PRARAMBH EXERCISE-1 (TOPICWISE)

ELECTROCHEMICAL CELLS AND 9. If a salt bridge is not used between two half cells, voltage:
GALVANIC CELL (1) Drops to zero (2) Does not change
(3) Increases gradually (4) Increases rapidly
1. In a galvanic cell, electron flow will be from:
(1) Negative electrode to positive electrode 10. The electro-chemical cell stops working after some time
because
(2) Positive electrode to negative electrode
(1) Electrode potentials of both electrodes become zero
(3) There will be no flow of electrons
(2) Electrode potentials of both electrodes become equal
(4) Cathode to anode in the external circuit
(3) Temperature of the cell increases
2. In a galvanic cell, the reactions taking place in the anodic
(4) The reaction starts proceeding in opposite direction
half cell and the cathodic half cell will be:
(1) Reduction 11. The reaction ½ H2(g) + AgCl(s) → H+(aq) + Ag(s) can be
represented in the galvanic cell as:
(2) Oxidation
(1) Ag|AgCl(s) | KCl(sol) || AgNO3(sol) | Ag
(3) Oxidation and reduction
(4) Reduction and oxidation (2) Pt, H2(g) | HCl(sol) || AgNO3(sol) | Ag

3. Which of the following is not true for a galvanic cell (3) Pt, H2(g) | HCl(sol) || AgCl(s) | Ag
represented in IUPAC system? (4) H2(g) | HCl(sol) || AgCl(s) | Ag
(1) Right hand electrode is a +ve terminal. 12. In a Daniell cell,
(2) Right hand electrode acts as cathode. (1) The chemical energy liberated during the redox reaction
(3) Electrons are given out in the external circuit from the is converted to electrical energy.
anode. (2) The electrical energy of the cell is converted to chemical
(4) Electrons are given out in the external circuit from the energy.
cathode. (3) The energy of the cell is utilised in conduction of the
4. A half cell reaction is one that: redox reaction.
(1) Involves only half a mole of electrolyte (4) The potential energy of the cell is converted into
(2) Goes only half way to completion electrical energy.
(3) Takes place at one electrode 13. For which of the following SOP and SRP are equal?
(4) Consumes half a unit of electricity (1) SHE (2) Mg Electrode
5. Which of the following energy changes occur in galvanic (3) Ni electrode (4) Cu electrode
cell? 14. Cathodic standard reduction potential minus anodic standard
(1) Electrical energy → Chemical energy reduction potential is equal to:
(2) Chemical energy → electrical energy (1) Faraday (2) Coulomb
(3) Chemical energy → Internal energy (3) Cell potential (4) Ampere
(4) Internal energy → electrical energy 15. What will be standard cell potential of galvanic cell with the
6. The purpose of the salt bridge in a galvanic cell is to following reaction?
(1) Prevent accumulation of charges around the electrodes 2Cr(s) + 3Cd2+(aq) → 2Cr3+(aq) + 3Cd(s)
(2) Facilitate continuity of the cell reaction [Given : E 0Cr3+ /Cr =
−0.74V and E 0Cd2+ /Cd =
−0.4V ]
(3) To produce current at a constant strength (1) 0.70 V (2) 1.14 V
(4) All the above (3) 0.34 V (4) –0.34 V
7. Agar–Agar is used in salt bridge since it is: 16. The electrode potential measures the:
(1) Electrolyte (2) Non–electrolyte (1) Tendency of the electrode to gain or lose electrons
(3) Inert electrolyte (4) A solid (2) Electron affinity of elements
8. The thermodynamic efficiency of cell is given by: (3) Difference in the ionization potential of electrode and
(1) –nFE (2) –nFE/∆G metal ion
(3) –nFE/∆H (4) None of these (4) Heat of combustion

