Important Questions for CBSE Class 12 Chemistry Electrochemistry
Previous Years’ Questions
2015 – Very Short Answer Type Questions [1 Mark]
1. How much charge in Faraday is required for the reduction of 1 mol of Ag+ to Ag?
Answer: Ag+ + e– ——> Ag(s)
1 Faraday of charge is required (charge on 1 mole of electrons).
2. How much charge is required for the reduction of 1 mole of Zn+2 to Zn?
Answer: Zn+2 + 2e– —–> Zn(s)
2 Faradays or 2 x 96500 C of charge is required.
3. How much charge in Faradays is required for the reduction of 1 mol of Al+3 to AI ?
Answer: 3 Faradays
Al3+ + 3e– ——-> Al(s)
3 moles of electron have charge = 3 Faradays.
4. Calculate the time to deposit 1.5 g of silver at cathode when a current of 1.5 A was passed through the solution of AgNO3. (Molar mass of Ag = 108 g mol-1, 1 F = 96500 C mol -1).
Answer:
Short Answer Type Questions [I] [2 Marks]
5. Calculate the time to deposit 1.17 g of Ni at cathode when a current of 5 A was passed through the solution of Ni(NO3)2.
(Molar mass of Ni = 58.5 g mol-1, 1 F = 96500 C mol-1).
Answer:
6. Accounts for the following
(i) Rusting of iron is quicker in saline water than in ordinary water.
(ii) Blocks of magnesium are straped to the steel hubs of ocean going ships.
Answer: (i) Electrolytes present in sea water favour the formation of more electrochemical cells on the surface of iron leading to increase in the rate of rusting.
(ii) Mg is more reactive than iron, therefore, prevents oxidation of steel (rusting of steel)
7. (a) Following reactions occur at cathode during the electrolysis of aqueous sodium chloride solution:
Na+(aq) + e-1 ——-> Na(s) E° = – 2.71 V
H+(aq) + e-1 ——–> 1/2H2(s) E° = 0.00 V
On the basis of their standard reduction electrode potential (E°) values, which reaction is feasible at the cathode and why?
(b) Why does the cell potential of mercury cell remain constant throughout its life?
Answer: (a) H+(aq) + e-1 —-> ~H2(g) will take place at cathode because it has higher reduction potential.
(b) It is because ions are not involved in net cell reaction.
8.(a) Following reactions occur at cathode during the electrolysis of aqueous silver chloride solution:
Ag+(aq) + e-1 ————> Ag(s) E° = +0.80 V
H+(aq) + e-1 ———–> 1/2H2(g) E° = 0.00 V
On the basis of their standard reduction electrode potential (E°) values, which reaction is feasible at the cathode and why?
(b) Define limiting molar conductivity. Why conductivity of an electrolyte solution decreases with the decrease in concentration?
Answer:
(b) Limiting molar conductivity is the maximum conductivity when solution is infinitely dilute, such that on further dilution there is no increase in Am. Conductivity decreases with decrease in concentration because number of ions per unit volume decrease.
9. Calculate the time to deposit 1.27 g of copper at cathode when a current of 2 A was passed through the solution of CuS04.
(Molar mass of Cu = 63.5 g mol-1, 1 F = 96500 C mol-1)
Answer:
10. (a) Following reactions occur at cathode during the electrolysis of aqueous copper (II) chloride solution:
Cu2+(aq) + 2 e-2 ———–> Cu(s) E° = + 0.34 V
H+(aq) + e-1 ————-> 1/2H2(g) E° = 0.00 V
On the basis of their standard reduction electrode potential (E°) values, which reaction is feasible at the cathode and why ?
(b) State Kohlrausch law of independent migration of ions. Write its one application.
Answer: (a) At cathode Cu2+ + 2e-1 ———-> Cu(s)
The above reaction is feasible at cathode because E°Cu2+/Cu = + 0.34, because reduction potential of Cu2+ is higher than H+ as E0H+/H2= 0.
(b) Kohlrausch law: It states that the limiting molar conductivity of an electrolyte is equal to the sum of contribution of cations as well as anions. Application: It helps in calculating A0 (limiting molar conductivities) of weak electrolytes.
