of the reaction, which is extensive property.
Answer:
(Δ_r G) Gibbs free energy of the reaction → Extensive property.
(E_cell) potential → Intensive property.

Question 19.
Give relation between standard Gibbs energy of the reaction and equilibrium constant.
Answer:
Δ_r G° = -2.303 RT log K
Δ_r G° → Standard Gibbs free energy of the reaction.
R → Gas constant (8.314 J K mol^-1).
T → Absolute temperature.
K → Equilibrium constant.

Question 20.
How electrical resistance varies along the length and area of cross section.
Answer:
Electrical resistance increases as its length increases (directly proportional to its length) and decreases as the area of cross section increases (inversely proportional to area of cross section).

Question 21.
Define the term conductivity (specific conductance). Give its unit
Answer:
Conductivity of a material is its conductance when it is 1 metre long and its area of cross section is one square metre (m²).
S.I unit – Siemen / metre (Sm^-1) / Ω^-1 m^-1 / mho m^-1.

Question 22.
What is electronic conductance. Name the factors on which this conductance depends?
Answer:
Electrical conductance through movement of electrons is called electronic conductance. Electronic conductance depends upon:

• The nature and structure of the metal.
• Number of valence electrons per atom.
• Temperature.

Question 23.
What is electrolytic or ionic conductance? Name the factors on which this conductance depends. [5M]
Answer:
The conductance of electricity by ions present in the solutions is called electrolytic or ionic conductance. Ionic conductance depends upon.

• Nature of the electrolyte.
• Nature of the solvent.
• Concentration of the electrolyte.
• Temperature.

Question 24.
Mention two factors which affects the conductivity of electrolytic solution.
Answer:
The conductance of electricity by ions present in the solutions is called electrolytic or ionic conductance. Ionic conductance depends upon.

• Nature of the electrolyte.
• Nature of the solvent.
• Concentration of the electrolyte.
• Temperature.

Question 25.
Give three differences between electronic and electrolytic conductance.
Answer:

Electronic conductance	Electrolytic conductance
1. Due to movement of electrons.	1. Due to movement of ions.
2. Decreases with increase in temperature.	2. Increases with increase in temperature.
3. There is no mass transfer.	3. There is mass transfer.

Question 26.
Explain the measurement of the conductivity of an ionic solutions.
Answer:
Conductivity of an ionic solution is given by the equation

K : is conductivity of the solution.
R : is resistance offered by the solution.
Cell constant (l/A) is determined by measuring the a resistance of the cell containing solution whose conductivity already known.

Resistance (R) of the ionic solution taken in the conductivity cell is determined using wheatstone bridge.

Substituting (R) and cell constant value in the above equation, conductivity of the solution can be determined.

Question 27.
What is molar conductivity? Give its unit.
Answer:
Molar conductivity is defined as the conducting power of all the ions produced by 1 mole of electrolyte in solution. It is denoted as ∧_m
∧_m (S.I unit) Sm² mol^-1.
(Non-S.I unit) S cm^2- mol^-1.

Question 28.
What is the S.I unit of molar conductivity.
Answer:
Sm² mol^-1

Question 29.
Give the relation between conductivity and molar conductivity.
Answer:

∧_m → Molar conductivity expressed in S cm^-1 mol^-1.
K → Conductivity expressed in S cm^-1.
C → Concentration in mol L^-1

Question 30.
What happens to the molar conductivity when 1 mole of KCl is dissolve in 1 litre and diluted to 5 litres.
Answer:
Molar conductivity increases.

Question 31.
What is the effect of change in concentration of an electrolyte solution on conductivity and molar conductivity?
Answer:
As concentration decreases (dilution increases) conductivity decreases and molar conductivity increases for both weak and strong electrolytes.

Question 32.
What is limiting molar conductivity?
Answer:
Molar conductivity of an electrolyte solution at infinite dilution (when concentration approches zero) is called limiting molar conductivity.

Question 33.
State Kohlrausch law of independent migration of ions. Give mathematical form of the law.
Answer:
Limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte.
If the molar conductivity of cation is denoted by λ₊° and that of anion λ_–°

Where v₊ and v_– are the number of cations and anions per formula unit of an electrolyte.

Question 34.
State Kohlrausch law.
Answer:
Limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte.
If the molar conductivity of cation is denoted by λ₊° and that of anion λ_–°

Where v₊ and v_– are the number of cations and anions per formula unit of an electrolyte.

Question 35.
Write the mathematical expression for limiting molar conductivity of sodium chloride. [NaCl]
Answer:

Question 36.
Give three applications of Kohlrausch law.
Answer:

1. Using this law it is possible to calculate ∧_m° for any electrolyte.
2. Using this law it is possible to calculate degree of dissociation of weak electrolyte.
3. Based on this law it is possible to calculate dissociation constant of a weak electrolyte.