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64 W NEET (XII) Module-1 CHEMISTRY
 =
17. If E oNi2+ |Ni 0.25V,
= E oCu 2+ |Cu 0.34V,
= E oAg + |Ag 0.8V and 26. For cell reaction, Zn + Cu2+ → Zn2+ + Cu, cell representation is:
E oZn 2+ |Zn = −0.76V, then which of the following reactions
(1) Zn | Zn2+ || Cu2+ | Cu (2) Cu | Cu2+ || Zn2+ | Zn
under standard condition will not take place in the specified (3) Cu | Znz+ || Zn | Cu2+ (4) Cu2+| Zn || Zn2+ | Cu
direction spontaneously? 27. The electrode potentials for:
(1) Cu(s) + Ni 2 + (aq) → Cu 2 + (aq) + Ni(s) Cu2+(aq) + e– → Cu+ (aq)
(2) Cu(s) + 2Ag + (aq) → Cu 2 + (aq) + 2Ag(s) and Cu+ (aq) + e– → Cu(aq) are 0.15 V
0
and + 0.50 V respectively. The value of E Cu 2+ |Cu will be:
(3) Cu(s) + Zn 2 + (aq) → Cu 2 + (aq) + Zn(s)
(1) 0.150 V (2) 0.500 V
(4) Both (1) and (3) (3) 0.325 V (4) 0.650 V
18. Fluorine is the best oxidising agent because it has: 28. When Zn metal is added to CuSO4 solution Cu is precipitated.
(1) Highest electron affinity It is due to:
(2) Highest reduction potential (1) Oxidation of Cu2+ (2) Reduction of Cu2+
(3) Highest oxidising potential (3) Hydrolysis of CuSO4 (4) Ionization of CuSO4
(4) Lowest electron affinity 29. Which of the following statements is true for the
19. Which one of the following metal will not reduce H2O? electrochemical Daniell cell?
(1) Ca (2) Fe (1) Electrons flows from copper electrode to zinc electrode
(3) Cu (4) Li (2) Currents flows from zinc electrode to copper electrode
(3) Cations moves towards copper electrode
20. The cell reaction for the given cell is:
(4) Cations moves towards zinc electrode
Pt(H2) | pH = 2 || pH = 3 | Pt(H2)
P1 = 1 atm P2 = 1 atm
NERNST EQUATION
(1) Spontaneous (2) Non-spontaneous
30. The relationship between free energy and electrode potential
(3) In equilibrium (4) Either of these
is:
21. Which of the following is the cell reaction that occurs when (1) ∆G = –nFE (2) ∆G = nFE
the following half – cells are combined?
∆H nFE
I2 + 2e– → 2I– (1M) ; E = + 0.54 V (3) ∆G = (4) ∆G =
nFE R
Br2 + 2e– → 2 Br– (1M) ; E = + 1.09 V
31. ΔG0 for the reaction, Cu2+ + Fe → Fe2+ + Cu is:
(1) 2Br– + I2 → Br2 + 2I– [given: E 0Cu /Cu = +0.34V , E 0Fe /Fe = –0.44V ]
2+ 2+

(2) I2 + Br2 → 2I– + 2Br–

(1) 11.44 KJ (2) 180.8 KJ
(3) 2I– + Br2 → I2 + 2Br– (3) 150.5 KJ (4) 28.5 KJ
(4) 2I– + 2Br– → I2 + Br2
32. E0c ell for the reaction, 2H2O → H3O+ + OH– at 25°C is
22. Arrange the following in the order of their decreasing –0.8277 V. The equilibrium constant for the reaction is:
electrode potentials: Mg, K, Ba, Ca (1) 10–14 (2) 10–23
(1) K > Ba > Ca > Mg (2) Ba > Ca > K > Mg (3) 10 –7 (4) 10–21
(3) Ca > Mg > K > Ba (4) Mg > Ca > Ba > K 33. The Nernst equation giving dependence of electrode
23. The emf of a galvanic cell, with electrode potential of reduction potential on concentration is:
Zn2+ = –0.76 V and that of Cu2+ = 0.34 V, is: 2.303 RT
E E0 +
(1) = log  M n + 
(1) 0.76 V (2) 1.1 V nF
(3) –1.1 V (4) 1.6 V
2.303 RT Mn + 
E E0 −
(2) = log  
24. If the half - cell reactions are given as: nF [M]
I. Fe2+(aq) + 2e– → Fe(s), E0 = –0.44 V
2.303 RT
1 E E0 −
(3) = log  M n + 
II. 2H+ (aq) + O 2 ( g ) + 2e − → H2O(d) , E0 = + 1.23 V nF
2
The E0 for the reaction, E E0 +
2.303 RT
log
[M]
(4) =
Fe(s) + 2H+ + 1/2 O2(g) → Fe2+ + H2O(l) nF Mn + 
 
