By passing 9.65 A current for 16 min 40 s, the volume of O2 

Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.
Advertisement

 Multiple Choice QuestionsMultiple Choice Questions

491.

In electrolysis of NaCl when Pt electrode is taken then H2 is liberated at cathode while with Hg electrode it forms sodium amalgam

  • Hg is more inert than Pt

  • more voltage is required to reduce H+ at Hg than at Pt

  • Na is dissolved in Hg while it does not dissolve in Pt

  • concentration of Hions is larger when Pt electrode is taken


492.

On the basis of the information available from the reaction,

43Al + O2  23 Al2O3

G = -827 kJ mol-1 of O2, the minimum emf required to carry out an electrolysis of Al2O3 is (F = 96500 mol-1)

  • 2.14 V

  • 4.28 V

  • 6.42 V

  • 8.56 V


493.

4.5 g of aluminium (atomic mass 27 u) is deposited at cathode from Al3+ solution by a certain quantity of electric charge. The volume of hydrogen produced at STP from H+ ions in solution by the same quantity of electric charge will be

  • 44.8 L

  • 11.2 L

  • 22.4 L

  • 5.6 L


494.

For the electrolysis of aqua CuSOsolution using inert Pt electrodes, the reaction on anode

  • 2SO42-  S2O32- + 212O2 + 2e-

  • Cu2+ + 2e- → Cu

  • 2H2O  → O2 + 4H++ 4e-

  • 2H+ + 2e- → H2


Advertisement
495.

If we mix a pentavalent impurity in a crystal lattice of Ge, what type of semiconductor formation will occur?

  • p-type

  • n-type

  • Both (a) and (b)

  • None of these


496.

If 'F' is Faraday and 'N' is Avogadro number, then charge of electron can be expressed as

  • F × N

  • FN

  • NF

  • F2N


Advertisement

497.

By passing 9.65 A current for 16 min 40 s, the volume of O2 liberated at STP will be

  • 280mL

  • 560mL

  • 1120mL

  • 2240mL


C.

1120mL

2 × 96500 C electncityis used to liberate= 22400mL O2 atSTP9.65 × 1000C electricity will liberate =22400 × 9.65 ×10002 × 96500= 1120 mL


Advertisement
498.

By diluting a weak electrolyte, specific conductivity (Kc) and equivalent conductivity (Λc) change as

  • both increase

  • Kc increase λdecrease

  • Kc decrease λc increase

  • both decrease


Advertisement
499.

In Daniel cell, anode and cathode are respectively

  • Zn| Zn2+ and  Cu2+|Cu 

  • Cu2+|Cu and Zn| Zn2+

  • Fe|Fe2+ and Cu2+|Cu

  • Cu2+|Cu and Fe|Fe2+


500.

Copper sulphate solution is electrolysed using copper electrode. The reaction taking place at anode is

  • H+ + e- → H

  • SO42- (aq) → SO4 + 2e-

  • Cu2+ + 2e- → Cu

  • Cu (s) → Cu2+ (aq) + 2e-


Advertisement