1
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
2
MCQ (Single Correct Answer)

### 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.
3
MCQ (Single Correct Answer)

### 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.
4
MCQ (Single Correct Answer)

### 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

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