Unit 3 Electrochemistry

Question 1

What is meant by ‘limiting molar conductivity’?

Answer 1

The molar conductivity of a solution at infinite dilution is called limiting molar conductivity and is represented by the symbol Λ_m.

Question 2

Express the relation between conductivity and molar conductivity of a solution held in a cell

Answer 2

Λ_m = K/C= Conductivity / Concentration

Question 3

What is the effect of catalyst on:

(i) Gibbs energy (ΔG) and
(ii) activation energy of a reaction?

Answer 3

(i) There will be no effect of catalyst on Gibbs .energy.
(ii) The catalyst provides an alternative pathway by decreasing the activation energy of a reaction.

Question 4

Two half cell reactions of an electrochemical cell are given below :
MnO₄^–(aq) + 8H⁺ (aq) + 5e^– → Mn²⁺ (aq) + 4H₂O (I), E° = + 1.51 V
Sn²⁺ (aq) → 4 Sn⁴⁺ (aq) + 2e^–, E° = + 0.15 V
Construct the redox equation from the two half cell reactions and predict if this reaction favours formation of reactants or product shown in the equation.

Answer 4

The reactions can be represented at anode and at cathode in the following ways :
At anode (oxidation) : Sn²⁺ → = Sn⁴⁺ (aq) + 2e^– ] × 5 E° = + 0.15 V
At cathode (reduction) :
MnO₄^–(aq) + 8H⁺ (aq) + 5e^– → Mn²⁺ (aq) + 4H₂O (I)] × 2 E° =+1.51 V
The Net R × M = 2MnO–4(aq) + 16H+ + 5Sn2+ → 2Mn2+ + 5Sn4+ + 8H2O
Now E°cell = E°cathode – E°anode
= 1.51 – 0.15 = + 1.36 V
∴ Positive value of E°cell favours formation of product.

Question 5

Express the relation among the cell constant, the resistance of the solution in the cell and the conductivity of the solution. How is the conductivity of a solution related to its molar conductivity?

Answer 5

1R×1a = Conductance (C) × Cell constant
Molar conductance : (Λm) = K×1000/C

Question 6

Given that the standard electrode potentials (E°) of metals are :
K+/K = -2.93 V, Ag+/Ag = 0.80 V, Cu2+/Cu = 0.34 V,
Mg2+/Mg = -2.37 V, Cr3+/Cr = -0.74 V, Fe2+/Fe = -0.44 V.
Arrange these metals in increasing order of their reducing power.

Answer 6

Ag+/Ag < Cu2+/Cu < Fe2+/Fe < Cr3+/Cr < Mg2+/ Mg < K+/K
More negative the value of standard electrode potentials of metals is, more will be the reducing power.

Question 7

Two half-reactions of an electrochemical cell are given below :
MnO–4 (aq) + 8H+ (aq) + 5e– → Mn2+ (aq) + 4H2O (I), E° = 1.51 V
Sn2+ (aq) → Sn4+ (aq) + 2e–, E° = + 0.15 V.
Construct the redox reaction equation from the two half-reactions and calculate the cell potential from the standard potentials and predict if the reaction is reactant or product favoured

Answer 7

The reactions can be represented at anode and at cathode in the following ways :
At anode (oxidation) :
Sn2+ → Sn4+ (aq) + 2e– ] × 5 E° = + 0.15 V
Af cathode (reduction) :
MnO–4(aq) + 8H+ (aq) + 5e– → Mn2+ (aq) + 4H2O (I)] × 2 E° = + 1.51 V
The Net R × M = 2MnO–4(aq) + 16H+ + 5Sn2+ → 2Mn2+ + 5Sn4+ + 8H2O
Now E°cell = E°cathode – E°anode
= 1.51 – 0.15 = + 1.36 V
∴ Positive value of E°cell favours formation of product.

Question 8

The chemistry of corrosion of iron is essentially an electrochemical phenomenon. Explain the reactions occurring during the corrosion of iron in the atmosphere.