Question 37.
State Faraday’s first law of clectrolvsis. Give its mathematical form.
Answer:
During electrolysis the amount of any substance deposited or liberated at any electrode is directly proportional to the quantity of electricity passed through the electrolytic solution.
W = ZQ; Q = It
W = ZIt
w → Mass of substance discharged at an electrode.
Z → Electrochemical equivalent.
I → Electric current in amperes.
t → Time in seconds.

Question 38.
State Faraday’s second law of electrolysis. Give its mathematical form.
Answer:
During electrolysis, when same quantity of electricity is passed through different electrolytic solutions connected in series, the mass of the substance produced at the electrodes are directly proportional to their equivalent masses.

Question 39.
How many coulombs of electricity is required to oxidise one mole of A1 I to A1³⁺ ?
Answer:
3 × 90500 coulombs

Question 40.
State Faraday’s first law of electrolysis. For the electrode reaction Zn²⁺ + 2e^– → Zn, what quantity of electricity in coulombs is required to deposit one mole of zinc.
Answer:
During electrolysis the amount of any substance deposited or liberated at any electrode is directly proportional to the quantity of electricity passed through the electrolytic solution.
W = ZQ; Q = It
W = ZIt
w → Mass of substance discharged at an electrode.
Z → Electrochemical equivalent.
I → Electric current in amperes.
t → Time in seconds.
To deposit 1 mole of Zinc from Zn²⁺ ion.
96500 × 2 coulombs are required.

Question 41.
What are the electrolysed product when molten sodium chloride is electrolysed?
Answer:
When molten sodium chloride is electrolysed sodium metal is discharged at cathode and chlorine gas is liberated at anode.
Anode: 2Cl^– → Cl₂ + 2e^–
Cathode: Na⁺ + e^– → Na.

Question 42.
What are the electrolysed product when aqueous sodium chloride is electrolysed?
Answer:
On electrolysis of aqueous sodium chloride H⁺ is discharged at cathode as hydrogen gas, due to its low discharge potential than Na⁺ and chlorine gas is liberated at anode.
Cathode: 2H⁺ + 2e^– → H₂
Anode: 2Cl^– → Cl₂ + 2e^–

Question 43.
Which gas is evolved at the cathode during the electrolysis of an aqueous solution.
Answer:
Hydrogen gas.

Question 44.
What is primary cell?
Answer:
The cell cannot be reused again is called primary cell.

Question 45.
With diagram explain dry cell (leclanche cell).
Answer:
Dry cell is a primary battery.

Anode: Zinc container of the cell.
Cathode: Carbon rod surrounded by MnO₂ powder.
The space between the anode and cathode is filled by paste of NH₄Cl and ZnCl₂ mixture.

Question 46.
Describe mercury cell
Answer:
Mercury cell is a primary cell. Suitable for low current devices.

Question 47.
What is secondary cell.
Answer:
A secondary cell after use can be recharged so that it can be used again.

Question 48.
Describe lead storage battery.
Answer:
It is a secondary cell, mainly used in auto mobiles.

Question 49.
What is secondary cell? Write the equation for the cathode reaction of lead storage battery.
Answer:
A secondary cell after use can be recharged so that it can be used again.
It is a secondary cell, mainly used in auto mobiles.

Question 50.
Give the overall reaction in Nickel cadmium coll.
Answer:

Question 51.
What are fuel cells?
Answer:
Galvanic cells that converts energy of combustion of fuels directly into electrical energy are called fuel cells.

Question 52.
Describe H₂ – O₂ fuel cell with diagram.
Answer:

In this cell, hydrogen and oxygen are bubbled through porous carbon electrodes into concentrated aqueous sodium hydroxide.
Hydrogen is fed into the anode compartment where it is oxidised.
The oxygen is fed into cathode compartment where it is reduced.

Question 53.
Write the cathodic and anodic cell reactions of hydrogen oxygen fuel cell.
Answer:

In this cell, hydrogen and oxygen are bubbled through porous carbon electrodes into concentrated aqueous sodium hydroxide.
Hydrogen is fed into the anode compartment where it is oxidised.
The oxygen is fed into cathode compartment where it is reduced.

Question 54.
What is corrosion?
Answer:
Destruction of metals by chemical and electrochemical agencies of atomsphere is called corrosion.

Question 55.
What is corrosion? Name a method to prevent it.
Answer:
Destruction of metals by chemical and electrochemical agencies of atomsphere is called corrosion.

• Painting the surface of the metal or covering the surface of the metal by a chemical bisphenol.
• Coat the surface of the metal that are inert and prevents corrosion.
• Coat the surface of the metal that are reactive and corrods itself and saves the object.