Short Answer Type Questions [II] [3 Marks]
11. Calculate E°eI1 and ArG° for the following reaction at 25 °C:
A2+(aq) + B+(aq) ——–> A5+(aq) + B(s)
Given Kc=1010, 1F=96500 C mol-1
Answer:
12. Calculate E°cell for the following reaction at 25 °C:
A + B2+ (0.001 M) ——–> A2+ (0.0001 M) + B
Given : Ecell = 2.6805 V, 1 F = 96500 C mol-1
Answer:
13. Calculate emf of the following cell at 25 °C:
Fe | Fe2+(0.001 M) 11 H+(0.01 M) | H2(g) (1 bar) | Pt(s)
E°(Fe2+|Fe) = -0.44V, E°(H+|H2) = 0.00 V
Answer:
14. Conductivity of 2.5 X 10-4M methanoic acid is 5.25 X 10-5 S cm-1 Calculate its molar conductivity and degree of dissociation.
Given: A°(H+) = 349.5 S cm2 mol-1 and A°(HCOO– ) = 50.5 S cm2
Answer:
15. Calculate the emf of the following cell at 25 °C:
Zn | Zn2+ (0.001 M) | | H+(0.01 M) | H2(g) (1 bar) | Pt(s)
Answer:
Long Answer Type Questions [5 Marks]
16. (a) The conductivity of 0.20 mol L-1 solution of KC1 is 2.48 X 10-2 S cm-1. Calculate its molar conductivity and degree of dissociation (a). Given X,°(K+) = 73.5 S cm2 mol-1 and Z°(cr) = 76.5 S cm2 mol-1.
(b) What type of battery is mercury cell? Why is it more advantageous than dry cell?
Answer:
17. (a) The conductivity of 0.1 mol L-1 solution of NaCl is 1.06 X 10-2 S cm-1. Calculate its molar conductivity and degree of dissociation (a).
Given Z°(Na+) = 50.1 S cm2mol-1 and 7°(C1-) = 76.5 S cm2 mol-1.
(b) What is the difference between primary battery and secondary battery? Give one example of each type.
Answer:
2014
Short Answer Type Questions [I] [2 Marks]
18. Define the following terms:
(t) Molar conductivity (Am)
(it) Secondary batteries
Answer: (i) Molar conductivity: It is defined as the conductance of solution containing 1 mole of electrolyte placed in a cell having electrodes unit distance apart having sufficient area of cross-section to hold electrolyte.
(ii) Secondary battery: Those batteries which can be recharged are called secondary batteries, e.g. lead storage battery, Nickel-cadmium battery.
19. A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5.0 ampere for 20 minutes. What mass of nickel will be deposited at the cathode? [Given: At. Mass of Ni = 58.7 g mol-1, IF = 96500 C mol-1]
Answer:
20. The resistance of 0.01 M NaCl solution at 25 °C is 200 Ω. The cell constant of conductivity cell is unity. Calculate the molar conductance.
Answer:
21. State Kohlrausch law of independent migration of ions. Why does the conductivity of a solution decrease with dilution?
Answer: It states, ‘the molar conductance at infinite dilution (A°, limiting molar conductivity) of an electrolyte is equal to the sum of contribution due to cation as well as anion’.
Λ0 NaCl = λ0Na+ + λ0Cl–
The conductivity of a solution decreases with dilution because number of ions per unit volume decreases.
22. Set up Nemst equation for the standard dry cell. Using this equation show that the voltage of dry cell has to decrease with use.
Answer:
23. Define the following terms:
(i) Limiting molar conductivity (A°m) (ii) Fuel cell
Answer: (i) Limiting molar conductivity: It is defined as the maximum molar conductance of an electrolyte when solution is infinitely dilute, i.e. concentration approaches zero when electrolyte solution is kept in cell having electrodes unit distance apart having large area of cross-section . to hold enough electrolyte.
(ii) Fuel cell: It is a cell in which chemical energy of fuel is converted into electrical energy, e.g. H2—02 fuel cell.
Short Answer Type Questions [II] [3 Marks]
24. (a) Calculate ArG° for the reaction:
Mg(s) + Cu22+(aq) > Mg2+(aq) + Cu(s)
Given: E°cell = + 2.71 V, 1 F = 96500 C mol-1
(b) Name the type of cell which was used in Apollo space programme for providing electrical power.
