1
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
2
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
3
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
4
MCQ (Single Correct Answer)

AIPMT 2003

The e.m.f. of a Daniell cell at 298 K is E1.

When the concentration of ZnSO4 is 1.0 M and that of CuSO4 is 0.01 M, the e.m.f. changed to E2. What is the relationship between E1 and E2?
A
E1 > E2
B
E1 < E2
C
E1 = E2
D
E2 = 0 $$ \ne $$ E1

Explanation

Cell reaction is,

Zn + Cu2+ $$ \to $$ Zn2+ + Cu

Ecell = Eocell - $${{RT} \over {nF}}\ln {{\left[ {Z{n^{2 + }}} \right]} \over {\left[ {C{u^{2 + }}} \right]}}$$

Greater the factor $${{\left[ {Z{n^{2 + }}} \right]} \over {\left[ {C{u^{2 + }}} \right]}}$$, less is the EMF.

Hence E1 > E2

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