(1) + 1.67 V (2) –1.67 V
34. The chlorate ion can disproportionate in basic solution
(3) + 0.79 V (4) –0.79 V according to reaction, 2ClO3−  ClO 2− + ClO 4−
25. The EMF of a galvanic cell is determined by using a: What is the equilibrium concentration of perchlorate
(1) Voltmeter (2) Spectrometer ions from a solution initially at 0.1 M in chlorate ions at
(3) Coulometer (4) Ammeter 298 K?

Electrochemistry 65
 Given 42. Alkali metals have high oxidation potential, hence, they
=
E 0 ClO− ClO− 0.36
= V and E 0 ClO− ClO− 0.33V at 298 K behave as good:
4 3 3 2

(1) 0.019 M (2) 0.024 M (1) Oxidising agent (2) Lewis bases
(3) 0.1 M (4) 0.19 M (3) Reducing agents (4) Electrolytes
35. The Gibb’s energy for the decomposition of Al2O3 at 500°C 43. Zn gives H2 gas with H2SO4 and HCl but not with HNO3
is as follows: because:
2 4
Al2 O3 → Al + O 2 ; ∆ r G = +966 kJ / mol (1) Zn act as oxidising agent when react with HNO3
3 3 (2) HNO3 is weaker acid than H2SO4 and HCl
The potential difference needed for electrolytic reduction
(3) In electrochemical series, Zn is above hydrogen
of Al2O3 at 500°C is atleast: –
(1) 5 V (2) 4.5 V (4) NO3 is reduced in preference to hydronium ion
(3) 3 V (4) 2.5 V 44. What is the electrode potential (in volt) of the following
36. Mark the correct Nernst equation for the given cell: electrode at 25°C?
Fe(s ) | Fe 2+ ( 0.001M ) || H + ( l M ) | H 2(g) (1bar ) | Pt ( s ) Ni(2+0.1M ) | Ni(s )

