1
MCQ (Single Correct Answer)

AIPMT 2005

The mass of carbon anode consumed (giving only carbon dioxide) in the production of 270 kg of aluminium metal from bauxite by the Hall process is
(Atomic mass : Al = 27)
A
270 kg
B
540 kg.
C
90 kg
D
180 kg

Explanation

2Al2O3 + 3C $$ \to $$ 4Al + 3CO2

From the above equation,

3 mol × 12 g mol–1 = 36 g of carbon is consumed

to give 4 mol × 27 g mol–1 = 108 g of Al

So, 108 g of Al is produced by 36 g of carbon

$$ \therefore $$ 270000 g of Al is produced by

= $${{36} \over {108}} \times 270000$$ of C

= 90000 g of C

= 90 kg of C
2
MCQ (Single Correct Answer)

AIPMT 2005

4.5 g of aluminium (at. mass 27 amu) 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
A
44.8 L
B
22.4 L
C
11.2 L
D
5.6 L

Explanation

Faraday second law of electrolysis

$${{{M_{Al}}} \over {{M_H}}} = {{{E_{Al}}} \over {{E_H}}}$$

$$ \Rightarrow $$ $${{4.5} \over {{M_H}}} = {{{{27} \over 3}} \over 1}$$

$$ \Rightarrow $$MH = 0.5 g

We know, Volume of 2 g H2 at STP = 22.4 L

$$ \therefore $$ Volume of 0.5 g H2 at STP

= $${{22.4 \times 0.5} \over 2}$$ = 5.6 L
3
MCQ (Single Correct Answer)

AIPMT 2004

The standard e.m.f. of a galvanic cell involving cell reaction with n = 2 is found to be 0.295 V at 25oC. The equilibrium constant of the reaction would be
A
2.0 $$ \times $$ 1011
B
4.0 $$ \times $$ 1012
C
1.0 $$ \times $$ 102
D
1.0 $$ \times $$ 1010

Explanation

We know, from Nernst Equation

Ecell = Eocell - $${{2.303RT} \over {nF}}{\log _{10}}K$$

At equilibrium Ecell = 0

$$ \therefore $$ 0 = Eocell - $${{2.303 \times 8.314 \times 298} \over {2 \times 96500}}{\log _{10}}K$$

$$ \Rightarrow $$ 0 = 0.295 - $${{2.303 \times 8.314 \times 298} \over {2 \times 96500}}{\log _{10}}K$$

$$ \Rightarrow $$ 0.295 = $${{0.0591} \over 2}{\log _{10}}K$$

$$ \Rightarrow $$ $${\log _{10}}K$$ = 10

$$ \Rightarrow $$ K = 1 $$ \times $$ 1010
4
MCQ (Single Correct Answer)

AIPMT 2003

On the basis of the information available from the reaction,

4/3Al + O2 $$ \to $$ 2/3Al2O3,   $$\Delta $$G = $$-$$ 827 kJ mol$$-$$1 of O2,

the minimum e.m.f. required to carry out an electrolysis of Al2O3 is
(F = 96500 C mol$$-$$1)
A
2.14 V
B
4.28 V
C
6.42 V
D
8.56 V

Explanation

4/3Al + O2 $$ \to $$ 2/3Al2O3,   $$\Delta $$G = $$-$$ 827 kJ mol$$-$$1

For 1 mol of Al, n = 3

$$ \therefore $$ For $${4 \over 3}$$ mol of Al, n = $$3 \times {4 \over 3} = 4$$

As $$\Delta $$G = - nFEo

$$ \Rightarrow $$ – 827 × 103 J = – 4 × E° × 96500

$$ \Rightarrow $$ Eo = $${{827 \times {{10}^3}} \over {4 \times 96500}}$$ = 2.14 V

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