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NEET Chemistry MCQ
Quiz 11
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Q.1
Which of the following is a strong electrolyte? [ Pb CET 1994]
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a) Ca(NO3)2
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b) HCN
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c)H2SO3
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d)NH4OH
Explanation
Ca(NO3)2 is salt , strong electrolyte Answer:(a)
Q.2
The standard EMF of Danielle cell is 1.1 V. The maximum electrical work obtained from the Danielle cell is [MP PMT 2002]
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a) 212.3 kJ
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b) 175.4 kJ
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c) 106.15 kJ
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d) 53.07 kJ
Explanation
For Danielle cell n=2 Now Work=nFE Work=2 × 96500 × 1.1=212.3 kJ Answer: (a)
Q.3
The oxidation potentials of A and B are +2.37 and +1.66V respectively. In chemical reactions ..[Pb CET 1997]
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a) A will be replaced by B
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b) A will replace B
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c)A will not replace by B
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d)A and b will not replace each other
Explanation
Higher the oxidation potential more reactivityAnswer: (b)
Q.4
On the basis of information available from the reaction (4/3)Al + O2 → (2/3)Al2O3, ΔG=-827 kJmol-1of O2 the minimum emf required to carry out an electrolysis of Al2O3 is [ F=96500 C mol-) [ CBSE PMT 2003]
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a) 2.14 V
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b) 4.28 V
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c)6.42 V
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d)8.56 V
Explanation
For electrolysis of Al2O3(2/3)Al2O3 → (4/3)Al + O2Oxidation state of O2 in Al2O3 is -2 and reduce to 0 In left hand side of above equation number of 'O' are 6 Thus (2/3)× 6 ×(-2)=4 electrons are accepted by Oxygen or n=4Now ΔG=-nFE-827,000=-4 × 96500 × VV=2.14 VAnswer: (a)
Q.5
Zn(s) + Cl2 (1atm) → Zn2+ + 2Cl-The E° of the cell is 2.12V. to increase E ..[ DPMT 2003]
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a) Zn2+ concentration should be increased
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b) Zn2+
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c)Cl- concentration should be increased
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d)partial pressure of Cl2 should be decreased
Explanation
from equation Thus if Zn2+ Answer:(b)
Q.6
Which one of the following cells can convert chemical energy of H2 and O2 directly into electrical energy? [ MP PMT 1998]
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a) Mercury cell
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b) Danielle cell
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c) Fuel cell
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d) lead storage cell
Explanation
Answer: (c)
Q.7
Pure water does not conduct electricity because it .. [ manipal PMT 1995]
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a)has low boiling point
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b) is almost unionised
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c)is neutral
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d)is readily decomposed
Explanation
Answer: (b)
Q.8
What is the potential of half-cell consisting of zinc electrode in 0.01m ZnSO4 solution at 25°C [ Eoxo=0.763V) [ manipal PMT 2001]
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a) 0.8221 V
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b) 8.221 V
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c) 0.5282V
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d)9.282V
Explanation
Zn2+ + 2e- → Zn Answer:(a)
Q.9
The equivalent conductance at infinite dilution of a weak acid such as HF [ Pb PMT 1998]
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a) Can be determined by extrapolation of measurements on dilute solutions of HCl, HBr and HI
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b) Can be determined by measurement on every dilute HF solutions
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c) Can best be determined from measurements on dilute solutions of NaF, NaCl and HCl
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d) Is an undefined quantity
Explanation
Option 'c' according to Kohlrausch's law Answer: (c)
Q.10
Reaction that takes place at graphite anode in dry cell is ..[MP PET 1996]
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a)Zn2+ + 2e- → Zn(s)
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b) Zn(s) → Zn2+ +2e-
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c)Mn2+ +2e- → Mn(s)
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d)Mn(s) → Mn+ + e- +1.5V
Explanation
Answer: (b)
Q.11
An gives H2 gas with H2SO4 and HCl but not with HNO3 because [ CBSE PMT2002]
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a) Zn acts as oxidizing agent when reacts with HNO3
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b) HNO3 is weak acid than H2SO4 and HCl