0 0.0591 [Fe 2 + ][H + ]2  0 2.303RT 
(1) E=
cell E cell − log  E ( Ni2+ |Ni) = − 0.25V, = 0.06 
2 [Fe][H 2 ]
 F 
0 0.0591 [Fe][H + ]2 (1) –0.28 V (2) –0.34 V
(2) E=
cell E cell − log
2 [Fe 2 + ][H 2 ] (3) –0.82 V (4) –0.22 V
2+
0.0591 [Fe ][H 2 ] 45. The relation between standard reduction potential of a cell
(3) E=
cell E 0cell − log
2 [Fe][H + ]2 and equilibrium constant is shown by:
0 0.0591 [Fe][H 2 ] n 0.059
(4) E=
cell E cell − log 0
(1) E cell = log K c (2) E 0cell = log K c
2 [Fe 2 + ][H + ]2 0.059 n
37. The oxidation potential of hydrogen half-cell will be negative log K c
0
if: (3) E cell = 0.059 n log K c (4) E 0cell =
n
(1) p(H2) = 1 atm and [H+] = 1 M
(2) p(H2) = 1 atm and [H+] = 2 M 46. Emf of hydrogen electrode in terms of pH is: [at 1atm
pressure]
(3) p(H2) = 0.2 atm and [H+] = 1 M
RT RT 1
(4) Both (2) and (3) (1) E H2 = pH (2) E H2 =
F F pH
38. The potential of a single electrode depends upon:
2.303RT
(1) The nature of the electrode (3) E H2 = pH (4) E H2 = − 0.0591pH
(2) Temperature F
(3) Concentration of the ion with respect to which it is 47. The equilibrium constant of the reaction,
reversible Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)
(4) All the above E0 = 0.46 V at 298 K is:
39. If the solution of the CuSO4 in which copper rod is immersed (1) 2.4 × 1010 (2) 2.0 × 1010
is diluted to 10 times, the electrode potential:
(3) 4.0 × 1010 (4) 4.0 × 1015
(1) Increases by 0.295V (2) Decreases by 0.0295V
(3) Increases by 0.059V (4) Decreases by 0.059V 48. For Cr2O72 – + 14H+ + 6e– → 2Cr3+ + 7H2O,
40. Calculate the emf of half cell: E0 = + 1.33V at [Cr2O72–] = 4.5 millimole per ltr, [Cr3+]
= 15 millimole per ltr, E is 1.06 V. The pH of the solution
; is nearly equal to:
(1) 2 (2) 3
(1) 1.45 V (2) 1.27 V
(3) –1.45 V (4) –1.27 V (3) 5 (4) 4

41. During electrochemical process: 49. For a cell reaction involving a two-electron exchange, the
(1) Gibbs free energy increases standard emf of the cell is found to be 0.295 V at 25°C. The
(2) Gibbs free energy remains constant equilibrium constant of the reaction at 25°C will be:
(3) No prediction can be made about Gibbs free energy (1) 1 × 10–10 (2) 29.5 × 10–2
(4) Gibbs free energy decreases (3) 10 (4) 1 × 1010

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66 W NEET (XII) Module-1 CHEMISTRY
CONDUCTANCE OF ELECTROLYTIC 58. Calculate molar conductivity of 0.15 M solution of KCl at
298 K if its conductivity is 0.0152 S cm–1:
SOLUTIONS
(1) 124 Ω–1 cm2 mol–1 (2) 204 Ω–1 cm2 mol–1
50. Molar conductance of KCl increases slowly with decrease
(3) 101 Ω cm mol
–1 2 –1 (4) 300 Ω–1 cm2 mol–1
in concentration because of:
(1) Increase in degree of ionisation 59. Specific conductance of 0.1 M NaCl solution is 1.01 × 10–2
(2) Increase in total number of current carrying species Ω–1 cm–1. Its molar conductance in Ω–1 cm2 mol–1 is:
(3) Weakening of interionic attractions and increase in ionic (1) 1.01 × 102 (2) 1.01 × 103
mobilities (3) 1.01 × 104 (4) 1.01
(4) Increase in hydration of ions 60. Limiting molar conductivity of NaBr is:
51. The reason for increase in electrical conduction of a weak (1) Λ0m NaBr = Λ0m NaCl + Λ0m KBr
electrolyte with increase in temperature: (2) Λ0m NaBr = Λ0m NaCl + Λ0m KBr – Λ0m KCl
(1) Increase in the number of ions
(3) Λ0m NaBr = Λ0m NaOH + Λ0m NaBr – Λ0m NaCl
(2) Increase in the speed of ions
(3) Increase in the degree of dissociation of electrolytes (4) Λ0m NaBr = Λ0m NaCl – Λ0m NaBr
(4) All the above 61. Which of the following is a poor conductor of electricity?
52. The molar conductance of acetic acid at infinite dilution is (1) CH3COONa (2) C2H5OH
∞. If the conductivity of 0.1M acetic acid is S, the apparent (3) NaCl (4) KOH
degree of ionisation is: 62. The units of conductivity of solution are:
10000S 10S (1) ohm–1 (2) ohms
(1) (2)
λ∞ λ∞ (3) ohm–1cm–1 (4) ohm–1eq–1
λ∞
(3) (4) 100000 63. Molar conductivity of 0.025 mol L–1 methanoic acid is
100S λ∞S 46.1 Scm2mol–1 the dissociation constant will be:
53. The variation of equivalent conductance of strong electrolyte (Given : λ0H+ = 349.6 Scm2 mol–1 and λ0HCOO– = 54.6 Scm2
with (concentration) is represented by: mol–1)
(1) 11.4% . 3.67 × 10–4 mol L–1
(1) (2) (2) 22.8% , 1.83 × 10–4 mol L–1
(3) 52.2% , 4.25 × 10–4 mol L–1
(4) 1.14 % , 3.67 × 10–4 mol L–1
64. Molar conductivity of NH4OH can be calculated by the
equation:
(3) (4) (1) Λ 0NH4OH = Λ 0Ba( OH ) + Λ 0NH4Cl − Λ 0BaCl2
2