Answer:
(b) H2—02 Fuel Cell was used in Apollo space programme for providing electrical power.
25. (i) Write two advantages of H2 – 02 fuel cell over ordinary cell.
(ii) Equilibrium constant (Kc) for the given cell reaction is 10. Calculate E°ell.
Answer: (i) (a) Its efficiency is high.
(b) It does not create pollution.
(c) H20 formed can be used by astronauts for drinking.
Long Answer Type Questions [5 Marks]
26. (a) Define tue following terms:
(i) Limiting molar conductivity (ii) Fuel cell
(b) Resistance of a conductivity cell filled with 0.1 mol L-1 KC1 solution is 100 Q. If the resistance of the same cell when filled with 0.02 mol L-1 KC1 solution is 520 Q, calculate the conductivity and molar conductivity of 0.02 mol L KC1 solution. The conductivity of 0.1 mol L 1.29 X 10-2 CT1 cm-1
Answer: (a) Refer Ans. to Q.23.
27. (a) State Faraday’s first law of electrolysis. How much charge in terms of Faraday is required for the reduction of 1 mol of Cu2+ to Cu?
(b) Calculate emf of the following cell at 298 K:
Mg(s) | Mg2+(0.1 M) 11 Cu2+ (0.01) | Cu(s)
[Given E°cell, = + 2.71 V, 1 F = 96500 C mol-1]
Answer:
28. (a) Predict the products of electrolysis in each of the following:
(i) An aqueous solution of AgNO3 with platinum electrode.
(ii) An aqueous solution of H2S04 using platinum electrode.
(b) Estimate the minimum potential differences needed to reduce Al2O3 at
500°C. The gibbs free energy change for the decomposition reaction
Answer:
2013
Very Short Answer Type Question [1 Mark]
29. Represent the galvanic cell in which the reactions is
Zn(s) + Cu2+(aq) > Zn2+(aq) + Cu(s)
Answer: Zn | ZnS04 (1M) | | CuS04 (1M) | Cu(s).
Short Answer Type Questions [l] [2 Marks]
30. The conductivity of 0.20 M solution of KC1 at 298 K is 0.025 S cm-1. Calculate its molar conductivity.
Answer:
31. The standard electrode potential (E°) for Daniell cell is + 1.1 V. Calculate the AG° for the reaction
Zn(s) + Cu2+(a<7) ——-> Zn2+(aq) + Cu(s) (1 F = 96500 C mol-1).
Answer:
Short Answer Type Questions [II] [3 Marks]
32. Calculate the emf of the following cell at 298 K:
Fe(s) | Fe2+(0.001 M) 11 H+ (1 M) | H2(g) (1 bar), Pt ($) (Given E°eU = – 0.44 V)
Answer:
33. Calculate the emf of the following cell at 25 °C:
Ag(s) | Ag+ (10-3 M) || Cu2+ (10-1 M) | Cu(s)
Given: E°cell = + 0.46 V and log 10n = n.
Answer:
Long Answer Type Questions [5 Marks]
34. (a) Predict the products of electrolysis in each of the following:
(i) An aqueous solution of AgNO3 with platinum electrode.
(it) An aqueous solution of H2S04 using platinum electrode.
(b) Estimate the minimum potential differences needed to reduce Al203 at 500°C. The gibbs free energy change for the decomposition reaction
2/3 Al2O3——>4/3 Al+O2
Answer: Refer Ans. to Q.28.
35. (a) State Kohlrausch’s law of independent migration of ions. Mention one application of Kohlrausch’s law.
(b) The resistance of a conductivity cell containing 10-3 M KC1 solution at 25 °C is
1500 Q. What is the cell constant if conductivity of 10-3 M KC1 solution at 25 °C is 1.5 X 10-4 S cm-1?
Answer: Refer Ans. to Q.12 (b).
36. Calculate emf and AG° for the following cell at 298 K:
Mg(s) |Mg2+(10-3M) || CU2+(10-3M)| CU(S)
Given:E0mg2+/mg = – 2.36 V andE0cu2+/cu = +0.34 V, [1 F = 96500 C mol-1]
Answer:
2012
Short Answer Type Questions [I] [2 Marks]
37. What type of a battery is lead storage battery? Write the anode and cathode reactions and the overall cell reaction occurring in the operation of a lead storage battery.
