NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1

### 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
2

### 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
3

### 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.
4

### 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.

### Joint Entrance Examination

JEE Main JEE Advanced WB JEE

### Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

NEET

Class 12