When a lead storage battery is discharged, then:

1. SO₂ is evolved.

2. Lead is formed.

3. Lead sulphate is consumed.

4. Sulphuric acid is consumed.

The ratio of masses of hydrogen and magnesium deposited by the same amount of electricity from H₂SO₄ and MgSO₄ in aqueous solution are:

1. 1:8

2. 1:12

3. 1:16

4. None of the above

The molar conductances of Ba²⁺ and Cl^- 127 and 76Ω^-1 cm^-1 mol^-1 respectively at infinite dilution. The equivalent conductance of BaCl₂ at infinite dilution will be

1. 139.52

2. 203

3. 279

4. 101.5

A depolariser used in dry cell batteries is: 

1. Ammonium chloride 

2. Manganese dioxide

3. Potassium hydroxide 

4. Sodium phosphate

The reaction taking place  at anode when an aqueous solution of CuSO₄ is electrolysed using inert Pt electrode:

1. 2${\mathrm{SO}}_4^{2-}$ $\to$ S₂${\mathrm{O}}_3^{2-}$ + 2e

2. Cu²⁺ + 2e $\to$Cu

3. 2H₂O $\to$ O₂ + 4H⁺ + 4e

4. 2H⁺ +2e $\to$ H₂

An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to

1. increase in ionic mobility of ions

2. 100% ionisation of electrolyte at normal dilution

3. increase in both, i.e. number of ions and ionic mobility of ions

4. increase in number of ions

For the cell, Ti|Ti⁺(0.001M)||Cu²⁺(0.1M)|Cu,E_cell at 25$°$C is 0.83 V. E_cell can be increased:

1. By increasing [Cu²⁺]

2. By increasing [Ti⁺]

3. By decreasing [Cu²⁺]

4. None of these

The atomic mass of Al is 27. When a current of 5 faraday is passed through a solution of Al³⁺ ions, the mass of Al deposited is:

1. 27 g

2. 36 g

3. 45 g

4. 9 g

Cu⁺(aq) is unstable in solution and undergoes simultaneous oxidation and reduction according to the reaction 

2Cu⁺(aq) $\rightleftharpoons$Cu²⁺(aq) + Cu(s)

Choose the correct E0 for above reaction if E⁰_cu²⁺/cu = 0.34 V and E⁰_cu²⁺/cu+ = 0.15 V 

1. -0.38 V

2. +4.9 V

3. +0.38 V

4. -0.19 V

On the basis of the information available from the reaction.

$\frac{4}{3}\mathrm{Al}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>{\mathrm{O}}_2<mspace\; width="1em"></mspace>\stackrel{<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>}{\to }\frac{2}{3}{\mathrm{Al}}_2{\mathrm{O}}_3,<mspace\; width="1em"></mspace>∆G<mspace\; width="1em"></mspace>=-827\mathrm{kJ}<mspace\; width="1em"></mspace>{\mathrm{mol}}^{-1}$ 

of O₂, the minimum EMF required to carry out the electrolysis of Al₂O₃ is : (F=96500 C mol^-1) 

1. 2.14 V

2. 4.28 V

3. 6.42 V

4. 8.56 V