Answer: Lead storage battery. It is a secondary cell. It consists of a lead anode and a grid of lead packed with lead dioxide as cathode. A solution of sulphuric acid (38% by mass) is used as an electrolyte.
The cell reactions when the battery is in use are given below:
Anode: Pb(s) + SO-2 (aq) ——–> PbS04(s) + e-2
Cathode: Pb02 (s) + S04-2(aq) + 4H+(aq) + 2 e– ——-> PbS04 (s) + 2H20 (/)
The overall cell reaction consisting of cathode and anode reactions is:
Pb(s) + Pb02(s) + 2H2SO4(aq) ——–> 2PbS04(s) + 2H20(/)
On recharging the battery, the reaction is reversed.
38. Zinc rod is dipped in 0.1 M solution of ZnS04. The salt is 95% dissociated at this dilution at 298 K. Calculate the electrode potential. Given E0zn2+/zn = – 0.76 V.
Answer:
39. Express the relation among the cell constant, the resistance of the solution in the cell and the conductivity of the solution. How is molar conductivity of a solution related to its conductivity?
Answer:
40. The molar conductivity of a 1.5 M solution of an electrolyte is found to be 138.9 S cm2mol-1. Calculate the conductivity of the solution.
Answer:
Short Answer Type Questions [II] [3 Marks]
41. A voltaic cell is set up at 25 °C with the following halfcells:
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.
[E0Ni2+/Ni= -0.25 V andE0al3+/al = – 1.66 V]
Answer:
42. The electrical resistance of a column of 0.05 mol L-1 NaOH solution of diameter 1 cm and length 50 cm is 5.55 X 103 ohm. Calculate its resistivity, conductivity and molar conductivity.
Answer:
43. A strip of nickel metal is placed in a 1 molar solution of Ni(NO3)2 and a strip of silver metal is placed in a 1-molar solution of AgNO3. An electrochemical cell is created when the two solutions are connected by a salt bridge and the two strips are connected by wires to a voltmeter.
(i) Write the balanced equation for the overall reaction occurring in the cell and calculate the cell potential.
(ii) Calculate the cell potential, E, at 25 °C for the cell if the initial concentration of Ni(NO3)2 is 0.100 molar and the initial concentration of AgNO3 is 1.00 molar. [E0Ni2+/Ni= -0.25 V; E0ag3+/ag = 0.80 V; log 10-1 = -1 ]
Answer:
44. When a certain electrolytic cell was filled with 0.1 M KC1, it has resistance of 85 ohms at 25 °C. When the same cell was filled with an aqueous solution of 0.052 M unknown electrolyte, the resistance was 96 ohms. Calculate the molar conductance of the electrolyte at this concentration.
[Specific conductance of 0.1 M KC1 = 1.29 X 10-2ohm-1 cm-1]
Answer:
Long Answer Type Questions [5 Marks]
45. (a) What type of battery is the lead storage battery? Write the anode and the cathode reactions and the overall reaction occuring in a lead storage battery when current is drawn from it.
(b) In the button cell, widely used in watches, the following reaction takes place:
Zn(s) + Ag20(s) + H20 (l) ——-> Zn2+(aq) + 2Ag(s) + 20H– (aq)
Determine E° and AG° for the reaction.
(Given: E0ag+/ag= + 0.80 V, E0zn+2/zn = – 0.76 V)
Answer:
46. (a) Define molar conductivity of a substance and describe how for weak and strong electrolytes, molar conductivity changes with concentration of solute. How is such change explained?
(b) A voltaic cell is set up at 25 °C with the following half-cells:
Ag+ (0.001 M) | Ag and Cu+2 (0.10 M) | Cu What would be the voltage of this cell?
(E0cell= O.4V)
Answer: (a) Molar Conductivity. It is defined as the conducting power of all the ions produced by one molar solution of an electrolyte. It is denoted by A .(or) It is defined as the conductivity of 1 M electrolytic solution placed between two electrodes 1 cm apart and have enough area of cross section to hold the entire volume.
Molar conductivity increases with decrease in concentration.
