1

### AIPMT 2012 Mains

The Gibb's energy for the decomposition of Al2O3 at 500oC is as follows
${2 \over 3}$ Al2O3 $\to$ ${4 \over 3}$ Al + O2
$\Delta$rG = +960 kJ mol$-$1
The potential difference needed for the electrolytic reduction of aluminium oxide (Al2O3) at 500oC is at least
A
4.5 V
B
3.0 V
C
2.5 V
D
5.0 V

## Explanation

We know,

$\Delta$Go = – nFEo

${2 \over 3}$ Al2O3 $\to$ ${4 \over 3}$ Al + O2

Total number of Al atoms in Al2O3

= ${2 \over 3} \times 2 = {4 \over 3}$

Al3+ + 3e $\to$ Al

As 3e change occur for each Al atom

$\therefore$ n = ${4 \over 3} \times 3 = 4$

Eo = - ${{\Delta G^\circ } \over {nF}}$

= - ${{960 \times 1000} \over {4 \times 96500}}$

= - 2.5 V
2

### AIPMT 2012 Mains

Standard reduction potentials of the half reactions are given below :
F2(g) + 2e$-$ $\to$ 2F$-$(aq) ;   Eo = + 2.85 V
Cl2(g) + 2e$-$ $\to$ 2Cl$-$(aq) ;   Eo = + 1.36 V
Br2(l) + 2e$-$ $\to$ 2Br$-$(aq) ;   Eo = + 1.06 V
I2(s) + 2e$-$ $\to$ 2I$-$(aq) ;  Eo = + 0.53 V

The strongest oxidising and reducing agents 23 respectively are
A
F2 and I$-$
B
Br2 and Cl$-$
C
Cl2 and Br$-$
D
Cl2 and I2

## Explanation

Higher the value of reduction potential higher will be the oxidising power whereas the lower the value of reduction potential higher will be the reducing power.
3

### AIPMT 2012 Prelims

Limiting molar conductivity of NH4OH
$\left[ {} \right.$i.e.  $\Lambda _{m\left( {N{H_4}OH} \right)}^0$$\left. {} \right]$ is equal to
A
$\Lambda _{m\left( {N{H_4}OH} \right)}^0 + \Lambda _{m\left( {NaCl} \right)}^0 - \Lambda _{m\left( {NaOH} \right)}^0$
B
$\Lambda _{m\left( {NaOH} \right)}^0 + \Lambda _{m\left( {NaCl} \right)}^0 - \Lambda _{m\left( {N{H_4}Cl} \right)}^0$
C
$\Lambda _{m\left( {N{H_4}OH} \right)}^0 + \Lambda _{m\left( {N{H_4}Cl} \right)}^0 - \Lambda _{m\left( {HCl} \right)}^0$
D
$\Lambda _{m\left( {N{H_4}Cl} \right)}^0 + \Lambda _{m\left( {NaOH} \right)}^0 - \Lambda _{m\left( {NaCl} \right)}^0$

## Explanation

According to Kohlrausch’s law, the molar conductivity of NH4OH

$\Lambda _{m\left( {N{H_4}OH} \right)}^0$ = $\Lambda _{m\left( {N{H_4}Cl} \right)}^0 + \Lambda _{m\left( {NaOH} \right)}^0 - \Lambda _{m\left( {NaCl} \right)}^0$
4

### AIPMT 2011 Prelims

The electrode potentials for Cu2+(aq) + e$-$ $\to$ Cu+(aq)
and Cu+(aq) + e$-$ $\to$ Cu(s) are + 0.15 V and + 0.50 V respectively.

The value of Eocu2+/cu will be
A
0.500 V
B
0.325 V
C
0.650 V
D
0.150 V

## Explanation

Cu2+(aq) + e$-$ $\to$ Cu+(aq) ; E1o = 0.15 V

Cu+(aq) + e$-$ $\to$ Cu(s) ; E2o = 0.50 V

Cu2+ + 2e $\to$ Cu ; Eo = ?

$\Delta$Go = $\Delta$G1° + $\Delta$G2°

$\Rightarrow$ – nFE° = – n1FE1° – n2FE2°

$\Rightarrow$ Eo = ${{1 \times 0.15 + 1 \times 0.50} \over 2}$ = 0.325 V