JEE Questions for Chemistry Electrochemistry Quiz 10 - MCQExams.com

Molar conductivity of a solution is 1.26 × 102Ω-1cm2mol-1. Its molarity is 0.01. Its specific conductivity will be
  • 1.26 × 10-5
  • 1.26 × 10-3
  • 1.26 × 10-4
  • 0.0063
Specific conductance of 0.1M Nitric acid is 6.3 × 10-2 Ohm-1 cm-1. The molar conductance of the solution is
  • 630 ohm-1 cm2
  • 315ohm-1 cm2
  • 100ohm-1 cm2
  • 6300 ohm-1 cm2
The ionic mobilities of the cation and the anion of a salt are 140 and 80 mhos respectively. The equivalent conductivity of the salt at infinite dilution will be
  • 160 mhos
  • 280 mhos
  • 60 mhos
  • 220 mhos
The equivalent conductance of NaCl, HCl and C2H5COONa at infinite dilution are 126.45, 426.16 and 91 ohm-1cm2. The eq. conductance of C2H5COOH is
  • 201.28 ohm-1cm2
  • 390.71 ohm-1cm2
  • 698.28 ohm-1cm2
  • 540.48 ohm-1cm2
On passing 1 Faraday of electricity through the electrolytic cells containing Ag+, Ni2+, Cr3+ ions solution, the deposited Ag (At. wt. = 108 ) , Ni(At. wt.=and Cr (At. wt. =are
Chemistry-Electrochemistry-3418.png
  • (a)
  • (b)
  • (c)
  • (d)
When 9650 coulombs of electricity is passed through a solution of copper sulphate, the amount of copper deposited is (Given At.wt. of cu = 63.6)
  • 0.318 g
  • 3.18 g
  • 31.8 g
  • 63.6g
Four moles of electrons were transferred from anode to cathode in an experiment on electrolysis of water. The total volume of the two gases (dry and at STP) produced will be approximately (in litres)
  • 22 .4
  • 44.8
  • 67.2
  • 89.4
Calculate the weight of copper that will be deposited at the cathode in the electrolysis of a 0.2 M solution of copper sulphate when quantity of electricity equal to that required to liberate 2.24l of hydrogen at NTP from a 0.1 M aqueous sulphuric acid, is passed. (Atomic mass of Cu = 63.
  • 1.59 g
  • 6.35 g
  • 3.18 g
  • 12.70 g
A 5 ampere is passed through a solution of zinc sulphate for 40 minutes. Find the amount of zinc deposited at the cathode
  • 40.65 g
  • 4.065 g
  • 0.4065 g
  • 65.04 g
On passing a current of 1.0 ampere for 16 min and 5 sec through one litre solution of CuCl2, all copper of the solution was deposited at cathode. The strength of CuCl2 solution was (Molar mass of Cu= 63.5, Faraday constant = 96500 C mol-1)
  • 0.07 M
  • 0.2 N
  • 0.005 M
  • 0.02 N
How many coulombs of electricity are required for reduction of 1 mole of MnO4− to Mn2+ ?
  • 96500 C
  • 1.93 × 105 C
  • 4.83 × 105 C
  • 9.65 × 106 C
A current of 9.65 A is drawn from a Daniell cell for exactly 1hour.The loss in mass of anode and gain in the mass of cathode will be (Given molar masses : Cu = 63.5, Zn = 65.4 mol-1)
  • 11.77 g, 11.43 g
  • 11.43 g, 11.77 g
  • 23.54 g,22.86 g
  • 22.86 g, 23.54 g
How many atoms of hydrogen are liberated at cathode , when 965 coulombs of charge is passed through water?
  • 6.02 × 1021
  • 6.02 × 1023
  • 6.02 × 1019

  • Chemistry-Electrochemistry-3426.png
An electric current is passed through silver voltameter connected to a water voltameter. The cathode of the silver voltameter weighed 0.108 g more at the end of the electrolysis. The volume of oxygen evolved at STP is
  • 56 cm3
  • 550 cm3
  • 5.6 cm3
  • 11.2 cm3
  • 22.4 cm3

Chemistry-Electrochemistry-3429.png
  • (2 × 0.‒ 0.= 1.006 volt
  • (0.771 ‒ 0.5 × 0.= 0.503 volt
  • 0.771 ‒ 0.536 = 0.235 volt
  • 0.536 ‒ 0.771 = ‒ 0.235 volt