This is because both number of ions as well as mobility of ions increase with dilution.
In the case of strong electrolyte, the number of ions does not increase appreciably, only the mobility of ions increases, therefore, Am increases a little as shown in graph as straight line.
In case of weak electrolyte both number of ions and mobility of ions increases, therefore, Affl increases sharply as shown by curve in the figure.
47. (a) How many moles of mercury will be produced by electrolysing 1.0 M Hg(NO3)2 solution with a current of 2.00 A for 3 hours?
[Hg(N03)2 = 200.6 g mol-1]
(b) A voltaic cell is set up at 25 °C with the following half-cells:
Al2+(0.001 M) and Ni2+(0.50 M)
Write an equation for the reaction that occurs when the cell generates an electric current and determine the cell potential.
(Given;ENi2+/Ni= – 0.25 V, Eal3+/al= – 1.66 V)
Answer:
2011
Very Short Answer Type Questions [1 Mark]
48. Express the relation between conductivity and molar conductivity of a solution held in a cell.
Answer: Am = 1000k/M , where m is molar conductivity, K is conductivity and M is molarity or concentration in mol L-1.
49. Express the relation among the conductivity of solution in the cell, the cell
constant and the resistance cf solution in the cell.
Answer: K =1 /R*l/A , where K is conductivity, R is resistance and l/A is cell constant.
Short Answer Type Questions [I] [2 Marks]
50. Express the relation among the cell constant, the resistance of the solution
in the cell and the conductivity of the solution. How is molar conductivity of a solution related to its conductivity?
Answer: Refer Ans. to Q.39.
51. Determine the values of equilibrium constant (Kc) and AG° for the following reaction:
Ni(s) + 2Ag+(aq) —–> Ni+2(aq) + 2Ag(s),
E° = 1.05 V (IF = 96500 C mol-1)
Answer:
52. The chemistry of corrosion of iron is essentially an electrochemical phenomenon. Explain the reactions occurring during the corrosion of iron in the atmosphere.
Answer:
53. Two half-reactions of an electrochemical cell are given below:
Mn04–(aq) + 8H+(aq) + 5e– > Mn2+(aq) + 4H20(l), 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 the reaction is reactant favoured or product favoured.
Answer:
54. the reactions occurring at (i) anode, (ii) cathode, during working of a mercury cell. Why does the voltage of a mercury cell remain constant during its operation?
Answer: Mercury cell. It consists of zinc mercury amalgam as anode, a paste of HgO and carbon as cathode. The electrolyte is a paste of KOH and ZnO. The reaction of the cell is *
At anode: Zn (amalgam) + 20H– ———> ZnO(s) + H20 + 2e–
At cathode: HgO(s) + H20 + 2 e-2——–> Hg(Z) + 20H–
The net reaction: Zn (amalgam) + HgO(s) ———-> ZnO(s) + Hg(l)
It gives constancy in voltage over long period because no ions are involved in net cell reaction. It is used in watches and hearing aids.
55. Define and express the relation between conductivity and molar conductivity
for the solution of an electrolyte.
Answer: Conductivity is defined as ease with which current flows through an electrolyte. It is the reciprocal of resistance.
Molar conductivity is the conductance of all the ions produced by 1 mole of electrolyte when electrodes are unit distance apart and have sufficient area of cross section to hold electrolyte.
AM =1000K /M
where K is conductivity, M is molarity and Am is molar conductivity.
56. Give an example of a fuel cell and write the cathode and anode reactions for it.
Answer: Fuel cells are those cells in which chemical energy of a fuel is converted into electrical energy.
57. Write the overall reaction that occurs during use (discharging) of nickel-cadmium cell. Is it a primary or a secondary cell? Mention its one merit over the lead storage cell.
Answer: Nickel-cadmium cell: It is a type of secondary cell which has longer life than lead storage cell but more expensive to manufacture. The overall reaction during discharge is
Cd(s) + 2 Ni(OH)3(s) > CdO(s) + 2Ni(OH)2(s) + H20(l).
Its merit over the lead storage cell is its longer life as compared to lead storage battery.
58. In the electrolysis of aqueous sodium bromide, there are two possible anodic reactions: *
2H2O(l) ——-> 02(g) + 4H+(aq) + 4e–, E° = 1.23V
2Br–(aq) ——–> Br2(g) + 2e-2, E° = 1.08 V
Which reaction occurs at anode and why?