Answer 8

The mechanism of corrosion is explained on the basis of electrochemical theory. By taking example of rusting of iron, we Refer tothe formation of small electrochemical cells on the surface of iron.
The redox reaction involves
At anode : Fe(S) → Fe2+ (aq) + 2e–
At cathode : H2O + CO2 ⇌ H2CO3 (Carbonic acid)
H2CO3 ⇌2H+ + CO22-
H2O ⇌ H+ + OH–
H+ + e– → H
4H + O2 → 2H2O
Then net resultant Redox reaction is
2Fe(s) + O2 (g) + 4H+ → 2Fe2+ + 2H2O

Question 9

Determine the values of equilibrium constant (Kc) and ΔG° for the following reaction :
Ni(s) + 2Ag+ (aq) → Ni2+ (aq) + 2Ag(s),
E° = 1.05 V
(1F = 96500 C mol-1)

Answer 9

According to the formula
ΔG° = -nFE° = – 2 × 96500 ×1.05
or ΔG° = -202650 J mol-1 = -202.65 KJ mol-1
Now ΔG° ⇒ -202650 J Mol-1
R = 8.314 J/Mol/K, T = 298 K

Question 10

Two half-reactions of an electrochemical cell are given below :
MnO–4 (aq) + 8H+ (aq) + 5e– → Mn2+ (aq) + 4H2O (I), E° = 1.51 V
Sn2+ (aq) → Sn4+ (aq) + 2e–, E° = + 0.15 V.
Construct the redox equation from the standard potential of the cell and predict if the reaction is reactant favoured or product favoured

Answer 10

The reactions can be represented at anode and at cathode in the following ways :
At anode (oxidation) :
Sn2+ → = Sn4+ (aq) + 2e– ] × 5 E° = + 0.15 V
At cathode (reduction) :
MnO–4(aq) + 8H+ (aq) + 5e– → Mn2+ (aq) + 4H2O (I)] × 2 E° = + 1.51 V
The Net R × M = 2MnO–4(aq) + 16H+ + 5Sn2+ → 2Mn2+ + 5Sn4+ + 8H2O
Now E°cell = E°cathode – E°anode
= 1.51 – 0.15 = + 1.36 V
∴ Positive value of E°cell favours formation of product.

Question 11

Express the relation among cell constant, resistance of the solution in the cell and conductivity of the solution. How is molar conductivity of a solution related to its conductivity?

Answer 11

GG* = K
where Q is conductance;
G * is cell constant;
K is conductivity
G* × 1/R = K ⇒ G* = RK
∴ Λm = K×1000/C S cm2 mol-1

Question 12

The molar conductivity of a 1.5 M solution of an electrolyte is found to be 138.9 S cm2 mol-1. Calculate the conductivity of this solution.

Answer 12

C= 1.5M, Λm = 138.9 S cm2 mol-1

Λm = K*1000/C

∴K = Λm×C/1000 =138.9×1.5/1000 = 0.20835 S cm-1

Question 13

A zinc rod is dipped in 0.1 M solution of ZnSO4. The salt is 95% dissociated at this dilution at 298 K. Calculate the electrode potential.
[E°Zn2+ /Zn = – 0.76 V]

Answer 13

The electode reaction is given as
Zn+2 + 2e → Zn
Using Nernest Equation

Question 14

The conductivity of 0.20 M solution of KCl at 298 K is 0.025 S cm-1. Calculate its molar conductivity.

Answer 14

Given: K=0.025 S cm-1

Molar conductivity Λm = 1000×κ/M

Hence, Λm = 0.025×1000 /0.20 ∴ Λm = 125 S cm2 mol-1

Question 15

State Kohlrausch law of independent migration of ions. Why does the conductivity of a solution decrease with dilution?

Answer 15

Kohlrausch law of independent migration of ions: The limiting molar conductivity of an electrolyte (i.e. molar conductivity at infinite dilution) is the sum of the limiting ionic conductivities of the cation and the anion each multiplied with the number of ions present in one formula unit of the electrolyte
Λ°m for AxBy = xλ°+ + yλ°–
For acetic acid Λ° (CH3COOH) = λ°CH3COO– + λ°H+
Λ°(CH3COOH) = Λ° (CH3COOK) + Λ° (HCl) – Λ° (KCl)

Question 16

Define the following terms :

(i) Fuel cell
(ii) Limiting molar conductivity (Λ°m)

Answer 16

(i) Fuel cells : These cells are the devices which convert the energy produced during combustion of fuels like H2, CH4, etc. directly into electrical energy.
(ii) The molar conductivity of a solution at infinite dilution is called limiting molar conductivity and is represented by the symbol Λ°m.

Question 17

Define the following terms :

(i) Molar conductivity (Λm)
(ii) Secondary batteries

Answer 17

Molar conductivity: Molar conductivity of a solution at a given concentration is the conductance of the volume ‘V’ of a solution containing one mole of electrolyte kept between two electrodes with area of cross section ‘A’ and distance of unit length. It is represented by Λm (lamda).
Λm = KAI/ l = 1 and A = V
∴ Λm = KV Unit = S cm2 mol-1
Secondary batteries : Those batteries which can be recharged by passing an electric current through them and can be used again and again are called secondary batteries.

