CBSE Class 12 Chemistry Electrochemistry Question Bank

Question. Which of the following is a secondary cell?
(a) Leclanche cell
(b) Lead storage battery
(c) Concentration cell
(d) All of these
Answer. B

Question. If limiting molar conductivity of Ca ²⁺ and Cl^– are 119.0 and 76.3 S cm² mol^-1, then the value of limiting molar conductivity of CaCl₂ will be
(a) 195.3 S cm² mol^-1
(b) 271.6 S cm² mol^-1
(c) 43.3 S cm² mol^-1
(d) 314.3 S cm² mol^-1.
Answer. B

Question. The reaction, 3ClO^– (aq) → ClO₃ (aq) + 2Cl^– (aq) is an example of
(a) Oxidation reaction
(b) Reduction reaction
(c) Disproportionation reaction
(d) Decomposition reaction
Answer. C

Question. Without losing its concentration ZnCl₂ solution cannot be kept in contact with
(a) Au
(b) Al
(c) Pb
(d) Ag
Answer. B 

SHORT ANSWER TYPE QUESTIONS

Question. The chemistry of corrosion of iron is essentially an electrochemical phenomenon. Explain the reactions occurring during the corrosion of iron in the atmosphere.
Answer. The redox reaction involves
At anode : Fe(S) → Fe²⁺ (aq) + 2e^–
At cathode : H₂O + CO₂ ⇌ H₂CO₃ (Carbonic acid)
H₂CO₃ ⇌2H+ + CO₂^2-
H⁺ + e^– → H
4H + O₂ → 2H₂O
Then net resultant Redox reaction is
2Fe(s) + O₂ (g) + 4H⁺ → 2Fe²⁺ + 2H₂O

Question. Write the reactions taking place at cathode and anode in lead storage battery when the battery is in use. What happens on charging the battery?
Answer. At Anode: Pb + SO₄^-2 → PbSO₄ + 2e
at Cathode: PbO₂ + SO₄^-2 + 4H⁺ + 2e → PbSO₄ + 2H₂O
On charging the battery, the reaction is reversed and PbSO₄ on anode and cathode is converted into Pb and PbO₂ respectively.

Question. State Kohlrausch law of independent migration of ions. Why does the conductivity of a solution decrease with dilution?
Answer. The limiting molar conductivity of an electrolyte 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. Using the E° values of A and B, predict which one is better for coating the surface of iron [E°(Fe²⁺/Fe) = – 0.44V] to prevent corrosion and why?
Given: E°(A²⁺|A) = -2.37 V and E°(B²⁺|B) = – 0.14 V 
Answer. ‘A’ will prevent iron from rusting. So, we can coat the iron surface with metal A because it has more negative value. 

 

1. Read the passage given below and answer the following questions:

The potential difference between the two electrodes of a galvanic cell is called the cell potential and is measured in volts. The cell potential is the difference between the electrode potentials (reduction potentials) of the cathode and anode. It is called the cell electromotive force (emf) of the cell when no current is drawn through the cell. It is now an accepted convention that we keep the anode on the left and the cathode on the right while representing the galvanic cell. A galvanic cell is generally represented by putting a vertical line between metal and electrolyte solution and putting a double vertical line between the two electrolytes connected by a salt bridge.
In a galvanic cell, the following cell reaction occurs:
Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s)
E°_cell = +1.56 V

Question. What is the direction of the flow of electrons?
(a) First from silver to zinc, then the direction reverses
(b) Silver to zinc
(c) First from zinc to silver, then the direction reverses
(d) Zinc to silver
Answer. D

Question. How will concentration of Zn²⁺ ions and Ag+ ions be affected when the cell functions?
(a) Concentration of both Zn and Ag+ ions increase
(b) Concentration of Zn²⁺ increases and Ag⁺ ions decreases
(c) Concentration of Zn²⁺ decreases and Ag⁺ ions increases
(d) Concentration of both Zn²⁺ and Ag⁺ ions decreases
Answer. B

Question. Name the cell which is generally used in inverters?
(a) Mercury cell
(b) Leclanche cell
(c) Lead storage battery
(d) Lithium ion battery
Answer. C

Question. Which cell used in hearing aids?
(a) Mercury cell
(b) Leclanche cell
(c) Dry cell
(d) Nickel Cadmium cell
Answer. A

Question. The standard electrode potential for Daniell cell is 1.1V. Calculate the standard Gibbs energy for the reaction:
Zn(s) + Cu²⁺ (aq) → Zn²⁺ (aq) + Cu(s).
(a) 215.36 kJ mol^-1
(b) -212. 27 kJ mol^-1
(b) 212.27 kJ mol^-1
(d) -218 kJ mol^-1
Answer. B

Question. 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. Mercury cell, Anode : Zn(Hg)+2OH^- →ZnO(s)+H₂O+2e^-
Cathode : HgO+H₂O+2e→Hg(l) + 2OH^-