Question 56.
Corrosion is a electrochemical phenomenon. Explain taking rusting of iron as example.
Answer:
When iron object exposed to moist air corrosion takes place. Iron being electropositive undergoes oxidation releasing electrons.
These electrons go to another spot of the metal and reduces oxygen in presence of H⁺.

The ferrous ions are further oxidised by atmospheric oxygen to ferric ions which come out as rust in the form of hydrated ferric oxide. Fe₂O₃. xH₂O.

Question 57.
Give three methods of prevention of rusting of iron.
Answer:

1. Painting the surface of the metal or covering the surface of the metal by a chemical bisphenol.
2. Coat the surface of the metal that are inert and prevents corrosion.
3. Coat the surface of the metal that are reactive and corrods itself and saves the object.

Question 58.
Write the equation for the anodic reaction during rusting of iron.
Answer:
When iron object exposed to moist air corrosion takes place. Iron being electropositive undergoes oxidation releasing electrons.
These electrons go to another spot of the metal and reduces oxygen in presence of H⁺.

The ferrous ions are further oxidised by atmospheric oxygen to ferric ions which come out as rust in the form of hydrated ferric oxide. Fe₂O₃. xH₂O.

Question 59.
Write the equations for the anodic and cathodic reactions occurring during rusting of iron.
Answer:
When iron object exposed to moist air corrosion takes place. Iron being electropositive undergoes oxidation releasing electrons.
These electrons go to another spot of the metal and reduces oxygen in presence of H⁺.

The ferrous ions are further oxidised by atmospheric oxygen to ferric ions which come out as rust in the form of hydrated ferric oxide. Fe₂O₃. xH₂O.

2nd PUC Chemistry Electrochemistry Problems and Solutions

Question 1.
E_cell° and E_zn° of electrochemical cell E_cell°.
Answer:

Question 2.
E°_cu = +0.34v and E°_Ag = +0.8V calculate E°_cell
Answer:

Question 3.

Answer:

Question 4.
Calculate the EMF of the cell reaction.
Ni(s) + 2Ag⁺(0.002M) → Ni²⁺(0.100M) + 2Ag(s)
Given E°_cell = 1.05 V.
Answer:
Data: [Ni²⁺] = 0.100 M [Ag⁺] = 0.002M
Ni(s) + 2Ag⁺ → Ni²⁺ + 2Ag(s)
n = 2
E°_cell = 1.05 V

Question 5.
Represent the cell in which the following reaction takes place.
Mg(s) + 2Ag⁺ (0.001M) → Mg²⁺ (0.130M) + 2Ag(s)
Calculate E_cell if E°_cell = 3.17 V.
Answer:
Representation of the cell
Mg/Mg²⁺(0.130M) | | Ag⁺ (0.01M) / Ag
Cell reaction Mg + 2 Ag⁺ → Mg²⁺ + 2Ag


Answer = 2.96 V.

Question 6.
Calculate EMF of the cell represents below
Zn/Zn²⁺ (C = 0.1M) | | Cu²⁺ (C = 1M)/ Cu at 25°C.
Given E°_cu = 0.34V, E°_Zn = – 0.76V.
Answer:
1.1295V.

Question 7.
For the standard cell Cu(s) | Cu²⁺(aq) | | Ag⁺ (aq) | Ag(s).

(i) Identify the cathode and the anode as the current is drawn from the cell.
(ii) Write the reaction taking place at the electrodes.
(iii) Calculate the standard cell potential.
Answer:
(i) Cathode is Ag⁺/Ag and anode is Cu²⁺/Cu
(ii) At cathode Ag⁺ + e^– → Ag
At anode Cu → Cu²⁺ + 2e^–

Question 8.
Write the Nernst equation and calculate the emf of the following cell at 298 K.
Mg (s) | Mg²⁺ (0.001M) | | Cu²⁺ (0.0001M) | Cu(s).

Answer:
2.67 V.

Question 9.
Calculate Δ_rG° for the reaction
Fe²⁺(aq) + Ag⁺(aq) Fe³⁺(aq) + Ag(s)
(E°_cell = 1.57V, F = 96487C).
Answer:
Data: E°_cell = 1.57 V
Fe²⁺ + Ag⁺ → Fe³⁺ + Ag(s)
No. of electrons involved in the reaction = n = 1
F = 96487 C
Formula: ΔG° = -nF E°_cell
= -1 × 96487 × 1.57 = 151.48 kJ

Question 10.
Find the value of ΔG° at 25°C for the following electrochemical cell.
Cu | Cu²⁺ (1M) | | Ag⁺ (1M) I Ag. [E°_cu = +0.34V, E°_Ag =+0.8V], F = 96487C.
Answer:

Question 11.
The electrode potential for the Daniel cell given below is 1.1V
Zn(s) | Zn²⁺ (aq) | | Cu²⁺ (aq) | Cu(s)
Write overall cell reaction and calculate the standard Gibb’s energy for the reaction. [F = 96487 C/mol].
Answer:
Data: E°_cell = 1.1V; F = 96487 V
Cell reaction: Zn(s) + Cu⁺(aq) → Zn²⁺(aq) + Cu(s)
n = 2
Formula:
ΔG° = -n E°_cell F = -2 × 1.1 × 96487
ΔG° = 212.27 kJ

Question 12.
Calculate standard free energy change for reaction.
Zn(s) + 2Ag⁺(aq) ⇌ Zn²⁺(aq) + 2Ag(s) ]
E°_cell = 1.56V; Given: 1F = 96500 C mol^-1.
Answer:
301.08 kJ.

