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Al2O3 is reduced by electrolysis at low potentials and high currents. If 4.0 × 104 amperes of current is passed through molten Al2O3 for 6 hours, what mass of aluminium is produced? (Assume 100% current efficiency. At. mass of Al = 27 g mol-1)
Given:
(i)Cu2+ + 2e– ⟶ Cu, E°= 0.337V
(ii)Cu2+ + e– ⟶ Cu+, E° = 0.153V Electrode potential,
E° for the reaction,Cu+ + e– ⟶ Cu, will be :
Standard free energies of formation (in kJ/mol) at 298 K are– 237.2, –394.4 and – 8.2 for H2O(l), CO2(g) and pentane (g),respectively. The value E°cell for the pentane-oxygen fuel cell is :
Kohlrausch’s law states that at :
Kohlrausch’s Law states that at infinite dilution, each ion migrates independent of its co-ion and contributes to the total equivalent conductance of an electrolyte a definite share which depends only on its own nature.From this definition we can see that option (d) is the correct answer.
On the basis of the following E° values, the strongest oxidizing agent is :
[Fe(CN)6]4- ⟶ [Fe(CN)6]3- + e– ; E° = -0.35V
Fe2+ ⟶ Fe3+ + e–; E° = -0.77V
From the given data we find Fe3+ is strongest oxidising agent.More the positive value of E°, more is the tendency to get oxidized. Thus correct option is (c).
The efficiency of a fuel cell is given by
Efficiency of a fuel cell (η) = \(\frac{\Delta G}{\Delta S}\)
The equilibrium constant of the reaction: 2Cu(s) + 2Ag+(aq) ⇌ Cu2+(aq) + 2Ag(s);
E° = 0.46 V at 298 K is
When a lead storage battery is discharged
Which colourless gas evolves, when NH4Cl reacts with zinc in a dry cell battery
2NH4Cl + Zn ⟶ 2NH3 + ZnCl2 + H2↑