Question. (a) The cell in which the following reaction occurs :2 Fe³⁺(aq) + 2I^- (aq) ⎯→ 2Fe²⁺(aq)+I₂(s) has E0cell = 0·236 V at 298 K. Calculate the standard Gibbs energy of the cell reaction. (Given : 1 F = 96,500 C mol−1 ) (2)
(b) How many electrons flow through a metallic wire if a
current of 0·5 A is passed for 2 hours ?
(Given : 1 F = 96,500 C mol−1) (3)
Answer. (a)ΔG⁰ = -nFE0 cell
n= 2 ,ΔG ⁰ = - 2 x 96500 C /mol x 0.236 V
= - 45548 J/mol
= -45.548 kJ/mol
(b) Q= I t = 0.5 x 2 x 60 x 60
= 3600 C
96500 C = 6.023 x 10²³ electrons
so 3600 C = 2.25 x 10²² electrons

Question. A current of 1.50 A was passed through an electrolytic cell containing AgNO3 solution with inert electrodes. The weight of silver deposited was 1.50 g. How long did the current flow ? (Molar mass of Ag = 108 g mol^–1, 1F = 96500 C mol^–1). 
Answer. Ag+ + 1e→Ag
96500C depositing 108 g of Ag
Quantity of electricity required for depositing
1.5g of Ag= 96500x1.5/108 =1340.27C
Q=Ixt so t=Q/I t=1340.27/1.5=893.5s.

Question.The conductivity of a 0.01 M solution of acetic acid at 298 K is 1.65 × 10^–4 S cm^–1 . Calculate molar conductivity (Λm ) of the solution. 
Answer. Λm =1000k/c, Λm=1000x1.65x10^-4/ 10^-2 =16.5Scm²mol^-1

Question. Consider the following reaction : Cu(s) + 2Ag⁺ (aq) → 2Ag(s) + Cu²⁺(aq)
(i) Depict the galvanic cell in which the given reaction takes place.
(ii) Give the direction of flow of current. (iii) Write the half-cell reactions taking place at cathode and anode.
Answer. (i) Cu(s)|Cu2+(aq)||Ag+(aq)|Ag(s)
(ii)Silver electrode to Copper electrode
(iii) Anode, Cu→ Cu²⁺ +2e
Cathode, Ag⁺ +1e→Ag

Question. E^o cell for the given redox reaction is 2.71 V
Answer. Mg(s) + Cu ²⁺ (0.01 M) –––→ Mg²⁺ (0.001 M) + Cu_(s)
(a)Calculate Ecell for the reaction.(b) Write the direction of flow of current when an external opposite potential applied is
(i) less than 2.71 V and (ii) greater than 2.71 V (3+2)
Answer. (a) E cell = E⁰ cell - 0.059 logK c
                                         n
= E^o cell - 0.059 log(10^-3)
2 (10^-2)
= 2.71+ 0.0295
Ecell = 2.7395 V
(b) i)Cu to Mg / Cathode to anode / Same direction
ii)Mg to Cu / Anode to cathode / Opposite direction

33. What is the cell potential for the cell at 250C
Cr / Cr³⁺(0.1 M) Fe²⁺ (0.01M) /Fe
E⁰Cr³⁺/Cr = -0.74V ;E0 Fe²⁺/Fe = - 0.44V.

34. Calculate ΔG⁰ for the reaction at 250
Zn (s)/ Zn²⁺ (0.0004M) Cd²⁺ (0.2m) 1 Cd (s)
E⁰Zn²⁺/Zn = -0.763V , E⁰Cd²⁺/Cd = -0.403V , F=96500 C Mol^-1 , R = 8.314 J/K.

35. Calculate Equilibrium constant K for the reaction at 298K
Zn (s) + Cu²⁺ (aq) →Zn²⁺/aq) +Cu
E⁰Zn²⁺Zn = -0.076V ; E⁰Cu²⁺Cu = +0.34V.

36. For what concentration of Ag⁺ (aq) will the emf of the given cell be zero at 250C ,if the concentration of Cu²⁺ (aq) is 0.1 M ?
Cu (s) / Cu²⁺ (0.1M)// Ag⁺ (aq) / Ag(s)
E⁰ Ag⁺ /Ag = +0.80V; E⁰ Cu²⁺ /Cu = 0.34 V

37. Calculate the standard free energy change for the cell- reaction.
Fe²⁺ (aq) + Ag+ (s) →Fe³⁺ (aq) + Ag(s)
How is it related to the equilibrium constant of the reaction?
E0Fe³⁺/Fe²⁺= + 0.77V , E⁰Ag⁺/Ag = +0.08V F= 96500 C/mol.

38. Predict the products of electrolyzing of the following
(a) A dil. Solution of H₂SO₄ with Pt. electrode
(b) An aqueous solution of AgNO₃ with silver electrode