Type – 3 Equilibrium constant

Question 13.
Calculate equilibrium constant of a reaction
Cu(s) + 2Ag⁺ (aq) → Cu²⁺ (aq) + 2Ag(s); E°_cell = 0.46 V.
Answer:
Date: E° = 0.46 V; n = 2, F = 96500 C mol^-1
Formula: E°_cell = \(\frac{2.303 R T}{n F}\) log k_c
At 298 K 0.46 = \(\frac{0.059}{2}\) log k_c
log k_c = \(\frac{0.46 \times 2}{0.059}\)
log k_c = 15.6
k_c = 3.92 × 10⁵

Question 14.
Calculate the equilibrium constant of the reaction at 298 K.
Mg(s) + 2Ag⁺(aq) → Mg²⁺(aq) + 2Ag(s); E°_cell = +3.16V.
Answer:
1.314 × 10¹⁰⁷

Question 15.
The cell in which the following reaction occurs:
2Fe^{3 +}(aq) + 2I (aq) → 2Fe²⁺(aq) + I₂(s)
has E°_cell = 0.236 V at 298 K.
Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.
Answer:

Type – 4 Conductivity, Molar Conductivity and Limiting Molar Conductivity

Question 16.
Resistance of a conductivity cell filled with 0.02 M KCl solution is 520 ohm. Calculate the conductivity and molar conductivity of that solution (cell constant 1.29 cm^-1).
Answer:
Data: Cell constant = 1.29 cm^-1;
Resistance = 520 Ohm
Molar concentration of KCl = 0.02 M

Question 17.
The resistance of solution of a salt occupying a volume between two platinum electrodes 1.8 cm apart and 5.4 cm² in area was found to be 32 ohm. Calculate the conductivity of the solution.
Answer:
Date: Resistance of a solution = R = 32 ohm
Distance between two electrodes = 1 = 1.8 cm
Area of the electrode = A = 5.4 cm²
Formula:

Question 18.
Resistance of a conductivity cell filled with 0.1 mol L^-1 KCl solution is 100Ω. If the resistance of the same cell when filled with 0.02 mol L^-1 KCl solution is 520 Ω. Calculate the conductivity and molar conductivity of 0.02 mol L^-1 KCl solution. The conductivitv of 0.1 mol L^-1 KCl solution is 1.29 S/m.
Answer:
Data: R of 0.1 M KCl = 100Ω (ohm);
K of 0.1 M KCl = 1.29 Sim^-1
R of 0.02 M KCl = 520 Ω (ohm)
Cell constant = Conductivity × Resistance = 1.29 × 100 = 129 m^-1
Concentration = 0.02 mol L^-1 = 20 mol m^-3

Question 19.

Answer:

Question 20.
Molar conductances of NaCI, HCl and CH₃COONa are 126.45, 426.16 and 91.0 ohm^-1 cm² mol at 298 K respectively. Calculate molar conductance at infinite dilution.
Answer:

Question 21.
Calculate the ∧°_m for MgCl₂, the limiting molar conductivities of Mg²⁺ and Cl^– ions are 106 S cm² mol^-1 and 76.3 S cm² mol^-1 respectively.
Answer:
258.6 S cm² mol^-1

Question 22.
A solution of CuSO₄ is electrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode?
Answer:
Data: I = 1.5 amp, t = 10 × 60600s
Cu²⁺ + 2e^– → Cu
I mole of Cu²⁺ requires 2 × 96500 C
Formula: I = \(\frac{Q}{t}\)
Q = It = 1.5 × 600 = 900C
2 × 96500 C deposits 63 g of Cu
900 C deposits ……. ?
= \(\frac{63 \times 900}{2 \times 96500}\) = 0.2938 g

Question 23.
If a current of 0.5 ampere flows through a metallic wire for 2 hours then how many electrons would flow through the wire.
Answer:
Data: I = 0.5 amp), t = 2 × 60 × 60 = 7200
Q = It
96500 Coulombs → 6.022 × 10²³ e^–s
= 0.5 × 2 × 60 × 60
3600 Coulombs → ?
= 3600 C.
= \(\frac{6.022 \times 10^{23} \times 3600}{96500}\) = 0.224 × 10²³ e^–s.