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1

AIPMT 2002

MCQ (Single Correct Answer)
In electrolysis of NaCl when Pt electrode is taken then H2 is liberated at cathode while with Hg cathode it forms sodium amalgam
A
Hg is more inert than Pt
B
More voltage is required to reduce H+ at Hg than at Pt
C
Na is dissolved in Hg while it does not dissolve in Pt
D
Conc. of H+ ions is larger when Pt electrode is taken.

Explanation

In electrolysis of NaCl when Pt electrode is taken then H2 liberated at cathode while with Hg cathode it forms sodium amalgam because more voltage is required to reduce H+ at Hg than Pt.
2

AIPMT 2001

MCQ (Single Correct Answer)
Standard electrode potentials are
Fe2+/Fe [Eo = $$-$$0.44] and
Fe3+/Fe2+[ Eo = 0.77];
If Fe2+, Fe3+ and Fe blocks are kept together, then
A
Fe3+ increases
B
Fe3+ decreases
C
Fe2+/Fe3+ remains unchanged
D
Fe2+ decreases.

Explanation

The metals having higher negative values of their electrode potential can displace metals having lower values from their salt solutions.
3

AIPMT 2000

MCQ (Single Correct Answer)
For the disproportionation of copper
2Cu+  $$ \to $$ Cu2+ + Cu, Eo is
(Given Eo for Cu2+/Cu is 0.34 V and
Eo for Cu2+/Cu+ is 0.15 V.)
A
0.49 V
B
$$-$$ 0.19 V
C
0.38 V
D
$$-$$0.38 V

Explanation

Cu2+ + 2e $$ \to $$ Cu; E°1 = 0.34 V .....(1)

Cu2+ + e $$ \to $$ Cu+ ; E°2 = 0.15 V.....(2)

Cu+ + e $$ \to $$ Cu; E°3 = ? .....(3)

$$ \therefore $$ $$\Delta $$Go1 = -2$$ \times $$0.34$$ \times $$F

and $$\Delta $$Go2 = -1$$ \times $$0.15$$ \times $$F

and $$\Delta $$Go3 = -1$$ \times $$E°3$$ \times $$F

Also, $$\Delta $$Go1 = $$\Delta $$Go2 + $$\Delta $$Go3

$$ \Rightarrow $$ -0.68F = -0.15F - E°3 $$ \times $$ F

$$ \Rightarrow $$ E°3 = 0 68 - 0 15 = 0 53 V

$$ \therefore $$ E°cell = 0.53 - 0.15 = 0.38 V
4

AIPMT 2000

MCQ (Single Correct Answer)
Equivalent conductances of Ba2+ and Cl$$-$$ ions are 127 and 76 ohm$$-$$1 cm$$-$$1 eq$$-$$1 respectively. Equivalent conductance of BaCl2 at infinite dilution is
A
139.5
B
101.5
C
203
D
279

Explanation

According to Kohlrausch’s law, the equivalent conductance of BaCl2 at infinite dilution,

$$\lambda $$$$\infty $$ of BaCl2 = $${1 \over 2}$$$$\lambda $$$$\infty $$ of Ba2+ + $$\lambda $$$$\infty $$ of Cl-

$$ \Rightarrow $$ $$\lambda $$$$\infty $$ of BaCl2 = $${1 \over 2} \times 127 + 76$$ = 139 5

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