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1

AIPMT 2006

MCQ (Single Correct Answer)
A hypothetical electrochemical cell is shown below.

$$A\left| {{A^ + }\left( {xM} \right)} \right|\left| {{B^ + }\left( {yM} \right)} \right|B$$

The emf measured is + 0.20 V. The cell reaction is
A
A + B+ $$ \to $$ A+ + B
B
A+ + B $$ \to $$ A + B+
C
A+ + e$$-$$ $$ \to $$ A;  B+ + e$$-$$ $$ \to $$ B
D
the cell reaction cannot be predicted.

Explanation

From the given expression:

At anode : A $$ \to $$ A+ + e

At cathode : B+ + e $$ \to $$ B

Overall reaction is : A + B+ $$ \to $$ A+ + B
2

AIPMT 2006

MCQ (Single Correct Answer)
EoFe2+/Fe = $$-$$ 0.441 V and EoFe3+/Fe2+ = 0.771 V, the standard EMF of the reaction Fe + 2Fe3+ $$ \to $$ 3Fe2+ will be
A
0.111 V
B
0.330 V
C
1.653 V
D
1.212 V

Explanation

At anode : Fe $$ \to $$ Fe2+ + 2e-; Eo = -0.441 V

At cathode : Fe3+ + e $$ \to $$ Fe2+; Eo = 0.771 V

Fe + 2Fe3+ $$ \to $$ 3Fe2+; Eo = ?

To get the above equation, (ii) × 2 – (i)

2Fe3+ + 2e $$ \to $$ 2Fe2+; Eo = 0.771 V

Fe $$ \to $$ Fe2+ + 2e-; Eo = - 0.441 V
------------------------------------------------
Fe + 2Fe3+ $$ \to $$ 3Fe2+

Eo = 0.771 + 0.441 = 1.212 V
3

AIPMT 2005

MCQ (Single Correct Answer)
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
4

AIPMT 2005

MCQ (Single Correct Answer)
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

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