Answer: 2Br–(aq) ———> Br2 + 2e~ occurs at anode because it has lower reduction potential.
Short Answer Type Questions [II] [3 Marks]
59. A voltaic cell is set up at 25 °C with the following halfcells:
Al/A13+(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.
[E°Ni2+/Ni= -0.25 V and E°Al3+/Al= — 1.66 V]
Answer: Refer Ans. to Q.41.
60. For the cell Zn(s) | Zn2+(2M) | | Cu2+(0.5 M) | Cu(s)
(a) Write equation for each half-reaction.
(b) Calculate the cell potential at 25 °C.
[Given: EZn2+/Zn= – 0.76 V; ECu2+/Cu= + 0 .34V]
Answer:
61. Calculate the degree of dissociation of acetic acid at 298 K, given that: Am (CH3COOH) = 11.7 S cm2 mol-1, Λ°m (CH3COO ) = 40.9 S cm2 mol-1,Λ°(H+) = 349.1 S cm2 mol-1
Answer:
62. At 298 K, the electrolytic conductivity of a 0.2 M KC1 solution is 2.50 X 10-2S cm-1. Calculate its molar conductivity.
Answer:
63. Calculate the equilibrium constant, K for the reaction at 298 K,
Zn (s) + Cu2+(aq) ——–> Zn2+(aq) + Cu(s)
Answer:
64. An aqueous solution of copper sulphate, CuS04 was electrolysed between platinum electrodes using a current of 0.1287 ampere for 50 minutes. [Atomic mass of Cu = 63.5 g mol-1]
(a) Write the cathodic reaction.
(b) Calculate:
(i) Electric charge passed during electrolysis (ii) Mass of copper deposited at the cathode [Given: 1F = 96,500 C mol-1]
Answer:
Long Answer Type Questions [5 Marks]
65. (a) What type of a battery is lead storage battery? Write the anode and cathode reactions and the overall cell reaction occurring in the operation of a lead storage battery.
(b) Calculate the potential for half-cell containing 0.10 M K2Cr207(aq), 0.20 M Cr3+(aq) and 1.0 X 104M H+(aq)
The half-cell reaction is K2Cr207–(aq) + 14 H+(aq) + 6e-2 ——–> 2Cr3+(aq) + 7H20(l) and the standard electrode potential is given as E°= 1.33 V.
Answer: (a) Refer Ans. to Q.37.
66. (a) Corrosion is essentially an electrochemical phenomenon. Explain the reactions occurring during corrosion of iron kept in an open atmosphere.
(b) Calculate the equilibrium constant for the equilibrium reaction:
Fe(s) + Cd+2 (aq) ———> Fe2+ (aq) + Cd(s)
(Given:Ecd2+/cd= — 0.40 V,EFe2+/Fe= — 0.44 V).
Answer: (a) In corrosion, a metal is oxidised by loss of electrons to oxygen and formation of oxides in presence of water, air and H+. It is basically an electrochemical phenomenon.
67. (a) How many moles of mercury will be produced by electrolysing 1.0 M Hg(N03)2 solution with a current of 2.00 A for 3 hours?
[Hg(N03)2 = 200.6 g mol-1]
(b) A voltaic cell is set up at 25 °C with the following half-cells:
Al3+(0.001 M) and Ni2+(0.50 M)
Write an equation for the reaction that occurs when the cell generates an electric current and determine the cell potential.
(Given;ENi2+/Ni = – 0.25 V,EAl3+/Al= – 1.66 V)
Answer: Refer Ans. to Q.47.
2010
Very Short Answer Type Question [1 Marks]
68. What is meant by ‘limiting molar conductivity’?
Answer: Limiting molar conductivity is equal to maximum conductance when solution
is very-very dilute such that there is no further increase in molar conductance on further dilution.
Short Answer Type Questions [I] [2 Marks]
69. Mention the reactions occurring at (i) anode, (ii) cathode, during working of a mercury cell. Why does the voltage of a mercury cell remain constant during its operation?
Answer: Refer Ans. to Q.56.
