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1

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.
2

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
3

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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