0 0
(2) Λ NH 4OH =Λ BaCl2 + Λ 0NH4Cl − Λ 0Ba( OH )
2

Λ 0Ba( OH ) + 2Λ 0NH4Cl − Λ 0BaCl2

54. The specific conductivity of N/10 KCl solution at 20°C is (3) Λ 0NH4OH = 2
0.0212 Ω–1 cm–1 and the resistance of the cell containing this 2
solution at 20°C is 55 Ω. The cell constant is: 2Λ 0NH4Cl − Λ 0Ba( OH )
0
(1) 3.324 cm–1 (2) 1.166 cm–1 (4) Λ NH4OH = 2

–1 2
(3) 2.372 cm (4) 3.682 cm–1 65. Units of the properties measured are given below, which of
55. Pure water does not conduct electricity because it is: the properties have not been matched correctly?
(1) Neutral (2) Readily decomposed (1) Molar conductance = Sm2 mol–1
(3) Almost unionised (4) Completely ionised (2) Cell constant = m–1
56. In aqueous solution, strong electrolytes ionize and yield: (3) Specific conductance = Sm2
(1) Ions (2) Electrons (4) Equivalent conductance = Sm2 (eq)–1
(3) Acids (4) Oxides 66. When water is added to an aqueous solution of an electrolyte,
57. NaCl, MgCl2 and CaSO4 are known as: what is the change in specific conductivity of the electrolyte?
(1) 1-1, 2-1, 2-2 types electrolyte respectively (1) Conductivity increases
(2) Strong, weak and strong electrolytes respectively (2) Conductivity decreases
(3) Electrolytes with different value of A (3) Conductivity remains same
(4) Electrolytes with same molar conductivity (4) Conductivity does not depends on number of ions

Electrochemistry 67
 67. The unit of cell constant is: 77. Mark the correct choice of electrolytes represented in the
(1) ohm–1 (2) ohm - cm graph:
(3) cm–1 (4) ohm–1cm2eq–1
68. The cell constant is the product of resistance and:
(1) Conductance (2) Molar conductance A
(3) Specific conductance (4) Specific resistance
69. What would be the equivalent conductivity of a cell in
which 0.5 M salt solution offers a resistance of 40 Ω whose
electrodes are 2 cm apart and 5 cm2 in area? (1) A → NH4OH, B → NaCl
(1) 10Ω–1 cm2 eq–1 (2) 20Ω–1 cm2 eq–1 (2) A → NH4OH, B → NH4Cl
(3) 30Ω–1 cm2 eq–1 (4) 25Ω–1 cm2 eq–1 (3) A → CH3COOH , B → CH3COONa
70. The molar conductance of Ba2+ and Cl– are respectively (4) A → KCl, B → NH4OH
127 and 76 Ω–1 cm2 eq–1 at infinite dilution. What will be 78. Fused NaCl has less electrical conductance than NaCl in the
the equivalent conductance of BaCl2 at Infinite dilution? aqueous solution. This is due to:
(1) 139.5 Ω–1 cm2 eq–1 (2) 203 Ω–1 cm2 eq­–1 (1) Fused NaCl has less number of ions
(3) 279 Ω–1 cm2 eq–1 (4) 101.5 Ω–1 cm2 eq–1 (2) Incomplete ionization occurs in the fused state
71. Electrical conductance through metals is called metallic (3) Na+, Cl– ions do not move freely in the fused salt
or electronic conductance and is due to the movement of (4) Fused NaCl has no ions
electrons. The electronic conductance depends on:
79. The degree of dissociation of an electrolyte does not depend
(1) The nature and structure of the metal on:
(2) The no. of valence electrons per atom (1) Nature of electrolyte (2) Catalytic action
(3) Change in temperature (3) Dilution (4) Temperature
(4) All of these 80. Match the list-I with list-II and mark the appropriate choice:
72. What happens at infinite dilution in a given solution?
List-I List-II
(1) The degree of dissociation is unity for the weak
A. Kohlrausch’s law P. Λ = Λ 0c + Λ a0
0
electrolytes eq

