1

### AIPMT 2004

The standard e.m.f. of a galvanic cell involving cell reaction with n = 2 is found to be 0.295 V at 25oC. The equilibrium constant of the reaction would be
A
2.0 $\times$ 1011
B
4.0 $\times$ 1012
C
1.0 $\times$ 102
D
1.0 $\times$ 1010

## Explanation

We know, from Nernst Equation

Ecell = Eocell - ${{2.303RT} \over {nF}}{\log _{10}}K$

At equilibrium Ecell = 0

$\therefore$ 0 = Eocell - ${{2.303 \times 8.314 \times 298} \over {2 \times 96500}}{\log _{10}}K$

$\Rightarrow$ 0 = 0.295 - ${{2.303 \times 8.314 \times 298} \over {2 \times 96500}}{\log _{10}}K$

$\Rightarrow$ 0.295 = ${{0.0591} \over 2}{\log _{10}}K$

$\Rightarrow$ ${\log _{10}}K$ = 10

$\Rightarrow$ K = 1 $\times$ 1010
2

### 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
3

### 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
4

### 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.