Question 18

Define the following terms :

(i) Rate constant (k)
(ii) Activation energy (Ea)

Answer 18

(i) Rate constant (k): It is a proportionality constant and is equal to the rate of reaction when the molar concentration of each of the reactants is unity.
(ii) Activation energy (Ea): The minimum extra amount of energy absorbed by the reactant molecules to form the activated complex is called activation energy.

Question 19

From the given cells: Lead storage cell, Mercury cell, Fuel cell and Dry cell
Answer the following:
(i) Which cell is used in hearing aids?
(ii) Which cell was used in Apollo Space Programme?
(iii) Which cell is used in automobiles and inverters?
(iv) Which cell does not have long life?

Answer 19

(i) Mercury cell is used in hearing aids.
(ii) Fuel cell was used in the Apollo Space Programme.
(iii) Lead storage cell is used in automobiles and inverters.
(iv) Dry cell does not have a long life.

Question 20

Calculate the degree of dissociation (a) of acetic acid if its molar conductivity (Λm) is 39.05 S cm2 mol-1.
Given: λ°(H+) = 349.6 S cm2 mol-1 and λ°(CH3COO–) = 40.9 S cm2 mol-1

Answer 20

Λ°m(HAC) = λ°H+ + λ°AC–
= λ°CH3COOH = λ°H+ + λ°CH3COO–
= 349.6 S cm2 mol-1 + 40.9 S cm2 mol-1
= 390.5 S cm2 mol-1

Question 21

Write the name of the cell which is generally used in hearing aids. Write the reactions taking place at the anode and the cathode of this cell.

Answer 21

Mercury cells are used in hearing aids.
Reaction at anode:
Zn (Hg) + 2OH^- → ZnO (s) + H₂O + 2e^-
Reaction at cathode:
HgO + H₂O + 2e^- → Hg (l) + 2OH^-

Question 22

Write the name of the cell which is generally used in transistors. Write the reactions taking place at the anode and the cathode of this cell.

Answer 22

Leclanche cells (Dry cell) is used in transistors.
Reaction at Anode: Zn(s) → Zn<sup>2+</sup> + 2e<sup>-</sup>

At Cathode:
MnO₂ + NH₄⁺+ e^- → MnO(OH) + NH₃

Question 23

A copper-silver cell is set up. The copper ion concentration in it is 0.10 M. The concentration of silver ion is not known. The cell potential is measured 0,422 V. Determine the concentration of silver ion in the cell.
Given : E°Ag+/Ag = + 0.80 V, E° Cu2+/Cu = + 0.34 V.

Answer 23

The reaction takes place at anode and cathode in the following ways :
At anode (oxidation) :
Cu(s) → Cu²⁺(aq) + 2e^-
At cathode (reduction) :
Cu(s) + 2Ag²⁺(aq) → Cu²⁺(aq) + 2Ag(s)
The complete cell reaction is

Question 24

The electrical resistance of a column of 0.05 M NaOH solution of diameter 1 cm and length 50 cm is 5.55 × 103 ohm. Calculate its resistivity, conductivity and molar conductivity.

Answer 24

Question 25

A voltaic cell is set up at 25°C with the following half cells :
Al/Al3+ (0.001 M) and Ni/Ni2+ (0.50 M)
Write an equation for the reaction that occurs when the cell generates an electric current and determine the cell potential.

E⁰_Ni²⁺/Ni=−0.25V and E⁰_Al³⁺/Al=−1.66V

(Log 8 × 10^-6 = -0.54)

Answer 25

Half cell reactions and overall cell reaction are

Question 26

What is corrosion? Explain the electrochemical theory of rusting of iron and write the reactions involved in the rusting of iron.

Answer 26

Corrosion: Corrosion is defined as the deterioration of a substance because of its reaction with its environment. Corrosion is an electrochemical phenomenon. At a particular spot of an object made of iron, oxidation takes place and that spot behaves as anode and the reaction is
At Anode : 2Fe → 2Fe⁺² + 4e^–

Electrons released at anodic spot move through the metal and go to another spot on the metal and reduce oxygen in presence of H⁺. This spot behaves as cathode
At Cathode : O₂ + 4H⁺ + 4e-
Overall reaction : 2Fe + O₂ + 4H⁺ → 2Fe⁺² + 2H₂O