70. Two half-reactions of an electrochemical cell are given below:
Mn04–(aq) + 8H+(aq) + 5e– ——-> Mn2+(aq) + 4H2O(l), E° = + 1.51 V
Sn-2(aq) ——–> Sn4+(aq) + 2e–, E° = + 0.15 V
Construct the redox equation from the standard potential of the cell and predict the reaction is reactant favoured or product favoured.
Answer: Refer Ans. to Q.55.
71. Express the relation among the cell constant, the resistance of the solution
in the cell and the conductivity of the solution. How is molar conductivity of a solution related to its conductivity?
Answer: Refer Ans. to Q.41.
72. Write the anode and cathode reactions occurring in a commonly used mercury cell. How is the overall reaction represented?
Answer: Mercury cell: It consists of zinc mercury amalgam as anode, a paste of HgO and carbon as cathode. The electrolyte is a paste of KOH and ZnO. The reaction of the cell is «
At anode: Zn (amalgam) + 20H– ——-> ZnO(s) + H20 + 2e–
At cathode: HgO(s) + H20 + 2e– ———> Hg(l) + 20H–
The net reaction: Zn (amalgam) + HgO(s) ——–> ZnO(s) + Hg(l)
It gives constancy in voltage over long period because no ions are involved in net cell reaction. It is used in watches and hearing aids.
73. 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 an increasing order of their reducing power.
Answer: Ag < Cu < Fe < Cr < Mg < K is the increasing order of reducing power.
Short Answer Type Questions [II] [3 Marks]
74. A copper-silver cell is set up. The copper ion concentration is 0.10 M. The concentration of silver ion is not known. The cell potential when measured was 0.422 V. Determine the concentration of silver ions in the cell.
(GivenE0Ag+/Ag= +0.80 E0Cu2+/Cu = +0.34 V)
Answer:
75. What is a nickel-cadmium cell? State its one merit and one demerit over lead storage cell. Write the overall reaction that occurs during discharging of this cell.
Answer: Nickel-cadmium cell: It is a type of secondary cell which has longer life than lead storage cell but more expensive to manufacture. The overall reaction during discharge is
Cd(s) + 2 Ni(OH)3(s) ——–> CdO(s) + 2 Ni(OH)2(s) + H20 (l).
Merit: It is easy to handle as it is less bulky than lead storage cell and has longer life than lead storage cell.
Demerit: It is more expensive than lead storage cell.
76. One half-cell in a voltaic cell is constructed from a silver wire dipped in silver nitrate solution of unknown concentration. The other half-cell consists of a zinc electrode in a 0.10 M solution of Zn (N03)2. A voltage of 1.48 V is measured for this cell. Use this information to calculate the concentration of silver nitrate solution. (Given E°Zn2+/Zn= 0.76 V , E°Ag+/Ag= 0-8 0 v)
Answer:
Long Answer Type Questions [5 Marks]
77. (a) Calculate the emf for the given cell at 25 °C:
Cr | Cr3+ (0.1 M)|| Fe2+ (0.01 M) | Fe
[Given:E0Cr3+/Cr= -0.74 V, E0Fe2+/Fe = -0. 44 V]
(b) Calculate the strength of the current required to deposit 1.2<g of magnesium from molten MgCl2 in 1 hour.
[1 F = 96,500 C mol-1; Atomic mass: Mg = 24.0]
Answer:
78. (a) Write the anode and cathode reactions and the overall reaction occurring in a lead storage battery.
(b) A copper-silver cell is set up. The copper ion concentration is 0.10 M. The concentration of silver ion is not known. The cell potential when measured was 0.422 V. Determine the concentration of silver ions in the cell. (Given E°Ag+/Ag= 0.80 V, E0Cu2+/Cu = 0.34 V).
Answer:
79. (a) State Kohlrausch law of independent migration of ions. Write an expression for the molar conductivity of acetic acid at infinite dilution according to Kohlrausch law.
(b) Calculate Λ ° for acetic acid.