Chemistry-Electrochemistry-3431.png
  • + 1.25 V
  • +1.75 V
  • ‒1.25 V
  • ‒1.75 V
The EMF of the cell Ni|Ni2+ || Cu2+ |Cu is 0.59 volt. The standard electrode potential (reduction potential) of copper electrode is 0.34 volt. The standard electrode potential of nickel electrode will be
  • 0.25 Volt
  • ‒ 0.25 Volt
  • 0.93 Volt
  • ‒ 0.93 Volt

Chemistry-Electrochemistry-3434.png
  • 1.56 V
  • 0.64 V
  • ‒ 1.56 V
  • 2.02 V
The standard EMF for the cell reaction. Zn + Cu2+ → Cu + Zn2+ is 1.1 volt at 25oC. The EMF for the cell reaction, when 0.1 M Cu2+ and 0.1 M Zn2+ solutions are used , at 25oC is.
  • 1.10 V
  • 0.10 V
  • ‒ 1.10V
  • ‒ 0.110 V
The potential of the cell containing two hydrogen electrodes as represented below
Chemistry-Electrochemistry-3437.png
  • ‒ 0.118 V
  • ‒ 0.0591V
  • 0.118 V
  • 0.0591 V

Chemistry-Electrochemistry-3439.png
  • E = ‒ 0.36 + (0.059/log (0.004/2)
  • E = + 0.36 + (0.059/log (0.004/2)
  • E = ‒ 0.36 + (0.059/log (0.2/0.004)
  • E = + 0.36 + (0.059/log (0.2/0.004)
What is the potential of half –cell consisting of zinc electrode in 0.01 m ZnSO4 solution at 25oC (E00x = 0.763 V)
  • 0.8221 V
  • 8.221 V
  • 0.5282 V
  • 9.282 V
The e.m.f of the following Daniell cell at 298 K is E1 Zn|ZnSO4 (0.01M)||CuSO4 (1.0M)|Cu. When the concentration of ZnSO4 is 1.0 M and that of CuSO4 is 0.01 M, the e.m.f changed to E2. What is the relationship between E1 and E2 ?
  • E1 > E2
  • E1 < E2
  • E1 = E2
  • E2 = 0 ≠ E1

Chemistry-Electrochemistry-3443.png
  • 1.80 volt
  • 1.07 volt
  • 0.82 volt
  • 2.14 volt

Chemistry-Electrochemistry-3445.png
  • 10‒ 37
  • 1037
  • 10‒17
  • 1017
Calculate the standard free energy change for the reaction,
2Ag + 2H+ → H2 + 2Ag+, Eo for Ag+ + e‒ → Ag is 0.80 V.
  • + 154.4 kJ
  • + 308.8 kJ
  • ‒ 154.4 kJ
  • ‒ 308.8 kJ

Chemistry-Electrochemistry-3448.png
  • 2.14 V
  • 4.28 V
  • 6.42 V
  • 8.56 V
An electrochemical cell is set up as follows:
Pt (H2, 1 atm.) |0.1MHCl | 0.1M Acetic Acid| (H2 ,1atm.) pt.
E.M.F. of this cell will not be zero because
  • The pH of 0.1 MHCl and 0.1 M acetic acid is not the same
  • Acids used in the two compartments are different
  • E.M.F of a cell depends on the molarities of acid used
  • The temperature is constant
Which of the following is incorrect about transport number?
  • it decreases with increase in concentration
  • it may increase or decrease with increase in temperature
  • it is never zero
  • it is different for Cl- ion in 0.1 MHCl and 0.1 M NaCl solution
Equal quantities of electricity are passed through three voltameters containing FeSO4, Fe2(SO4)3 and Fe(NO3)3 Consider the following statements in this regard:
1. The amount of iron deposited in FeSO4 and Fe2 (SO4)3 are equal.
2. The amount of iron deposited in Fe(NO3)3 is two third of the amount of iron deposited in FeSO4
3. The amount of iron deposited in Fe2(SO4)3 and Fe(NO3)3 is equal of these statements
  • 1 alone is correct
  • 1 and 2 are correct
  • 2 and 3 are correct
  • 3 alone is correct
0:0:1


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