(2) The electrolyte is 100% ionised B. Molar conductivity Q. K × 1000

(3) All interionic attractions disappear Λm =
Molarity
(4) All the above
73. With increase in temperature, the electrical conduction of
C. Degree of dissociation R. Λm
α=
metallic conductor: Λ 0m
(1) Increases D. Dissociation constant S. cα 2
(2) Decreases Ka =
(3) Remains the same 1− α
(4) Changes irregularly
(1) A-(R); B-(S); C-(P); D-(Q)
74. During electric conduction, the composition of which of the (2) A-(P); B-(Q); C-(R); D-(S)
following is changed?
(3) A-(S); B-(P); C-(Q); D-(R)
(1) Graphite (2) Zinc wire
(4) A-(Q); B-(R); C-(S); D-(P)
(3) Copper wire (4) H2SO4
81. Conductivity of a saturated solution of a sparingly soluble
75. What will be the molar conductivity of Al3+ ions at infinite salt AB at 298 K is 1.85 × 10–5 Sm–1. Solubility product of
dilution if molar conductivity of Al2 (SO4)3 is 858 Scm2 the salt AB at 298 K is:
mol–1 and ionic conductance of SO42– is 160 Scm2 mol–1 [Given; Λ0m (AB) = 140 × 10–4 Sm–2 mol–1]
at infinite dilution? (1) 5.7 ×10–12 (2) 1.32 ×10–12
(1) 189 Scm2 mol–1 (2) 698 Scm2 mol–1 (3) 7.5 ×10 –12 (4) 1.74 ×10–12
(3) 1018 Scm2 mol–1 (4) 429 Scm2 mol–1 82. At 18°C, the conductance of H+ and CH3COO– at infinite
76. How much will the reduction potential of a hydrogen dilution are 315 and 35 mho cm2 eq–1 respectively. The
electrode change when its solution initially at pH = 0 is equivalent conductivity of CH3COOH at infinite dilution
neutralised to pH = 7? is_____ (mho cm2 eq–1)
(1) Increase by 0.059 V (2) Decrease by 0.059 V (1) 350 (2) 280
(3) Increase by 0.41 V (4) Decrease by 0.41 V (3) 30 (4) 315