Given that Λ° (HC1) = 426 S cm2 mol-1
Λ° (NaCl) = 126 S cm2 mol-1
Λ° (CH.COONa) = 91 S cm2 mol-1
Answer: (a) Kohlrausch’s law of independent migration of ions: According to this law, molar conductivity of an electrolyte, at infinite dilution, can be expressed as the sum of the contributions from its individual ions. If the molar conductivity of the cations is denoted by Λ,+ and that of the anions by Λ, then the law of independent migration of ions is
80. (a) How many moles of mercury will be produced by electrolysing 1.0 M Hg(NO3)2 solution with a current of 2.00 A for 3 hours?
[Hg(N03)2 = 200.6 g mol-1]
(b) A voltaic cell is set up at 25 °C with the following half-cells:
Al3+(0.001 M) and Ni2+ (0.50 M)
Write an equation for the reaction that occurs when the cell generates an electric,, current and determine the cell potential.
(Given;ENi2+/Ni= – 0.25 V, EAl3+/Al = – 1.66 V)
Answer: Refer Ans. to Q.47.
2009
Short Answer Type Questions [I] [2 Marks]
81. Define and express the relation between conductivity and molar conductivity for the solution of an electrolyte.
Answer: Refer Ans. to Q.55.
82. What type of a battery is lead storage battery ? Write the anode and cathode reactions and the overall cell reaction occurring in the operation of a lead storage battery.
Answer:Refer Ans. to Q.37.
83. Two half-reactions of an electrochemical cell are given below:
Mn04–(aq) + 8H+(aq) + 5e– ——–> Mn2+(aq) + 4H20(l), 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 the reaction is reactant favoured or product favoured.
Answer: Refer Ans. to Q.53.
84. Express the relation among the cell constant, the resistance of the solution in the cell and the conductivity of the solution. How is molar conductivity of a solution related to its conductivity?
Answer: Refer Ans. to Q.39.
Short Answer Type Questions [II] [3 Marks]
85. A voltaic cell is set up at 25 °C with the following halfcells:
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.
[ENi2+/Ni = – 0.25 V and EAl3+/Al= – 1.66 V]
Answer: Refer Ans. to Q.41.
86. A copper-silver cell is set up. The copper ion concentration is 0.10 M. The concentration of silver ion is not known. The cell potential when measured was 0.422 V. Determine the concentration of silver ions in the cell.
(Given EAg+/Ag = +0.80 V, ECu2+/Cu = +0.34 V)
Answer: Refer Ans. to Q.74.
Long Answer Type Questions [5 Marks]
87. (a) Corrosion is essentially an electrochemical phenomenon. Explain the reactions occurring during corrosion of iron kept in an open atmosphere.
(b) Calculate the equilibrium constant for the equilibrium reaction:
Fe(s) + Cd2+ (aq) ——–> Fe2+ (aq) + Cd(s)
(Given: E0Cd2+/Cd= – 0.40 V, E0Fe2+/Fe = – 0.44 V).
Answer: Refer Ans. to Q.66.
88. (a) How many moles of mercury will be produced by electrolysing 1.0 M Hg(NO3)2 solution with a current of 2.00 A for 3 hours?
[Hg(NO3)2 = 200.6 g mol-1]
(b) A voltaic cell is set up at 25 °C with the following half-cells:
Al3+(0.001 M) and Ni2+(0.50 M)
Write an equation for the reaction that occurs when the cell generates an electric current and determine the cell potential.
(Given; ENi2+/Ni = – 0.25 V, EAl3+/Al = – 1.66 V)
Answer: Refer Ans. to Q.47.
89. (a) Calculate the emf for the given cell at 25 °C:
Cr | Cr3+(0.1 M) || Fe2+ (0.01 M) | Fe [Given: E°Cr3+/Cr = -0.74 V, E°Fe2+/Fe = -0.44 V]
(b) Calculate the strength of the current required to deposit 1.2 g of magnesium from molten MgCl2 in 1 hour.
[1 F = 96,500 C mol-1; Atomic mass: Mg = 24.0]
Answer: Refer Ans. to Q.77.
90. (a) Define molar conductivity of a substance and describe how for weak and
strong electrolytes, molar conductivity changes with concentration of solute. How is such change explained?
(b) A voltaic cell is set up at 25 °C with the following half-cells:
Ag+(0.001 M) | Ag and Cu2+(0.10 M) | Cu What would be the voltage of this cell?
(E0cell = 0 .46 V)
Answer: Refer Ans. to Q.46.