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 H⁺ ions is larger when Pt electrode is taken

On the basis of the information available from the reaction,

$\frac{4}{3}\mathrm{Al} + {\mathrm{O}}_2 \to \frac{2}{3} {\mathrm{Al}}_2{\mathrm{O}}_3$

$∆$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

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

For the electrolysis of aqua CuSO₄ solution using inert Pt electrodes, the reaction on anode

• $2{\mathrm{SO}}_4^{2-} \to {\mathrm{S}}_2{\mathrm{O}}_3^{2-} + 2\frac{1}{2}{\mathrm{O}}_2 + 2{\mathrm{e}}^{-}$

• Cu²⁺ + 2e^- → Cu

• 2H₂O  → O₂ + 4H⁺+ 4e^-

• 2H⁺ + 2e^- → H₂

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

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

• F × N

• $\frac{\mathrm{F}}{\mathrm{N}}$

• $\frac{\mathrm{N}}{\mathrm{F}}$

• F²N

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

• 280mL

• 560mL

• 1120mL

• 2240mL

By diluting a weak electrolyte, specific conductivity (K_c) and equivalent conductivity (Λ_c) change as

• both increase

• K_c increase λ_c decrease

• K_c decrease λ_c increase

• both decrease

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⁺

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

• H⁺ + e^- → H

• SO${}_4^{2-}$ (aq) → SO₄ + 2e^-

• Cu²⁺ + 2e^- → Cu

• Cu (s) → Cu²⁺ (aq) + 2e-