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68 W NEET (XII) Module-1 CHEMISTRY
 83. The extent of ionization of weak electrolyte increases: 91. The number of faradays required to liberate 1 mole of any
(1) With increase in concentration of the solute element indicates:
(2) On addition of excess of water (1) Weight of the element
(3) On decreasing the temperature (2) Conductance of the electrolyte
(4) On stirring the solution vigorously (3) Charge on the ion of the element
(4) Isotopic number
84. The equivalent conductance of a 1 N solution of an
electrolyte is nearly: 92. 6.023 × 1023 electrons are equal approximately to:
(1) 1000 times its specific conductance (1) 10 coulombs (2) 96500 coulombs
(2) 10 times its specific conductance (3) 1electron volt (4) 0.1 Faraday
(3) 100 times its specific conductance 93. Dilute nitric acid on electrolysis using platinum electrodes
(4) The same as its specific conductance yields:
(1) Both oxygen & hydrogen at cathode
85. The specific conductance (κ) of an electrolyte of 0.1 N
concentration is related to equivalent conductance Λ by the (2) Both oxygen & hydrogen at anode
following formula: (3) H2 at cathode and O2 at anode
(1) Λ = κ (2) Λ =10 κ (4) O2 at cathode and H2 at anode
(3) Λ = 100 κ (4) Λ = 10000 κ 94. Which of the following occurs at cathode?
1
86. A current of 1.40 ampere is passed through 500 mL of (1) 2OH − → H 2 O + O 2 + 2e −
0.180 M solution of zinc sulphate for 200 seconds. What 2
will be the molarity of Zn2+ ions after deposition of Zinc? (2) Ag → Ag+ + e–
(1) 0.154 M (2) 0.177 M (3) Fe2+ → Fe3+ + e–
(3) 2 M (4) 0.180 M (4) Cu2+ + 2e– → Cu
87. The equivalent conductance at infinite dilution of a weak 95. How much metal will be deposited when a current of
acid such as HF: 12 ampere with 75% efficiency is passed through the cell
for 3 h?
(1) Can be determined by extrapolation of measurements
on dilute solutions of HCl, HBr and HI [Given ; Z(electrochemical equivalent) = 4 × 10–4]
(2) Can be determined by measurement on very dilute HF (1) 32.4 g (2) 38.8 g
solutions (3) 36.0 g (4) 22.4 g
(3) Can be best determined from measurements on dilute 96. If 54 g of silver is deposited during an electrolysis reaction,
solutions of NaF, NaCl and HCl how much aluminium will be deposited by the same amount
(4) Is an undefined quantity. of electric current?
(1) 2.7 g (2) 4.5 g
88. Which of the following statements is correct for an
electrolytic solution upon dilution? (3) 27 g (4) 5.4 g
(1) Conductivity increases 97. The quantity of electricity needed to separately electrolyse
(2) Conductivity decreases 1 M solution of ZnSO4, AlCl3, and AgNO3 completely is in
the ratio of:
(3) Molar conductance decreases but equivalent conductance
(1) 2 : 3 : 1 (2) 2 : 1 : 1
increases
(3) 2 : 1 : 3 (4) 2 : 2 : 1
(4) Molar conductance increases while equivalent
conductance decreases. 98. Electrolysis of salt solution is due to the formation of:
(1) Electron (2) Ions
ELECTROLYTIC CELL AND ELECTROLYSIS (3) Oxides (4) Acids
99. During the electrolysis of fused NaCl, which reaction occurs
89. Faraday’s laws of electrolysis are related to the: at anode?
(1) Equivalent mass of the electrolyte (1) Chloride ions are oxidised
(2) Equivalent weight of the cation /anion (2) Chloride ions are reduced
(3) Atomic weight of the electrolyte (3) Sodium ions are oxidised
(4) Atomic number of the cation/anion (4) Sodium ions are reduced
90. The unit of electrochemical equivalent is: 100. What is the time required (in seconds) for depositing all the
(1) Gram silver present in 125 mL of 1 M AgNO3 solution by passing
(2) Gram/Ampere sec a current of 241.25 A [1F = 96500 C]?
(3) Gram/Coulomb (1) 10 sec (2) 50 sec
(4) Coulomb/Gram (3) 1000 sec (4) 100 sec

