What is the emf of the cell, whose half cells are given below

Mg²⁺ + 2e^- → Mg(s) E =-2.37 V

Cu²⁺ + 2e^- → Cu(s) E =+0.34 V

• -2.30V

• 1.336V

• 2.17V

• 2.93V

Zn | $\underset{\mathrm{C} = 1}{{\mathrm{Zn}}^{2+}} \left|\right| \underset{\mathrm{C} = 1}{{\mathrm{Cu}}^{2+}}$ | Cu

If the standard reduction potential of zinc electrode and copper half cell is -0.76 V and 0.34 V respectively then the emf will be

• 1.1 V

• 1.4 V

• 1.34 V

• None of these

Rusting on iron needs

• dry air

• air and water

• distilled water and carbon dioxide

• oxygen and carbon dioxide

In galvanic cell, the salt bridge is used to

• complete the circuit

• reduce the electric resistance in the cell

• separate cathode from anode

• carry salts for the chemical reaction

155.

A cell, with cell constant 0.4 cm^-1, has the resistance of 40 ohm of a 0.01 M solution of an electrolyte, then the molar conductivity in ohm^-1 cm² mol^-1 will be

• 10⁴

• 10³

• 10²

• 1

If the standard elelctrode potential for the cell Zn | Zn²⁺ (aq) || Cu²⁺ (aq) | Cu is 1.10 V, then the maximum work done by this cell will be

• -106.15 kJ

• -212.30 kJ

• -318.45 kJ

• -424.60 kJ

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