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1

### AIPMT 2002

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

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

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

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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