Electrochemistry 69
101. On passing 3A of electricity for 50 min, 1.8 g metal deposits, 111. Lithium is generally used as an electrode in high energy
the equivalent mass of metal is: density batteries. This is because
(1) 9.3 (2) 19.3 (1) Lithium is the lightest element
(3) 38.3 (4) 39.9 (2) Lithium has quite high negative reduction potential
102. 1C electricity deposits: (3) Lithium is quite reactive
(1) 10.8 g of Ag (4) Lithium does not corrode easily
(2) 96500 g of Ag
112. The reaction which is taking place in nickel-cadmium battery
(3) Electrochemical equivalent of Ag can be represented by which of the following equation?
(4) Half of electro-chemical equivalent of Ag (1) Cd + NiO(OH) + 2H2O → Cd(OH)2 + 2Ni(OH)2
103. In an electrolytic cell, current flows from: (2) Cd + NiO2 + 2OH → Ni + Cd(OH)2
(1) Cathode to anode in outer circuit (3) Ni + Cd(OH)2 → Cd + Ni(OH)2
(2) Anode to cathode outside the cell
(4) Ni(OH)2 + Cd(OH)2 → Ni + Cd + 2H2O
(3) Cathode to anode inside the cell
113. In dry cell cathode is:
(4) Current does not flow
(1) Zn (2) Carbon rod
104. During electrolysis, electrons flow from:
(3) Zn+NH4Cl (4) C+MnO2
(1) Cations to cathode (2) Anode to cathode
(3) Cathode to anode (4) Anions to anode 114. In Leclanche cell, Zinc rod is placed in:
105. When 9.65 coulomb of electricity is passed through a solution (1) 10%NH4Cl (2) 20%NH4Cl
of silver nitrate (atomic mass of Ag 108 g mol–1). The amount (3) 30%NH4Cl (4) 40%NH4Cl
of silver deposited is: 115. Which of the following does not conduct electricity?
(1) 16.2 mg (2) 21.2 mg (1) Fused NaCl (2) Solid NaCl
(3) 10.8 mg (4) 6.4 mg
(3) Brine solution (4) Copper
106. A certain current liberated 0.50 gm of hydrogen in
2 hours. How many grams of copper can be liberated by the
FUEL CELLS
same current flowing for the same time in a copper sulphate
solution? 116. In fuel cell, oxidants used are:
(1) 12.7 gm (2) 15.9 gm (1) O2 (2) H2O2
(3) 31.8 gm (4) 63.5 gm (3) HNO3 (4) All
107. The cathode of an electrolytic cell and a reducing agent are 117. Theoretical efficiency of fuel cell is:
similar because both: (1) Nearly 60%
(1) Are metals (2) Supply electrons
(2) 50%
(3) Remove electrons (4) Absorb electrons
(3) 33%
108. The cathode reaction in electrolysis of dilute sulphuric acid
(4) Nearly 100%
with platinum electrode is:
(1) Oxidation 118. Which of the following statements is true for fuel cells?
(2) Reduction (1) They are more efficient
(3) Both oxidation and reduction (2) They are free from pollution
(4) Neutralization (3) They run till reactants are active
109. In electrolysis of dil. H2SO4 using platinum electrodes: (4) All of the above
(1) H2 is evolved at cathode 119. The efficiency of a fuel cell is given by:
(1) ∆H (2) ∆G
(2) SO2 is produced at anode
(3) O2 is obtained at cathode ∆G ∆S
(4) SO2 is produced at cathode
(3) ∆G (4) ∆S
∆H ∆G
BATTERIES
120. The overall reaction of a hydrogen - oxygen fuel cell is:
110. When lead storages battery discharges
(1) 2H2(g) + O2(g) → 2H2O(l)
(1) SO2 is evolved
(2) 2H2(g) + 4OH–(aq) → 4H2O(2)+ 4e–
(2) PbSO4 is consumed
(3) Lead is formed (3) O2(g) + 2H2O(l) + 4e– → 4OH–(eq)
(4) H2SO4 is consumed (4) 4OH–(aq) + 4e– → 2H2O(l)

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70 W NEET (XII) Module-1 CHEMISTRY