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c)In electrochemical series Zn is above hydrogen
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d)NO3- is reduced in preference to hydronium ion
Explanation
Answer: (d)
Q.12
Calculate the standard free energy change for reaction, 2Ag + 2H+ → H2 + 2Ag+ E° for Ag+ + e- → Ag is 0.80V [ ISM Dhanbad 1994]
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a) +154.4 kJ
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b) +308.8 kJ
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c)-154.4 kJ
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d)-308.8 kJ
Explanation
Given potential of Ag is reduction potential, Hydrogen electrode is cathode(0Volt) and Ag is Anode(0.8Volt) Cell potential = E cell=reduction potential of (cathode - Anode) Cell potential E0 = 0-0.8 = -0.8Volts ΔG=-nFe°ΔG - -2 ×96500 ×(-0.80)=154.4kJ Answer:(a)
Q.13
For the cell reaction,Cu2+ [C1(aq)] + Zn(s) → Zn2+ [C2(aq)] + Cu(s)of an electrochemical cell, the change in free energy, ΔG at a given temperature is a function of ..[CBSE PMT 1998]
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a) ln(C1)
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d) ln(C2)
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b) ln(C₂/C₁)
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c) ln(C₁ + C₂)
Explanation
from the equations Hence ΔG is the function of ln(C2/C1) Answer: (b)
Q.14
Specific conductance of 0.1M sodium chloride solution is 1.06×10⁻² ohm-1 cm-Its molar conductance in ohm-1cm2 mol-1 is [ kerala MEE 2001]
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a) 1.06 ×102
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b) 1.06 ×103
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c) 1.06 ×104
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d) 5.3 ×102
Explanation
Use formula Answer: (a)
Q.15
Specific conductance of 0.1M Nitric acid is 6.3 × 10⁻² ohm-1cmthe molar conductance of the solution is
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a) 630 ohm-1 cm2
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b) 315 ohm-1 cm2
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c) 100 ohm-cm2
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d) 6300 ohm-1cm2
Explanation
Use formula Answer: (a)
Q.16
Ionic mobility of Ag+ is...λAg=5 ×10⁻⁴ ohm-1cm2 eq-1)[ orissa JEE 2003]
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a) 5.2 ×10⁻¹⁹
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b) 2.4 ×10⁻⁹
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c)1.52 ×10⁻⁹
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d)8.52 × 10⁻⁹
Explanation
Ionic mobility=Ionic conductance / 96500 Answer:(a)
Q.17
electrolysis of dilute aqueous NaCl solution was carried out by passing 10 mili ampere current. The time required to liberate 0.01 mol of H2 gas at the cathode is ( 1 Faraday=96500 C mol-1){IIT 2008]
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a)9.650 × 104 sec
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b) 19.3 ×104 sec
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c)28.95 × 104 sec
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d)38.6 × 104 sec
Explanation
Q=I × t Q=10 × 10⁻³ × t ---(1) 2 H2O + 2e- → H2 + 2OH- 0.1 mole of H2 is liberated by 0.02 Faraday of charge i.e Q=0.02 × 96500 .... (2) From (1) and (2) we get 10 × 10⁻³ × t=0.02 × 96500 t=19.3 × 104 secAnswer: (b)
Q.18
A solution of CuSO4 is electrolysed for 7 minutes with a current of 0.6amp. The amount of electricity passed is equal to [ Pb CET 1991]
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a) 4.2 coulombs
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c)126 coulombs
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d)35 coulombs
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b) 2.6 ×10⁻³ faraday
Explanation
Charge Q=It Q=0.6 ×(7×60)=252 C In faraday=252/96500=2.6 ×10⁻³ FaradayAnswer: (b)
Q.19
How many coulombs of electricity are consumed when 100 mA current is passed through a solution of AgNO3 for half an hour during an electrolysis experiment? [ Pb CET 1989]
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a) 108
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b) 180
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c) 1800
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d) 18000
Explanation
Electricity Q - It Q=(100 × 10⁻³) × 1800 Q=180 Answer: (b)
Q.20
The standard reduction potential for two reaction are given belowAgCl(s) + e- → Ag(s) + Cl-(aq), E°=0.22VAg+(aq) + e- → Ag(s) , E°=0.80Vthe solubility product of AgCl under standard condition of temperature (298K) is given by ..[kerala MEE 2003]
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a) 1.6 ×10-
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b) 1.5 ×10⁻¹⁰
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c)3.2 ×10⁻⁸
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d)3.2 ×10⁻¹⁰
Explanation
Subtracting second equation from first we getAgCl(s) → Ag+(aq) + Cl-(aq) , E°=-0.50V From Nernst equation As AgCl is solid its concentration is 1 and equilibrium E=0 Thus E°=0.0591 log[Ag+][Cl-] E°=0.0591 log Ksp -0.5=0.0591 log Ksp log Ksp=-9.8139 Or Ksp1.535 ×10⁻¹⁰Answer: (b)
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